Chemistry College - Learn Chemistry

Redox Reactions or Oxidation Reduction Reactions

2012-01-14T02:39:00.000-08:00

Redox Reactions / Oxidation and reduction reactions

These are reactions where electrons are transferred from one species (atom, molecule or ion) to another. We can write 'half' equations to show only what happens to the species losing electrons or a different 'half' equation to show the species gaining electrons.
The whole equation is put together by making sure that the numbers of electrons are balanced in each half equation and adding them together (when the electrons will cancel out)
Oxidation
This is the name given to removal of electrons from a species - the reagent causing the loss of electrons is called the oxidising agent

Example:

Mg_(s) Mg²⁺ + 2e

In this (half) equation the magnesium atom loses electrons and becomes an ion.

Reduction
This is the gain of electrons - the species donating the electrons is called the reducing agent

Example

Fe³⁺ + 3e Fe_(s)

In this (half) equation the iron(III) ion gains three electrons to become an atom.

Redox reactions
Obviously the electrons leave one species and go to another. Consequently reduction has to be accompanied by oxidation and vice versa. For this reason reactions involving transfer of electrons are called reduction and oxidation or redox for short

Example

3Mg_(s) + 2Fe³⁺ 2Fe_(s)+ 3Mg²⁺

The electrons from the magnesium are transferred to the iron(III) ions

Summary

Loss of electrons = Oxidation Gain of electrons = Reduction	Mnemonic (memory aid) OIL-RIG
			Oxidation Is Loss Reduction Is Gain

10.1.2: Calculate the oxidation number of an element in a compound. Oxidation numbers should be shown by a sign (+ or -) and a number, eg +7 for Mn in KMnO4.

Oxidation number
This is the apparent valency of an atom within a compound. It is usually considered as if the element were bonded ionically to allow the apparent number of electrons gained or lost to be assessed.
The sum of all the oxidation numbers in a compound must add up to 0. By convention, the oxidation number is written as a Roman numeral in the name, eg. iron II sulphate, sulphur VI oxide.
The oxidation number of an uncombined element is always zero (0)
Calculating the oxidation number
There are some elements that virtually always have the same oxidation number and these can be used to calculate the oxidation numbers of the atoms in question.
Hydrogen, for example always has an oxidation number of -1 when bonded to a metal (more electropositive element) and +1 when bonded to a more electronegative element (non-metal). Oxygen is always -2 (except when in the form of the peroxide ion when it has an O-O bond giving it an oxidation number of -1). Group 1 and 2 metals usually have an oxidation number of 1+ and 2+ respectively.

Example - Calculate the oxidation number of sulphur in sulphuric acid H₂SO₄
Hydrogen = +1 oxidation number
Oxygen = -2 oxidation number
Therefore:

(2 x H) + S + (4 x O) = 0

2 + S -8 = 0

S = 6

Example - Calculate the oxidation number of nitrogen in calcium nitrate Ca(NO₃)₂
Calcium is in group 2 = +2 oxidation number
Oxygen = -2 oxidation number
Therefore:

(+2) + [(2 x N) + (6 x -2)] = 0

+2 + 2N -12 = 0

2N = 10

N = +5

10.1.3: State and explain the relationship between oxidation numbers and the names of compounds. Oxidation numbers in names of compounds are represented by Roman numerals, eg iron(II) oxide, iron(III) oxide.

Names of compounds
Where there is any doubt about the oxidation state of an element within a compound it is stated using Roman numerals immediately after the ambiguous element. For example Iron compounds may be iron in the oxidation state +2 or +3 - it must therefore be stated as iron II or iron III in the compound name.
In the examples above the full systematic name for sulphuric acid is sulphuric(VI) acid and calcium nitrate is calcium nitrate(V)

Example - Name the following compound - FeSO₄
Oxidation state of the oxygen = -2 Oxidation state of the sulphur = +6
Therefore oxidation state of the iron = - (+6 - 8) = +2
The name of the compound FeSO₄ is iron(II) sulphate

Example - Name the following compound - TiCl₄
Oxidation state of the chloride = -1
Therefore oxidation state of the titanium = - (- 4) = +4
The name of the compound TiCl₄ is titanium(IV) chloride

10.1.4: Identify whether an element is oxidised or reduced in simple redox reactions, using oxidation numbers. Appropriate reactions to illustrate this can be found in topics 3 and 11. Possible examples include: iron(II) and (III), manganese(II) and (VII), chromium(III) and (VI), copper(I) and (II), oxides of sulphur and oxyacids, halogens and halide ions.

Oxidation and reduction
As stated above, for the purposes of oxidation and reduction the oxidation number can be thought of as the apparent ionic charge of an atom within a compound. For example, in sulphuric acid the sulphur is in the VI (6+) oxidation state. For the purposes of redox we can consider that it has an ionic charge of +6 (even though it is clearly covalently bonded). This makes it easier to follow any transfer of electrons.
If the sulphur changes to an oxidation state of IV during a chemical reaction then it has gone from an apparent ionic charge of +6 to a charge of +4, i.e. it has gained two electrons (negative charges). It has therefore been reduced (gain of electrons) in the process.

Examples

2FeCl₂ + Cl₂ 2FeCl3

The iron changes from 2+ to 3+ and is therefore oxidised (removal of electrons)

The chlorine gains an electron to go from 0 to -1 and is therefore reduced (addition of electrons)

Zn + CuSO₄ Cu + ZnSO₄

The zinc changes from oxidation state 0 to +2 (removal of electrons) it is oxidised

The copper changes from 2+ to 0 and is oxidised and is therefore reduced (addition of electrons)

Cr₂O₇^2- + 3SO₂ + 2H+ 2Cr³⁺ + 3SO₄^2- + H₂O

The chromium changes from +6 to +3 and is therefore reduced (gain of electrons)

The sulphur changes from +4 to +6 and therefore loses electrons = oxidation (loss of electrons)

2KI + Br₂ 2KBr + I₂

The iodide ions (oxidation number = -1) change to iodine (oxidation number = 0) : oxidation

Bromine (element, oxidation number = 0) changes to bromide ions (oxidation number = -1) : reduction

5Fe²⁺ + MnO₄^- + 8H⁺ 5Fe³⁺ + Mn²⁺ + 4H₂O

The iron changes from 2+ to 3+ and is oxidised (removal of electrons)

The manganese atom changes from +7 to +2 and is therefore reduced (addition of electrons)

10.1.5: Define the terms oxidising agent and reducing agent.

Oxidising agents
These are the chemicals that cause the oxidation in a redox reaction. We call the reacting compounds in a reaction the reagents (short form of the words reacting agents).
We consider that the removal of electrons from a species is oxidation and these electrons have to be taken away by another compound or species. This species that attracts the electrons is said to be the oxidising agent i.e. the reagent that causes the oxidation.
Reducing agents
Similarly the reagent that causes reduction in a redox reaction is said to be the reducing agent.
The oxidising agent takes the electron and is itself reduced, the reducing agent loses the electrons and is itself oxidised.

2KI	+ Br₂	2KBr	+ I₂
Iodide ions get oxidised	Bromine gets reduced
Iodide - reducing agent	Bromine - oxidising agent

Cr₂O₇^2-	+ 3SO₂ + 2H+	2Cr³⁺	+ 3SO₄^2- + H₂O
Chromium VI gets reduced	Sulphur IV gets oxidised
Chromium VI oxidising agent	Sulphur IV reducing agent

Fundamentals of Acid Base Chemistry

2012-01-14T02:20:00.000-08:00

Fundamentals of Acid-Base Chemistry

Definitions of Acidity and Basicity

For more than two-hundred years, chemists have struggled to come up with a way to describe acid-base reactions that is at the same time physically relevant, specific enough to be accurate, and general enough to include everything that should be considered an acid-base relationship.

Svante Arrhenius first defined acids to be proton (H⁺) donors and bases to be hydroxide ion (OH^-) donors in aqueous solution. The Arrhenius model of acids and bases is summarized by the following two reactions:

Figure %: The Arrhenius model of acids and bases, where A = acid and B = base

At the time that Arrhenius proposed these definitions, water was virtually the only solvent used in chemistry, and nearly all known acids and bases contained protons (H⁺) and hydroxyl groups (OH), respectively. His definition was sufficient for the chemistry that was understood then. But progress in chemistry necessitated new definitions: it was discovered that ammonia behaves like a base, and HCl donates protons in non-aqueous solvents. The Bronsted-Lowry model of acids and bases serves that need by describing acids as proton donors and bases as proton acceptors. These definitions remove the role of solvent and allow bases like ammonia and fluoride ion to be classified as bases, so long as they bond to protons. The Bronsted-Lowry model implies that there is a relationship between acids and bases (acids transfer protons to bases) and allows us to define conjugate acids and conjugate bases, as seen in .

Figure %: The Bronsted model of acids and bases

You should note in the figure above that the conjugate acid of the base, BH⁺, acts as an acid in the reverse reaction by donating a proton to the conjugate base, A^-, of the acid HA.
Despite the usefulness of the Bronsted-Lowry definition, there is an even more general definition of acids and bases provided by G. N. Lewis. The Lewis model of acids and bases proposes that an acid is an electron pair acceptor while a base is an electron pair donor. This model of acidity and basicity broadens the characterization of acid-base reactions to include reactions like the following which do not involve any hydrogen transfers. The nitrogen atom in ammonia donates an electron pair to complete the valence octet of boron.

Figure %: Example of a Lewis acid-base reaction

Because we are more interested now in describing terms and processes that involve proton transfers (pH, titration), we will focus on the Bronsted-Lowry definitions of acids and bases. We will leave consideration of the Lewis model of acids and bases for studying reactions in organic chemistry.

Reactions of Acids and Bases with Water

As you may have noted already in the acid-base reactions above, we use arrows in both reaction directions to indicate that these are equilibrium processes. Proportions of reagents and products at equilibrium can be described by an equilibrium constant. The equilibrium constant given in is for the reaction of an acid, HA, with water as shown.

Although water is a reactant in the above reaction and belongs in the equilibrium constant, its value of 55.6 M in aqueous solution is so large in comparison with the change in water concentration at equilibrium that we will assume that the value of [H₂O] is constant. Using that assumption, we will define the acid dissociation constant, K _a, in to be the following:

Figure %: Definition of the acid dissociation constant

From the form of the above equation we can see that stronger acids, those that dissociate to a greater extent, will have larger values of K _a whereas weaker acids will have smaller values of K _a. A practical range for K _a values runs from 10^-12 for very weak acids to 10¹³ for the strongest acids. Knowing this practical range of acidity constants will aid in judging how reasonable your answers are when you calculate values for K _a in problems.
In an analogous way, we define K _b, the base constant in to be the following:

Figure %: Definition of the base dissociation constant

Stronger bases have larger values of K _b while weaker bases have smaller values of K _b. K _b's of typical bases in inorganic chemistry tend to have a range of values between 10^-11 and 10³.
As you may have discovered in our above discussion, water can act as both an acid and as a base. For this reason water is said to be amphiprotic. Water is often incorrectly termed amphoteric. An amphiprotic species like water can either donate or accept a proton. Amphoteric species can both donate and accept hydroxide ions, as water cannot. The following reaction in , called the autoionization of water, has the equilibrium constant K _w defined in the manner of K _a and K _b. The dissociation constant K _w for water is 1 x 10^-14 at room temperature (298 K), and tends to rise with higher temperatures.

Figure %: The autoionization of water

Knowing that K _w = 1 x 10^-14 is useful because the relationship between K _a and K _b for a conjugate acid-base pair is K _a * K _b = K _w. Therefore, you can calculate the K _a of the conjugate acid of a base when given its K _b. This point is proved in below.

Figure %: The autoionization constant of water equals the product of the acid and base dissociation constants of a conjugate acid-base pair.

From the proven expression, we see that strong acids (large K _a's) form weak conjugate bases and weak acids form strong conjugate bases.

Mole Calculations

2012-01-13T18:46:00.000-08:00

One of the main problems that beginning chemistry students have is in doing conversions between grams, moles, and molecules (or atoms). Usually, a question will be asked of you in the following form:
How many moles are in 22 grams of copper metal?
If you're confused by this problem, don't worry. Most people are when they start doing this kind of problem. To make life easier for you, I put together a "road map" which tells you exactly what you need to do to convert between atoms (or molecules), grams, and moles.

You should read this picture the same as you would read a subway map. For example, if you want to go from King Street in Alexandria to National Airport on the Blue line of the Metro, you need to first go to the Braddock Road station. The same thing is true here - if a problem tells you to go from atoms to grams, you need to first go through moles before you do anything else. In our example that we're discussing, though, we are making a one-stop trip.
Example: How many moles are in 22 grams of copper metal?
In all problems like this, you need to go through four steps to find a solution.

The Four Steps to Solving Mole Problems:

Step 1: Figure out how many parts your calculation will have by using the diagram
Looking at the diagram above, we can see that we are going between grams and moles, which is a one-step conversion. Furthermore, we can see that we need to use the atomic mass of copper as our conversion factor.
Step 2: Make a T-chart, and put whatever information the problem gave you in the top left. After that, put the units of whatever you were given in the bottom right of the T, and the units of what you want to find in the top right.
In this case, the problem gave you "22 grams of copper" as the starting information. Because this is what you were given, put "22 grams of copper" in the top left of the T. Since "grams of copper" is the unit of what you were given, put this in the bottom right of the T. Since you want to find out how many moles of copper are going to be made, put "moles of copper" as your unit in the top right. When you've done this, your calculation should look like this:

Step 3: Put the conversion factors into the T-chart in front of the units on the right.
As we saw from the "map", the conversion factor between grams and moles is the atomic mass of copper. Because we measure atomic mass in grams, you need to put the atomic mass in front of the unit "grams of copper". What do you put in front of moles? Whenever you do a calculation of this kind, you need to put "1" in front of moles, like you see here:

Step 4: Cancel out the units from the top left and bottom right, then find the answer by multiplying all the stuff on the top together and dividing it by the stuff on the bottom.
In this case, you'd multiply 22 by one and divide the result by 63.5. Your answer, 0.35 moles of copper:

And that's how you do a one-step problem of this kind!

Solving Two-Step Mole Calculation Problems:

What happens if we need to solve a problem that requires we not just go from one box in the next in our diagram, but across the entire diagram? Well, it means that we need to do two steps in our calculation. Let's see that "map" again to see what I mean:

If we were asked to convert 22 grams of copper to atoms of copper, we'd have to go from one end of the map to the other. Instead of doing a simple one step calculation, we'd need to do a two-step calculation, with the first step going from grams to moles and the second step going from moles to atoms.
How can we solve this kind of problem? Well, we start off by doing the same thing that we did in our last example: We had to convert grams to moles before, and we can see from the map that we have to convert grams to moles now, too. To refresh your memory, here's the calculation from last time:

In the next step, we do the same thing over again, except that we need to add another T to the T-chart. When you do this, take the units of the thing at the new top left and put them on the bottom right (in this case, moles). Then take the units of what you want (in this case, atoms) and put it in the top right. Finally, put in your conversion factors, which from the chart above is Avogadro's number, or 6.02E23. Since this number refers to the number of atoms in a mole of a substance, we put this in front of "atoms of copper". Again, put the number "1" in front of moles, because we're saying that there are 6.02E23 atoms in ONE mole of an element.
When we add all these terms in, we can cross out the units that cancel out, as shown. To get the answer, multiply all the numbers on the top together and divide by the numbers on the bottom. Your answer should then be set up like this:

And that's how you do mole problems!

Atomic Structure and Quantum Mechanics

2011-11-03T07:28:00.000-07:00

ELECTROMAGNETIC RADIATION

Electromagnetic radiation is one of the ways energy travels through space. This includes X-rays, gamma rays, ultraviolet light (or UV), infrared light, microwaves and radio waves. Visible light also makes up a small fraction of what is known as the electromagnetic spectrum. The electromagnetic spectrum is the spectrum of all possible types of electromagnetic radiation.

The Visible Spectrum

Although gamma rays may be vastly different from visible light, they both exhibit the same type of wave behavior. All types of electromagnetic radiation travel at the speed of light in a vacuum too.
The three main characteristics of a wave are wavelength, frequency, and speed. Wavelength is the distance between two consecutive peaks or troughs. It is symbolized by the Greek letter lambda, l . Frequency is the number of waves (cycles) that pass a given point in one second. It is symbolized by the Greek letter nu, n . Speed is defined as cycles per second, or hertz, abbreviated Hz.
Wavelength and frequency are inversely related. This is shown by the formula

c = (l)(n)

In which n is the frequency, l is the wavelength, and c is the speed of light (2.9979 ´ 10⁸ m/s).

THE NATURE OF MATTER

Around 1900, everyone believed that energy and matter were two different things altogether. Matter was thought to be made of particles (atoms) and energy was thought to be made of light (electromagnetic radiation). Particles had mass while energy occurred in waves and was massless.
That changed in 1900 when Max Plank, a German physicist, studied the emission of radiation from solid objects. He observed that the physics of his day could not describe the results he was obtaining. He postulated the theory that energy can be only gained or lost in whole number multiples of the quantity hn , where h is Plank’s constant. Experiments determined the constant to be equal to 6.626 ´ 10^-34 J× s. Therefore, the change in energy for a system can be shown by the equation

DE = nhn

where n is an integer (1, 2, 3, …).
With this, a new set of thinking came about. Before it was thought that a transfer of energy was continuous. Now it was realized that energy could only occur in "packets" called quantum.
Albert Einstein was next, proposing that even electromagnetic radiation was quantized. He suggested that electromagnetic radiation could be viewed as a stream of particles called photons. The energy of a photon could be described by the equation

E_photon = hn = hc/l

where h is Plank’s constant, n is the frequency, and h is the wavelength.
In his special theory of relativity, he showed that energy has mass. This is shown in the formula

E = mc²

where c is the speed of light, m is the mass, and E is the energy.
The proof of energy having mass is shown if you rearrange the formula in the following form:

m = E / c²

Since you know that

m = E / c² = (hc/l) / c² = h / lc

one can clearly see de Broglie’s equation,

l = h / mv

v = velocity, not frequency
This allows us to calculate the wavelength for a particle. It also proves that particles do behave as waves.
In summary, it was now found that energy is quantized, meaning that it can occur only in discrete units called quanta. It was also discovered that energy does contain mass. This is known as the dual nature of light.

BOHR’S ATOMIC MODEL

In 1913, Niels Bohr, a Danish physicist, developed a quantum model for the hydrogen atom. By combining classical physics and some theories of his own Bohr was able introduce a revolutionary model for all atoms.
Previous atomic models had the negatively charged electrons orbiting the positively charged nucleus. However, this is inaccurate because this would cause the electrons to eventually "fly off" the atom. The man who invented this model of the atom, Ernst Rutherford, knew that this was true, but couldn’t think of a better way to describe an atom.
Using the spectrum of hydrogen, Bohr found that when a prism diffracted the light color was displayed at only four discrete increments.

He saw that these results would fit his model if he assumed that the angular momentum of the electron could only occur in certain increments. Angular momentum equals the product of mass, velocity, and orbital radius. Using this he assigned the hydrogen atom energy levels consistent with the hydrogen emission spectrum.
This gave way to Bohr historic equation for the energy levels available to the electron in the hydrogen atom. The n is an integer. The larger the value of n, the larger is the orbital radius. Z is the nuclear charge.

Bohr’s equation was able to accurately calculate the energy levels for the hydrogen atom. Each energy level pertains to an electron in an exited state. It will move up to higher energy levels, but when it goes back to its ground state (where n=1) or to a lower level orbital it emits energy. That is what Bohr viewed when he saw the spectrum of hydrogen.
To find the total change in energy once subtracts the energy in the initial state minus the energy in the final state.

DE = E_final - E_initial

If a negative sign appears as the result, that means that the atom lost energy and thus is in a more stable state. However, if you insert this value into a separate equation, use the absolute value of the change in energy.
The next equation’s purpose is derived from Bohr’s formula. It determines the energy used for an electron moving from one level (n_initial) to another level (n_final).

-2.178 * 10^-18 J ( 1/(n_final)²) - (1/(n_initial)²))

However, it was soon found that Bohr’s model does not work for other atoms. Later models, however, would explain atoms other than hydrogen.

QUANTUM MECHANICS

It was now known that Bohr’s model would work for hydrogen and hydrogen only. A new model was needed. This became known as quantum mechanics. As de Broglie’s equation showed that electrons acted as waves. Many physicists now tried to attack the problem of atomic structure by using the wave properties of an electron. To them the electron bound to the nucleus seemed similar to a standing wave.
Take a stringed instrument. The string vibrates to produce a musical tone. The waves are "standing," because they are stationary; the waves do not travel along the string. Now put this into the hydrogen atom, with an electron acting as a standing wave. One physicist, Erwin Sgrödinger, used the formula

H y = E y

to illustrate the wave properties of an atom. The formula is in that form because the math is too complicated to be detailed here. The y is called the wave function. It is a function of the x, y, and z points on a three-dimensional graph. The H represents a set of mathematical operations called an operator. The E represents the total energy of the atom. When this equation is analyzed in its entirety, each solution consists of a wave function (y ) that is characterized by a particular value of E. A specific wave function is often referred to as an orbital.
To show this, let us concentrate on the hydrogen atom. The wave function corresponding to the lowest energy is called the 1s orbital. Remember that an orbital IS NOT a Bohr orbit. Th electron is the 1s orbital is not moving around the nucleus in a circular motion.
Here is a quick point. According to Werner Heisenberg "there is a fundamental limitation to just how precisely we can know both the position and the momentum of a particle at a given time. This is known as the Heisenberg uncertainty principle. In the formula

Dx * D(mv) > h/4p

where Dx is the uncertainty in a particles position and D(mv) is the uncertainty of a particle’s momentum. This all means that one cannot simultaneously find a particle's position and momentum.
Using this it was determined that for the 1s orbital, one has the greatest probability of finding an electron near the nucleus. That is because of the smaller radius and volume of the atom as a whole. Thus, the farther away from the nucleus you look, the less of a chance there is to find an electron.
To better understand electron orbitals as a whole, electron numbers have been made to identify the electrons of an atom.
The principal quantum number (n) has values of 1,2,3… It pertains to the period that the atom is in. Periods are represented by rows. For example, period two has a principal quantum number of 2.
The angular momentum quantum number (l) has values of 0 to n-1. This is related to the shape of an orbital. If l=0, the letter is s; l=1 is called p; l=2 is called d; and l=3 is known as f. This is explained later in this reading.
The magnetic quantum number (m_l) has integral values between l and –l, including zero. For example, if the value of l is 2, then the possible values of m_l are -2, -1, 0, 1 and 2.
The electron spin quantum number (m_s) can have one of two values: +(1/2) and –(1/2). This is because only two electrons can occupy any orbital, and they must have opposite spins.
But what do these have to do with anything? There will be more on that next.

AUFBAU PRINCIPLE

The s orbitals are represented by the Alkali and Alkali Earth Metal groups (groups 1A and 2A, respectively). The p orbitals are the other representative elements (3A through 8A). The d orbitals are the transition metals, excluding the inner transition metals. Finally, the f orbitals are the inner transition metals.
According to the rules stated previously, hydrogen is in the 1s orbital. This is because it is in the s orbital of the first period. And since it is the first atom of the 1s orbital, it is given the distinction 1s¹. Thus helium, even though it is technically a Noble Gas, is classified as 1s². That is because helium is the second atom of the 1s orbital.
Lithium is in the 2s orbital. It is identified as 2s¹. Since you have to add on the previous orbitals to make it correct, the correct from is 1s²2s¹. Beryllium is 1s²2s². Are you beginning to see the pattern? Boron, which is the p orbital is also the first atom in the 2p group. Thus it is 2p¹. However, you must add the previous orbitals on to make it right, so the correct form is 1s²2s²2p¹. Carbon is in the form 1s²2s²2p². Neon, which is at the end of the 2p orbital, is 1s²2s²2p⁶.
The d orbital is somewhat different. Since the transition metals begin at the fourth period, the first orbital is the 4s orbital. However, when writing the transition metals, one must start with 3d. Why isn’t it 4d? This is because the 3d orbitals are part of the same orbital as the 3s and 3p orbitals, but they are at a higher energy level than 4s. That is why you have to fill the 4s orbitals before the 3d orbitals.
Therefore, the configuration of scandium is 1s²2s²2p⁶3s²3p⁶4s²3d¹. After filling the 3d orbitals, the 4p orbitals are filled. The two exceptions to the rule are chromium and copper. Instead of normal configurations, they are 4s¹3d⁵ and 4s¹3d¹⁰ respectively. No one really knows why this happens.
Also, a shortcut to writing these configurations is substituting the highest filled Noble Gas for its configuration. For example, the configuration of potassium is normally 1s²2s²2p⁶3s²3d¹⁰4s¹. However one can substitute the configuration of argon with [Ar]. Thus the shorthand way to write an electron configuration for potassium is [Ar]4s¹.
For the f orbitals the rules change again. The f orbitals represent the inner transition metals. Since it is often very difficult to explain, use the chart below to figure out the filling order for all electrons.

Take the chart below.

--Image not done yet--

You can clearly see that the 3d orbital is higher than 4s. That is a visual representation of the fill order. The first two electrons, representing hydrogen and helium, respectively. It now looks like this:

--Image not done yet--

Notice that there are two arrows used. The up arrow describes the +(1/2) electron and the down arrow describes the –(1/2) electron.
The 2s elements are lithium beryllium. Add them and you get.

--Image not done yet--

Then fill up the 2p orbitals one by one. Remember that you fill from left to right. Therefore, there shouldn’t be any 2p orbitals with two electrons until oxygen. When you reach neon, the whole 2p orbital is filled. That is why the Noble Gasses are so unreactive. Their orbital subshells are completely filled, thus leaving no spots for elements to bond with it.
This process goes on and on. Use the fill chart given above to remember what order the electrons fill in.

HYBRIDIZATION

Hybridization is the mixing of normal atomic orbitals to form special orbitals for bonding. It is a type of covalent bonding. For instance, the first type of hybrid orbital is the sp³ orbital. It is named that way because it is a combination of one 2s and three 2p orbitals. From the figure below, you can see how the hybridization occurs.

--Image not done yet--

From the figure below for carbon we see that there are the 2s and 2p orbitals to work with. Since carbon is sp³ hybridized, you take both the 2s and 2p orbitals and create a new set of orbitals, sp³. This means you now have a set of four orbitals. Since you all four electrons with you, you now have an electron in each orbital (fill left to right). Thus, carbon can bond with up to four other elements that can donate an electron to fill each orbital. An example of such an element is hydrogen. The electron of hydrogen can bond with the lone electron for each of carbon’s four hybridized orbitals. Thus, CH₄ (methane) is formed.

Ethylene (C₂H₄) is an example of sp² hybridization. The two carbons are double bonded to each other. For each carbon atom three sp² hybrid orbitals are created. Because of this, one 2p orbital must be left behind. If you looked at this geometrically, you would see the three sp² orbitals coming out of hydrogen with the remaining p orbital perpendicular to the sp² orbitals. See the figure below. The sp² orbitals, when bonded, share a pair of electrons. As a result, two sp² orbitals are each bonded to a hydrogen atom and the other produces a carbon-carbon bond. These bonds are known as sigma (s) bonds.

However, carbon is double bonded. That means that another bond needs to be made to form the double bond. This job is left to the remaining 2p orbital. As shown in the figure below, the p orbital creates what is known as a pi (p ) bond.

For sp hybridization, let us look again at carbon. You should already know by now that sp means that there are two hybridized orbitals. This also means that there are two unchanged 2p orbitals remaining. Let us take the CO₂ molecule. We know that the carbon is double bonded to both oxygen atoms and that each oxygen atom contains two lone pairs of electrons. As stated with the sp² hybridization, sigma bonds must link up the carbon and oxygen atoms. However, since oxygen has two lone pairs of its own, it must undergo hybridization of its own. Oxygen is sp² hybridized. Then, since carbon has two unchanged 2p orbitals, it creates two pi bonds, one with each oxygen atom. And since oxygen is sp² hybridized, it will form a p bond with the carbon atom.
By now you should realize that every double bond is made of a sigma and a pi bond.
Another example of sp hybridization is the N₂ molecule. The nitrogen molecule has a triple bond, with each atom containing a lone pair of electrons. Thus, it must have a sigma bond and two pi bonds. The unused sp orbital contains the lone pair of electrons.
The octet rule states that an atom must always have eight electrons in its valence shell to be stable. However, there are exceptions to this rule. One exception is phosphorous pentachloride (PCl₅). When looking at the molecules’ structure, you notice that there are ten electrons in the valence shell. To do this, the molecule must be dsp³ hybridized. This contains a one d orbital, one s orbital, and three p orbitals. Each dsp³ orbital creates a sigma bond with a sp³ orbital from the chlorine atom. The other three sp³ orbitals create the atom’s lone pairs.
The last example is d²sp³ hybridization. Sulfur hexafluoride (SF₆) is an example. By now, you should pretty much know what will happen.

BOND ORDER & PARAMAGNETISM

Bond order is an indicator of bond strength. It is the number of bonding electrons minus the number of antibonding electrons, all divided by two.

For any molecule, for example O₂, you take the total number of electrons involved, which is 16, and place them across the chart starting at the bottom. Remember to go left to right. Then total up the bonding and antibonding electrons and calculate the bond order. The greater the bond order, the greater the bond strength. Also, while bond order increases, bond energy increases and bond length decreases.
Paramagnetism is when a substance is attracted into an inducing magnetic field. Diamagnetism is when the substance is repelled by an inducing magnetic field.
Looking at the stack chart can identify paramagnetism. If the chart shows that the molecule has any unpaired electrons, as shown above with O₂, then it is paramagnetic.

Non-Ideal Behaviour of Gases

2011-11-01T20:38:00.000-07:00

Deviations from Ideal Behavior

All real gasses fail to obey the ideal gas law to varying degrees

The ideal gas law can be written as:

For a sample of 1.0 mol of gas, n = 1.0 and therefore:

Plotting PV/RT for various gasses as a function of pressure, P:

The deviation from ideal behavior is large at high pressure
The deviation varies from gas to gas
At lower pressures (<10 atm) the deviation from ideal behavior is typically small, and the ideal gas law can be used to predict behavior with little error

Deviation from ideal behavior is also a function of temperature:

As temperature increases the deviation from ideal behavior decreases
As temperature decreases the deviation increases, with a maximum deviation near the temperature at which the gas becomes a liquid

Two of the characteristics of ideal gases included:

The gas molecules themselves occupy no appreciable volume
The gas molecules have no attraction or repulsion for each other

Real molecules, however, do have a finite volume and do attract one another

At high pressures, and low volumes, the intermolecular distances can become quite short, and attractive forces between molecules becomes significant

Neighboring molecules exert an attractive force, which will minimize the interaction of molecules with the container walls. And the apparent pressure will be less than ideal (PV/RT will thus be less than ideal).

As pressures increase, and volume decreases, the volume of the gas molecules becomes significant in relationship to the container volume

In an extreme example, the volume can decrease below the molecular volume, thus PV/RT will be higher than ideal (V is higher)

At high temperatures, the kinetic energy of the molecules can overcome the attractive influence and the gasses behave more ideal

At higher pressures, and lower volumes, the volume of the molecules influences PV/RT and its value, again, is higher than ideal

The van der Waals Equation

The ideal gas equation is not much use at high pressures
One of the most useful equations to predict the behavior of real gases was developed by Johannes van der Waals (1837-1923)
He modified the ideal gas law to account for:

The finite volume of gas molecules
The attractive forces between gas molecules

van der Waals equation:

The van der Waals constants a and b are different for different gasses
They generally increase with an increase in mass of the molecule and with an increase in the complexity of the gas molecule (i.e. volume and number of atoms)

Substance	a (L² atm/mol²)	b(L/mol)
He	0.0341	0.0237
H₂	0.244	0.0266
O₂	1.36	0.0318
H₂O	5.46	0.0305
CCl₄	20.4	0.1383

Example
Use the van der Waals equation to calculate the pressure exerted by 100.0 mol of oxygen gas in 22.41 L at 0.0°C
V = 22.41 L
T = (0.0 + 273) = 273°K
a (O₂) = 1.36 L² atm/mol²
b (O₂) = 0.0318 L /mol

P = 117atm - 27.1atm

P = 90atm

The pressure will be 90 atm, whereas if it was an ideal gas, the pressure would be 100 atm
The 90 atm represents the pressure correction due to the molecular volume. In other words the volume is somewhat less than 22.41 L due to the molecular volume. Therefore the molecules must collide a bit more frequently with the walls of the container, thus the pressure must be slightly higher. The -27.1 atm represents the effects of the molecular attraction. The pressure is reduced due to this attraction.

Kinetics

2011-11-01T03:12:00.000-07:00

Chemical kinetics- the area of chemistry that studies the rates of reactions
Reaction mechanism- the steps involved in a chemical reaction.

With the understanding of the speed and the process of a reaction, chemists can facilitate reactions to their needs.
Reaction rate is simply the change in the amount of product or reactant over time. The amount is measured in molarity while the change in time is in seconds. Therefore the reaction rate is:

The triangle in front of the letters is called "delta" and represents "change of."
A is the reactant or product, and the enclosed brackets stands for the molarity. Think of the reaction rate as the slope of the concentration vs. time graph.

The reaction rate of the reactant is always negative because the reactant is decreasing to form the product.
The reaction rate of the product is always positive because the product is being formed.

Ex. N_2(g)+ 3H_2(g) Þ 2NH_3(g)
The reaction rate is: -D [N₂]/D T. The rate of consumption of hydrogen is 1.5 times as fast as the formation of ammonia because 3 moles of hydrogen will produce 2 moles of ammonia.
A reaction can proceed forward as well as backwards. For now the reverse reaction is considered negligible because the reactions in this tutorial are studied under conditions where the reverse reaction is insignificant.

The rate law, another form of is: Rate = k[A]ⁿ

Where k is the rate constant and n is the order, both values must be determined through experiments.
For the rate law to hold true:

We assume that the reverse reaction is insignificant.
The order, n, is an experimental value.

The rate law of N_2(g)+ 3H_2(g) Þ 2NH_3(g)
Rate = k[N₂]ⁿ[3H₂]^m
The overall reaction order is the total order in the equation: m + n
Again, n and m are experimental figures.

There are two types of rate laws.
Differential rate law- relates the rate law with the concentration of reactant. Ex. Rate = k[A]ⁿ
Integrated rate law- relates the rate law with time. For those of you who have studied calculus, this law is the integration of the differential rate law. Ex. ln[A]=-kt+[A]₀
Three types of rate orders.
0 order
the rate of the reaction is independent of the concentration. The rate is constant.

Half-life is the amount of time it takes for something to decompose to half of its original amount. With a half-life of 5 days, 1kg of road kill would become 0.5 kg in 5 days.

First order
the concentration of reaction is directly proportional to the rate. If you triple the concentration, the reaction rate will triple.

Second order
the reaction rate grows exponentially with concentration. If you double the concentration, the reaction rate quadruples.
Since a reaction occurs in a series of steps, names are given to each step along with the substances involved in the reaction.
O_3(g) Þ O_2(g) + O_(g)
O_3(g) + O_(g) Þ 2 O_2(g)
O_(g)in called an intermediate.

An intermediate does not qualify as a reactant or product, but its consumed or formed.
Each step of reaction is called an elementary step.

Molecularity- minimum number of molecules that must collide for the reaction in that step to take place. Here is how you name them:

Number of reactants	Name	Rate Law
1;[A]Þ products	Unimolecular	k[A]
2;2[A]Þ products	Bimolecular	k[A]²
2;[A]+[B]Þ products	Bimolecular	k[A][B]
3;2[A]+[B]Þ products	Termolecular	k[A]² [B]
3;[A]+[B]+[C]Þ products	Termolecular	k[A][B][C]

So what does temperature have on the reaction rate? For a reaction to occur, molecules must collide, and raising the temperature increases the movement of molecules. Therefore, as temperature increases, the reaction rate also increases.
But the for a reaction to proceed, a specific amount of energy must first be achieved. This energy is called activation energy. A rise in temperature also lowers the activation energy.
Pg. 572 fig. 12.9
Pg. 573 fig. 12.10
For a reaction to occur:

Molecules must produce enough energy to overcome the activation energy.

The molecules in the reactants must be orientated in such a manner that old bonds are broken so new bonds in the product can form.

The most common equation for activation energy is:
Learn this equation, it’s very useful. After all, anybody can plug in numbers and use a calculator.
R=8.3145
A catalyst is a substance that speeds up chemical reaction without being consumed in the process. A prime example is the enzymes in the human body. Chemists use catalysts to speed up slow reactions. An example of this is the production of ammonia, know as the Haber Process.
A catalyst helps a reaction by lowering its activation energy; therefor, speeding up the reaction. Since the reaction is very slow on its own, chemists add a catalyst to speed it up.
Two types of catalyst:
Homogeneous-the catalyst exists in the same phase during the reaction.
O_3(g) Þ O_2(g) + O_(g)
O_3(g) + O_(g) Þ 2 O_2(g)
For this reaction, the O_(g) is the catalyst.
Heterogeneous-the catalyst enters and exists in different phases of the reaction. The catalyst usually provided the area where gaseous molecules can adsorb on its surface, then it forms the product.
The term adsorption is usually related with heterogeneous catalysts. Adsorption is the collection of a substance on the surface of another.

Thermochemistry

2011-10-31T04:33:00.000-07:00

Topics Covered

Physics 'n' Energy
Chemistry 'n' Energy
Enthalpy
Heat and Temperature
Hess's Law
Standard Enthalpies of Formation

Physics 'n' Energy

Thermochemistry is really all about energy. But first we've got to know what the hell energy is. The way that is used to describe it is mostly helpful in Physics. Therefore, I'm gonna describe it in physics first, and then the next section will be energy as it relates to our subject, chemistry. Heh...
Energy is the ability to do work. We've all heard that before. In actuality, it's sort of hard to define energy exactly. Even my normally adequate book admits that. Even though that's sorta fuzzy, you've used the term 'energy' before in everyday speech. It's... well... energy!!!
There are two flavors of energy; kinetic and potential. Let's start with kinetic energy. All matter that is moving has a certain amount of kinetic energy. In fact, if you actually visit my pages, I gave the formula for kinetic energy before:

It is dependant on mass and velocity. Note that it is proportional to the mass and the square of velocity. So if Ball A of some mass and velocity has a certain energy, and you have Ball B, with twice the velocity but only half the mass of Ball A, Ball B will still have twice as much kinetic energy, because velocity is squared. Kinetic energy can be changed, usually by a change in velocity (objects don't just change their mass, they're more likely to drop or gain in speed. Always exceptions, like if a rocket is losing fuel, then it's dropping in mass.) This change in kinetic energy is what is work. Like if a rocket is speeding up, then work is being done on it, because it's gaining kinetic energy. Likewise, if a rocket is slowing down, then the rocket is doing work, because it's losing kinetic energy. Yep, kinetic energy is some good stuff to have.
And then there's potential energy. It's harder to see, because it's the energy you can't see. You can 'see' the energy of a ball being thrown, you feel the energy transfer as you catch a ball. Potential energy is the energy available to the ball to convert to kinetic. In physics, the potential energy you talk about the most in mechanics is gravitational potential energy. This is simply the potential energy caused by gravity. Like if we take that same ball, and hold it off the side of the Empire State Building, it has the "potential" of gaining a lot of kinetic energy (by dropping it). It has a lot of potential energy, but no kinetic (it's not moving...yet.) But once you do, all that potential energy will be converted to kinetic, not all at once, but gradually. The ball speeds up faster and faster as it falls. A rule; the higher an object is, the more gravitational potential energy it has.
Now you don't have to know anything I have just said to do good in chemistry, just take away with you the Law of Energy Conservation. When you add the kinetic and potential energies of that ball at any time during its fall, they will be the same. Energy can't be gained or lost; it is always converted. Even when that ball hits the ground, and has lost all its kinetic and potential energy, that energy didn't really die, it just got converted to something else. The pavement (or car, wherever it landed doesn't matter) will have gained a slight increase in temperature, and that's where the energy went. It was dissipated as heat. The energy in a closed system always remains constant. I say closed system because if you considered the Earth a closed system, and then an outside source of energy (such as the falling of a great meteorite) came in, then the closed system will have more energy.
Energy comes in all sorts of shapes and sizes. There's heat (probably most important energy in chemistry), light, sound, mechanical, electrical, nuclear, matter/anti-matter reactions, the list goes on...
Let's talk chemistry.

Chemistry 'n' Energy

Finally, back to chemistry. In chemistry, reactions can give off heat, or absorb heat. Those that give off heat are called exothermic, and those that take in heat are called endothermic. In exothermic reactions, heat is a product (it's being formed), so a reaction of this kind might look like this:

A + B ---> C + D + heat

And similarly, if a reaction is endo, then it acts like a reactant (goes on the left side):

A + B + heat ---> C + D

Where does this heat flow from and to? In chem, you consider the entire Universe to be divided into two parts: the system, and the surroundings. The reaction is the system, and the surroundings is everything else. For the sake of pictures, pretty pictures, let's say the system is pink and the surroundings are purple. Here are illustrations of an exothermic and endothermic reaction.

Where is this heat coming from? I said that the energy in a closed system is constant, so in an exothermic reaction, how come this heat is coming from nowhere? It's not really. Before the reaction, there was some potential energy stored in the bonds that made up the chemicals. When the bonds were broken and new bonds were formed to make new things, the energy of the new bonds was less than the energy they had previously. So that potential energy that was 'lost' was actually converted into heat, and that's where the heat came from. So the total energy level of the system and surroundings has in fact remained equal. This is in fact the first law of thermodynamics.

In chemistry, you always look at the system's point-of-view. The energy of the system can be changed in two ways: either change the heat of the system, or make it do work or work done on it. The energy change of the system is equal to the amount of heat added to it, plus the amount of work done on it. In other words (or symbols):

DE is the change of energy that happens to the system. q is the heat added, and w is the work done to it.
In thermodynamics, there is the concept of + and - signs. Since we are talking about point-of-view of the system, let's see how a + or - q/w means. If q is positive, that means the system will gain heat. If it's minus, then heat is being removed. That's simple enough. If w is positive, then work must be done on something as to make the energy of the system gain. If you do work on the system, then w will be positive. If you are expending energy, doing work, to make the system happy, then the system will gain energy. Similarly, if you're tired of doing work for this system that has done nothing for you, you can let the system do work for you. In which case w is negative, since it will be losing energy.

The unit of energy is the joule (Symbol: J). It is equal to the amount of kinetic energy a 2 kilogram ball has when traveling at 1 m/s. If bigger units are needed, the kilojoule (kJ) is used, and it's obviously 1000 joules.
I think we can do a problem or two here.

Example:

Calculate the change in energy of a system if it did 14.3 kJ of work, while giving off 34.5 kJ of heat. Answer
The most important part of this problem is to figure out the sign of w and q. It's giving off heat, therefore losing energy, so you can expect q to be negative. If it is doing work, then it is also losing energy in that way too, so w is negative as well. All you gotta do is add them up (-14.3 kJ + -34.5 kJ) and you will learn that the system has lost a total of 48.8 kJ.

So the change of energy of a system is only dependant on the work a system does and the heat flow. Note the difference between heat and temperature: THEY ARE NOT THE SAME!!! Heat is energy. Temperature, is... well... how hot something is. It is the heat flow that causes a change in the temperature. If you add heat (energy), then the object's temperature will go up; by how much, you cannot tell by just knowing the heat involved. You'll learn more of this later.
Let's talk about a way some system can do work. There's many ways, so we'll just cover one; gases. Think about what you do if you squeeze a balloon. Aren't you doing work to it? You are using your hard-earned energy to push this balloon in, and since energy is always transferred, it has to go somewhere, so it goes into the system of the gas in the balloon. So, if a gas is being crushed (volume is getting smaller), then work is being done to it, and w in that case is positive.
And if the volume is getting bigger, then w is negative. Why? Because the gas is pushing against the walls of the container, to give itself more room. So it's doing work, or losing energy.
The work being done is the pressure times the change of volume, like this:

w = PDV

This equation is not done yet. If volume is increasing (DV is positive) , then by equation it says that w is positive. But we just said that the system will be expending work in increasing its volume, so w should be negative when the change in volume is positive. A simple negative sign should fix that. And your final equation for the work done by a gas changing volume is:

You can see how I could incorporate this into a problem involving the first and second equations, but I won't. You get the idea.

Enthalpy

Aaah, yes, enthalpy. If you've just heard of enthalpy, you will have no idea what it is. When you're done with enthalpy, you'll still not have a clear good idea of what it is, but you will know how to do problems with it, and that's what is important.
The enthalpy of a system, H, is simply defined as:

H = E + PV

Enthalpy is equal to the total energy of the system, plus the pressure of the system times the volume of the system. It's sort of hard to grasp what enthalpy is, from that definition. Instead of enthalpy (H) itself, you will usually deal with a change of enthalpy (DH). And if pressure is constant, and the only work allowed to work on the system is through volume, then:

DH = q

Yep, you should think of enthalpy as sort of like heat. It's not heat exactly, but if those two conditions are met, then it is heat.
So if the change of enthalpy is increasing, that means it is gaining an increase of energy, and therefore is endothermic. If the change of enthalpy is decreasing, that means it is losing heat to the surroundings, or exothermic. Final thought: +DH = Endo, -DH = Exo.
How do you find the change of H? It's the enthalpy of the final products, minus the enthalpy of the reactants. Or...

DH = H_products - H_reactants

So if the products have less energy than reactants, then DH will be negative, indicating energy was lost. And vice versa.

Heat and Temperature

I hope by now you realize the difference between heat and temperature. If you add heat (energy) to something, it will eventually start getting hotter and hotter. But the difference is, some substances will get hotter than others, if given the same energy. For example, on a hot summer day, everything around you might seem very hot (like the pavement, or your car), but if you go into a pool of water, it will be much cooler. Both the pavement and the water have received about the same amount of energy from the sun, but the pavement has gone through a much bigger temperature change than the water. Water is resistant to a change of temperature, in comparison to the pavement. Let's assume the temperature change is proportional to the energy it has received, so we can come up with an equation relating temperature and energy.

C is called the heat capacity and is different for different substances. This number C will tell you how hard is it for this substance to get hot. As you can figure, the bigger C is for some substance, the more energy is needed to get it hot.
We are forgetting one important factor; how much of the substance there is. The more there is, the more energy is needed to change its temperature. Putting all three things into account, we can come up with a more useful equation.

This is the same equation as the one before, but notice there's an m in the bottom. This is the definition for specific heat capacity. It's equal to the heat added divided by the mass of it and the change in temperature. As with the normal heat capacity, the bigger this number for some substancee, the harder it is to change its temperature. Water's specific heat capacity (or simply specific heat) is around 4.18 J/^oC g.
Important! The mass in the above equation must be in grams. Well, if you use kilograms consistently, I guess it doesn't matter, but since specific heats in tables in books are given with grams, you should use grams.
Rearranging the above equation, you can solve for energy, since you can figure out the other three by measuring but not always energy.

This starts a whole new area called constant-pressure calorimetry. You can figure out how much energy is given off by certain reactions, as long as the pressure is constant. Otherwise, some of the energy would be lost/gained by changing the pressure and your results would be off. You will do a lab with this, probably doing the reaction with styrofoam cups. Why? Because you don't want the energy escaping into the atmosphere or to the walls of the container; you want accurate results. Styrofoam is an insulator, doesn't absorb much energy. The basic setup is as follows:

You use two styrofoam cups to make sure all the energy involved in the reaction is directly related to the temperature change of the chemicals, not the container/air. The cover on top is to make sure energy isn't leaving from the top. The thermometer is to measure the temp change.
How exactly do you do this? Let's say you're going to measure the energy given off by adding 50.0 mL of Chemical A and 20.0 mL of Chemical B. You can start with either one, let's start with A. Measure the weight of the cups beforehand. Put the 50.0 mL of A into the cup. Measure the temperature before. And then add B to it, and cover it and mix periodically. Keep on look at the temp, it should be changing. When it's done changing and stays the same, record the new temp. That's all there is to it. (And then measure the mass after the two things were put together.)
Then in q = s x m x T, you can subtract the masses to find the mass of the chemicals (in grams!). You can subtract the final temp minus the initial temp to change delta-T. But what about s? You can use water's specific temp, 4.18 J / K m. Why? Because most chemicals you'll be using won't be pure chemicals, but rather solutions. In case u forgot, that means they're dissolved in water, and it's not enough to significantly change the 4.18. So now you can find out how much heat is gained or lost.
There is one more concept here. Since pressure is constant, we can figure that the heat lost/gained is directly related to the enthalpy change. You could say that q = DH, but you'd be WRONG! In actuality, q = -DH. Why? Remember that enthalpy looks at energy from the point of the system. If heat was released (indicated by positive q), that means the system itself has lost it. Or if heat was absorbed (negative q), that means the system has gained it. So heat evolved and enthalpy change are just opposite of each other.
We've talked about constant-pressure cal. in detail; I'll just touch on constant-volume calorimetry here. Instead of keeping constant pressure, it tries to keep volume constant. That means you need a container that won't change volume. They're usually big strong metal cubes called a bomb calorimeter. I won't go through the specifics, because you probably won't be doing a lab with it; just be aware.
One last thing; in addition to specific heat capacity, there's the molar heat capacity. Instead of using grams to measure amount of substance, it uses moles. I won't give you equation, or any examples, because using specific heat is by far what everyone uses. Of course it has different units; instead of J/ K g, it's J/K mol.

Hess's Law

This has to do with enthalpy again. Yay. Remember that reactions have a certain enthalpy change with it: if it's positive, then the reaction will absorb energy; if negative, the reaction will give off energy. We can write the equation, and then the delta-H associated with it to the right of that. Like for example, to show the enthalpy change for the boiling of one mole of water to water vapor, you can write like this:

H₂O_(l) ---> H₂O_(g) DH = 44 kJ

So before one mole of water evaporates, 44 kJ of energy must be given to it.
I had to look up the 44 kJ in a table or something; there's no way anyone can just look at the equation and come up with it. But how about reactions that can't be found in a table? There are tons of reactions; you just can't carry a 30 pound book with you and look them all up. Hess's Law exists to make this easier for you. Technically, it means that the enthalpy change between two states is not dependant on the pathway it takes to get there. At first, this doesn't seem to help you one bit; but it means if you can add two or more equations to get the desired equation, then you can add their respective enthalpy changes to get the enthalpy change of this equation. There are two rules you will need to use in these problems. We'll use the equation above to illustrate these two rules:

If you have to multiply each side of a reaction by some number X, then multiply the respective enthalpy change of that reaction by X also. So, if we needed to use not

H₂O_(l) ---> H₂O_(g), but twice that:

2H₂O_(l) ---> 2H₂O_(g)
Then we just multiply the delta-H of the original reaction by the same factor, two. Therefore the enthalpy change of 2 moles of water evaporating is 88 kJ.
If you have to flip the equation around (the right side on the left and the left side on the right; switching the products and reactants; you get the idea) then just take the negative of the delta-H to get the delta-H of your new reaction. So if we wanted to find the enthalpy change of one mole of water vapor condensing to liquid water, like this:

H₂O_(g) ---> H₂O_(l)
Then you just take the negative of the original delta-H, so the enthalpy change of the above equation is -44 kJ.

So now we can find enthalpies of certain reactions without a table. For example, if we needed to know the enthalpy change when 3 moles of water vapor condenses to water, all you do is flip the original equation and multiply by 3, or -132 kJ.
This is all you're going to learn in this section. But there are more involved problems. For the sake of simplicity, I'm not going to use real chemicals and enthalpies, because I don't know them. I'll just use chemicals "A", "B", and so forth. Have fun!

Example:

Please find the enthalpy change of

A + 2B ---> C

using the following information:
#1: E ---> D + 2A, DH = 43.22 kJ
#2: B ---> F + [1/4]D, DH = 342 kJ
#3: 4F + E ---> 2C, DH = 2.2 kJ
Answer
This is about the hardest problem you might get. They can get harder, but I think they're past the scope of this course (I hope!) You have to add the equations below it to get the final equation. Where do we start? It's actually quite simple if you start with the beginning equation. We need a single A on the left. Well, look at the first given equation. It has a 2A in it, and no where else will you find an A. So you KNOW that the A on the left hand side MUST come from Equation #1.
You have to modify this equation so that the A's here look like what you want it to (the beginning equation). There's 2A's on the right. You want it to look like a single A on the left. So we can divide by two, and flip the equation.

A + [1/2]D ---> [1/2]E

Good! Now we have 1 A on the left. But remember that if we modify the equation, we have to modify the delta-H on this one too. You flipped it, so you can take the negative. So delta-H = -43.22 kJ. But we also divided by two, so divide this by two, and our new delta-H for #1 is -21.61 kJ. Keep this number in mind.
Let's move on to the next figure in the beginning equation: 2B on the left. Well, out of the three givens, only #2 has a B anywhere. So that's where we must get our B's from. Problem is, the B is on the left like we want it, but there's only one. So multiply by two:

2B ---> 2F + [1/2]D

Coolio. Now we got 2B on the left. That's it. But remember to multiply original delta-H by 2, so the new one is 684 kJ.
And finally, we need one C on the right. Since we used #1 and #2, we might suspect that C lies in #3. And of course, there it is, on the right of #3 like we want it. But there are 2 instead of one. What to do? Of course you divide by 2:
2F + [1/2]E ---> C
And then divide 2.2 kJ by 2, or 1.1 kJ for the new reaction.
Now what? Add the three new equations together (put all lefts on left, and all rights on right):
A + [1/2]D + 2B + 2F + [1/2]E ---> [1/2]E + 2F + [1/2]D + C
It looks like it's more fukt up than before, but look carefully and something magical happens; don't you have a [1/2]D on left and [1/2]D on right? You can subtract [1/2]D from both sides and cancel them. Same with 2F, and [1/2]E! So cancelling them, you have:

A + 2B ---> C

Wow, it's amazing, it's the beginning equation! Since all you did was add the modified equations to get the desired equation, all you have to do to find the desired enthalpy change is add the modified enthalpy changes:
DH = -21.61 kJ + 684 kJ + 1.1 kJ = 663 kJ.
Woohoo! That's the enthalpy change for A + 2B ---> C. It's strange how the given equations just coincidentally cancelled out to make exactly the equation we were looking for. It's not coincidence; these type of problems must be planned out. It's a LOT harder making one of these problems than it is to solve them!

Standard Enthalpies of Formation

This has to do with enthalpy once again, as you might have brilliantly deduced from the title. Basically, the standard enthalpy change of formation of something is the enthalpy change of the reaction of the elements in their natural state coming together to form it under standard conditions. For example, the reaction of the formation of water is:

H_{2 (g)} + [1/2]O_{2 (g)} ---> H₂O_(l)

Note that hydrogen and oxygen must be in diatomic form and gaseous, because that is how they exist in a standard state. What is this standard state? In thermodynamics, the standard state is at 1 atmosphere and 25 ^oC (about room temperature).
The standard enthalpies of formation for many substances can be found in a table. What is so useful about this? Let's say you have a reaction that you have to find the enthalpy change in the standard state. But instead of before where we had all those given equations to mess around with, you are just given a table of standard enthalpies of formation. How will you do this? Here's the equation:

DH^o = Sn_pDH^o_{f p} - Sn_rDH^o_{f r}

You might be staring at this mess and wondering what the hell it means, but basically it says that you add up the standard enthalpies of formation on the right side (multiplying with respective coefficients) and subtracting the standard enthalpies of formation on the left side (ditto.) You know, it just might be easier if you saw this one in an example. Once again, I am using A's, B's, C's, and D's, but in a real problem you'll get real things to use.

Example:

Find the change of enthalpy of the following equation (under standard conditions):

A + 2B ---> 2C + 3D

Using the following standard enthalpies of each substance:
A: 34 kJ/mol
B: 45 kJ/mol
C: 22.3 kJ/mol
D: 56 kJ/mol
Answer
Firstly, I should explain the kJ/mol. Enthalpies are supposed to be in joules, or kilo joules, right? But for formation, the enthalpy depends on how much you are forming. So, for A, it takes 34 kJ to form one mole of A. So if you were forming 2 moles of A, the enthalpy would be twice as much, or 68 kJ. If you were making 3.45 moles of B, the enthalpy of formation would be 155.25 kJ (3.45 mol x 45 kJ/mol).
Ok, let's start. The formula says to take the stuff on the right, and subtract the stuff on the left. So start with the first thing on the right, 2C. The standard enthalpy for C is 22.3 kJ/mol, but we have 2 of them. So the enthalpy for 2C is 44.6 kJ.
Next up on right; 3D. D's enthalpy of formation is 56 kJ/mol, and 3 of them makes 168 kJ.
So you add up to get total enthalpy on right, which is 44.6 kJ + 168 kJ = 212.6 kJ. Keep this in mind.
Now we go to the left. First thing on left: A. There's only one of A, so total enthalpy is 34 kJ.
Lastly, there are 2 B's. Enthalpy of 1 B: 45 kJ, but there are two, so total is 90 kJ.
Now you add up all the enthalpies on the left. So that is 34 kJ + 90 kJ = 124 kJ.
Last step: Subtract total of left from total on right. 212.6 kJ - 124 kJ = 88.6 kJ.
And there's your answer!
One note: You can only do this if the reaction is to take place in standard conditions. I don't know if I mentioned this before, but enthalpy is dependant on temperature and pressure. So if the reaction above took place at 10 degrees Celsius and at 30 atm's, you couldn't do it this way. Of course, if they gave you enthalpies of formation at those conditions, then go right ahead. (They have to match.)

Types of Chemical Reactions

2011-10-31T04:19:00.000-07:00

Most reactions in chemistry take place with water as the solvent. Solutions where water is the solvent are called aqueous solutions. Water is known to be a polar molecule because the unequal distribution of charge caused by the electronegative oxygen. This polarity gives the water the ability to dissolve ions and other polar substance.
A useful method for characterizing an aqueous solution is its electrical conductivity. If a solution conducts electricity well, it is considered a strong electrolyte. If it only conducts slightly, it is considered a weak electrolyte; if it doesn’t conduct, it’s a nonelectrolyte.
Strong electrolytes are substances that are completely ionized in water. Such examples are strong acids, strong bases, and soluble salts. Acids ionize into H⁺ and A^-; bases ionize into OH^- and X⁺.
A weak electrolyte is a substance that only slightly ionizes when added to water. These substances include weak acids, weak bases, and slight soluble salts. Weak acids are those that only dissociate slightly into H⁺ and A^-. Similarly, weak bases are those that dissociate only slightly into OH^- and X⁺.
Nonelectrolytes are substances that dissolve in water, but that don’t break up into ions. These substances are mostly polar molecules; the reason they don’t conduct electricity is that no ions are formed
Solution Reactions
One type of reaction is called a precipitation reaction. This occurs when two solutions are mixed resulting, and a solid or precipitate forms. The precipitate contains ions that when combined are insoluble with water. However, these individual ions do dissolve in water. So, the result of mixing these ions is an insoluble solid. One example of a precipitation reaction is when a solution containing Ca²⁺ and a solution containing SO₄^2-; the result of mixing these the formation of solid CaSO₄.
Another type of reaction is an acid-base reaction. An acid-base reaction is one when the net reaction is the combination of a proton and hydroxide to form water. The acid base reaction is also called a neutralization reaction. When performing calculations fo an acid-base reaction follow these simple steps:

List all the species before any reaction occurs and decide what reaction will occur.
Write the balance net-ionic equation
Calculate the moles of reactants
Determine the limiting reactant where appropriate
Calculate the moles of the required reactant or product.
Convert to grams or volume, as required.

The final type of aqueous reaction is an oxidation-reduction reaction. An oxidation-reaction is characterized by the transfer of one or more electrons. These reactions are often used for energy production; in fact, these reactions are often used in the human body to provide energy.
The concept of oxidation states provides a means for keeping track of the movement of electrons in a redox reaction. The oxidation states can be assigned according to these rules:

The oxidation state of an atom in an element is 0.
The oxidation state for a single atom ion is the charge of the ion.
Oxygen is assigned the oxidation state of –2 in covalent compounds, except in peroxides where oxygen is assigned a –1 state.
In covalent compound hydrogen is assigned a +1 state.
In a covalent compound, fluorine is always –1 state.
The sum of the oxidation states must equal to overall charge of the molecule.

When balancing oxidation it is convenient to divide the reaction into two half-reactions; one reaction involves the oxidation, the other the reduction. Then, follow these steps if it’s in acid:

write the half reactions
for each half reaction balance the element except H and O, balance the O with water, balance the H with H+, and balance the charge with electrons.
If necessary, multiple the reaction by an integer to equalize the number of electrons.
Add the half reaction

If the reaction occurs in base repeat the above steps, but after balancing the hydrogen with H⁺, add hydroxides to cancel out the H⁺’s. Then, continue the above steps as before.

Periodic Trends

2011-10-31T04:15:00.000-07:00

Ionization energy is defined as the energy required to remove an electron from a neutral atom.
Electronegativity is a measure of the tendency of an atom to attract or gain an electron. Nonmetals are usually said to be electronegative; metals, electropositive.
Atomic size for the reactive elements is based on atomic radii that are based on bonding distances in compounds.

The inert gas sizes are determined in a different manner, so are not included. They do show the same vertical trend in size and ionization energy.Elements in the last horizontal row-elements 87 - 118 - are not included, although those for which data is available follow the same general trends.
Based on the observed trends, francium and cesium (on the bottom left), are the most metallic elements (i.e. the most likely to lose an electron) while fluorine (on the upper right) is the most nonmetallic (i.e. the most likely to gain an electron).

The above periodic trends are caused by the interactions of three factors: nuclear charge (the number of protons in the nucleus), the electron shell(s), and the shielding (the effect of the electrons between the outer electrons and the nucleus). These effects are summarized in Table 1 below. Similar arguments can be used to explain the trends in ionization energy and electronegativity.

Table 1. Factors affecting horizontal and vertical trends in atomic size:

	Horizontal Behavior		Vertical Behavior
Factors:	Change	Effect on Size	Change	Effect on Size
Atomic number: number of protons	increases	decreases	increases	decreases
Electron shell(s)	stays same	no effect	shell added	increases
Shielding (intervening electrons)	stays same	no effect	increases	increases
Net Effect:		decreases*		increases*

* the increased shielding and the added shell(s) override the effect of the increasing nuclear charge.

Another helpful bit of information is a list of the common nonmetals in decreasing order of electronegativity - F, O, Cl, N, Br, I, S, C……H - (pronounced fossil n brisk to help memorize).
This, along with geometric considerations, will help to determine the polarity (another lesson) of small covalent compounds.

Solutions

2011-10-29T22:26:00.000-07:00

A solution is defined as a homogeneous mixture or a mixture where the components are uniformly mingled.

Normally we thing of a solution is the liquid state, but this is a common misconception; a solution can be in any state. For example, air is a solution of oxygen, nitrogen, and a variety of other gases all in the gas state. Steel is also solution of carbon and iron, but this solution exists in the solid state. However, the most important solutions in chemistry involve water, so this section will focus of aqueous solution.
Before we begin discussing the properties of solutions, we must define ways to describe the concentrations of various mixtures:

Molarity (M): the number of moles per liter (moles/Liter)

Mole fraction (c ): the ratio of the number of moles of a compound to the number of moles of solution (c A = moles A / (Moles A + Moles B))

Molality (m): the number of moles solute1 per kilogram solvent (m = moles solute/ kilograms solvent)

Mass percent or weight percent: the percent by mass of a solute1 in a solution (Mass percent = mass solute / mass solution * 100)

1 A solute is the compound being dissolved in medium. The solvent is the medium in which the compound in being dissolved.

Forming a solution

Dissolving compound in liquids, such as water, is very common; so it would be valuable to understand how and why compound form solutions. To address these questions we must examine what happens when you add a compound to a liquid:
Step 1: the solute particles separate (overcoming the intermolecular attractive forces).
Step 2: the solvent particles separate (overcoming the intermolecular attractive forces).
Step 3: the solute and solvent interact to form the solution.
Each of these steps results in a change of energy. We can sum these energy changes to find the energy change of the solution, called enthalpy (heat) of solution (D H_soln). Therefore, the D H_soln is:

D H_soln = D H₁ + D H₂ + D H₃

The first two steps usually result in a large positive D H, but the third step is usually negative. If the first two enthalpies are greater that the third, the overall D Hsoln will be positive and vice versa. It is important to note, that process with a large positive D Hsoln tend not to occur; however, process with small positive D Hsoln may occur because nature strives for disorder (entropy), and a solution is more disordered than its individual components.

The above concepts can be applied to understand structure can effect solubility; let’s use water mixed with oil for an example. The first two steps from above result in a positive D H, because it requires energy to break water’s hydrogen bonding, and oil’s dispersion forces. The third step does not counter this with a large negative D H, because there is little interaction between oil and water. Thus the D Hsoln is large and positive, so the reaction is not favored. Salts do dissolve in water, because they have a significant interaction with water resulting in a low D Hsoln. Salts are describes as hydrophilic (water loving) where oils are called hydrophobic (water fearing).
Pressure can also effect the solubility of gasses; the relationship between pressure and the concentration of dissolved gas is described in Henry’s law; the amount of gas dissolved is directly proportional to the pressure of the gas above the solution:

P=kC

Temperature effects the solubility of solutions. One normally thinks that by heating the solution, more solute can be dissolved: this is a misconception. There is come correlation between D Hosoln of a solution and the solubility (see Le Chatelier’s Principle), but the temperature dependence can only truly be seen by experiments.

Gases dissolved in a solution follow a simple rule; the higher the temperature, the less gas dissolved.
Effects of solutions
When a nonvolatile solute is added to a solvent, the vapor pressure of the solvent is lowered. Fancois M. Raoult studied phenomenon, and his results are desribes by Raoult’s Law:

P_soln = C_solventP^o_solvent

One is able to use a simple model to understand the phenomenon; think of the dissolved solute molecules evenly distributed in the solution, even at the surface. Then understand that this caused less molecules of volatile solvent to be at the surface and less molecules evaporating.

This decrease in vapor pressure as a result of the solution can help to explain the following colligative properties, properties that depend only on the number of molecules not he molecules themselves.
When a nonvolatile solute is added to a solvent, the boiling point of the solution is higher than that of the pure solvent. This can be explained by the decreased vapor pressure of the solution; the solution boils when the vapor pressure matches the outer pressure (usually 1 atm), but since the vapor pressure is lowered, a higher temperature is needed to match that outer pressure. This change is given by the equation:

D T = Kbm_solute

Similarly, when a nonvolatile solute is added to a solvent, the freezing point of the solution is lower than that of the pure solvent. This is also explained by the decreased vapor pressure; the solution freezes when the vapor pressure for the liquid and solid phases is the same. This occurs at a lower temperature because of the decreased vapor pressure. This change is given by the equation;

D T = Kfm_solute

The osmotic pressure of a solution is also a colligative property. Osmotic pressure (p ) is the pressure that must be applied to stop osmosis, or the flow of solvent down a concentration gradient. The solvent attempts to move to an area where there is a lesser concentration; the driving force of this movement is an increase in disorder. Experimentally, the osmotic pressure has been show to follow the equation:

p = MRT

where p is the osmotic pressure in atmospheres, M is the molarity of the solute, R is the universal gas law constant, and T is the temperature.
All these colligative properties must be slightly changed for electrolytes, which breakup into multiple pieces. The van’t Hoff factor i must be inserted into the equations. The van’t Hoff factor I is:

i = moles of particle in solution / moles of solute dissolved

Therefore, the new modified freezing point and boiling point equation is:

D T = iKm

The modifies osmotic pressure equation is:

p = iMRT

One should note that ion pairing may result is slightly lowered results than calculated. For example I for NaCl is 2, but the observed i for a .10 m solution is 1.87. This is caused by a few ions pairing together making a fewer number of molecules.

Colloids

A colloid is similar to a solution, except that a colloid is a suspension of single large molecule or clumps of small molecules. Mud is an example of a colloid; the mud particles stick together and remain suspended in water for an extremely long period of time. The reason these particles remain in solution is that they have layers of oppositely charged articles. When the charges on the outside of these particles are the same, they experience electrostatic repulsion and remain suspended.

Colloids can be destroyed, called coagulation, by heating or adding an electrolyte.

Gases

2011-10-29T21:08:00.000-07:00

Following concepts will be discussed in this post.

Pressure
Temperature
R, the Gas Constant
A Few Gas Laws
Back to Stoich
Kinetic Molecular Theory of Gases
Effusion and Diffusion
An Alternative to the Ideal Gas Law

This is a different section. The sections before this just built upon each other, all having to do with chemicals and reactions and formulas. The first part of this chapter will have almost nothing to do with that. Then of course, we must tie it back in. But it's a good break from stoich, you must agree.
Before anything, I hope I don't have to define "gas" for you. Well, just in case, a gas is a substance that doesn't have a definite shape nor volume. Obviously gases will form the shape of their container. And their volume is simply the volume of the container they're in.
Pressure
Gases push and expand. It's in their nature. That's why they will evetually take the shape of their container: they push and expand, hitting the walls and being stopped. That's because gases have particles that have lots of motion, flying around, in comparison to solids, whose molecules are almost fixed. Gases push against the walls of their container, with pressure. Pressure is defined as a force per unit area. It is simply not enough to say how much it pushes, but also how much surface area it is pushing on.

For example, if you press a book against your hand with a relatively strong force, it doesn't hurt. That force is acting on a large area, so you don't feel much pressure. But what would happen if you push a needle pointwise first into your hand, with the same force? You will definitely feel a pressure. Because that force is acting on a very small area, the pressure will be greater.
You can easily see the pressure by a gas in a balloon. The reason it stays round is because the gas inside is under pressure, pushing at the walls in every direction. If you try to squeeze it, you can feel the gases' pressure increasing, pushing against you. (That's one of the gas laws.)
So what are units of pressure? Pressure is force divided by area, or newtons per square meter. The SI system doesn't like to have different units' names in other units, so they call a newton per square meter a pascal (Symbol: Pa). So 1 pascal of pressure is equal to a force of 1 newton pressing down on a square meter. If you don't know force, a newton is pretty small. Imagine a small force delivered to a relatively big area of 1 m². A pascal is very small. A more common unit is the kilopascal (Symbol: kPa). Obviously 1 kPa = 1000 Pa.
That's the SI version. Obviously, here in the great United States, we like to use different units that run different to the rest of the world, and pressure is no exception. We use pounds per square inch (Symbol: psi), which is the same concept as the pascal, just with different units. More common units to use in chemistry are the torr and the atmosphere. To discuss any of these requires a discussion of atmospheric pressure:
Air doesn't weigh much. Yet it does have a little weight. Look at all the air above us. There's a ton of it in the atmosphere. It's hard to imagine all that air, and that we don't feel it crushing us down. Air does weigh a lot. But since it's force is spread out over such a large surface as the earth, it's not as much. Still, the pressure is equal to about 101325 pascals, or almost 15 pounds per square inch. That means that a force on a 8.5 by 11 inch piece of paper should be almost 1400 pounds! How come we don't feel all this atmosphere pressure? Because it acts in all directions. That piece of paper feels 15 psi pushing down, pushing up, pushing left, pushing right, pushing in all directions. So the net force, and pressure, is zero. The point is, wherever there is air, it will be pushing at 15 psi in all directions. So, let's take a can. There is air inside and outside a can, and the air inside is pushing outward at 15 psi, and the air outside is pushing inward at 15 psi, and the can doesn't implode nor explode.
Would the atmospheric pressure be greater or less at higher altitudes than at lower altitudes? Well, as you go up, there is less and less air pushing down on you, since a lot of it is now below you. So air pressure will decrease as you go up. That's why they pressurize air inside an airplane, to keep the pressure at normal 15 psi inside, while it's much lower outside. If you thought someone could open the door of the plane, it would take a couple of tons of force to get that door open. Plane doors swing inward, and the much greater pressure on the inside would oppose the door moving in.
One more thing about atmospheric pressure: why vacuums suck. If you have a hollow cube, and the air is taken out of the middle, you have an enormous force crushing all sides of the cube. If you make a tiny hole in this cube, then all this air will try to rush in, because it pushes, and you get a sucking effect.
Anyway, I hope you get what atmospheric pressure is. They named a unit of pressure after it, appropriately called the atmosphere (Symbol: atm). 1 atm is equal to the pressure of the atmosphere, at sea level. So, 1 atm = 101325 Pa. To explain torrs (or mmHg, same thing), we must set up a lab.
Fill a dish with some mercury. Fill a tube completely with mercury. Turn this tube really quickly upside down into this dish. Let's see what happens:

This is how it looks like. Immediately after you flip it, the mercury will want to fall out of the tube, due to gravity. But, there is one force acting on the liquid that will oppose that: atmospheric pressure. It is pushing on the surface of the liquid, and pushing the column back up, like so:

Those arrows are pushing on the liquid, and forcing the column to go back inside the tube. These two forces on the column will come to an equilibrium (they will balance). The result: the distance between the top to the tube and the top of the mercury is always 76.0 cm (or more commonly in mm, 760 mm), at least at sea level. That is the concept of the mm Hg (pronounced millimeters of mercury): 1 mm Hg is the amount of pressure that would make that column of vacuum above the mercury be 1 mm tall. Therefore, the atmospheric pressure is 760 mm Hg. Also, another name for mm Hg is the torr. Just another way of saying it. So 1 mmHg = 1 torr.
There are other units of pressure, like bars or mm H₂O, but they aren't used commonly, at least not in chemistry. So that's about it for pressure. It will come back to haunt you later, in fact, probably in the next section. So beware!
Temperature
You might feel silly to be reading about temperature. You know what it is. It's how hot or cold something is. But in terms of particles and chemistry and such, what exactly is temperature? It's a measure of the average kinetic energy of the particles in a gas. What is kinetic energy? The definition is as follows:

m stands for mass, and v stands for velocity. So if something is moving faster than before, it will have more kinetic energy than before. And so temperature is really the measure of how fast the particles are moving around. (You generally ignore the mass part of it in temperature, since particles are known to change speed but usually not mass... :)
So now you can see why if you heat a liquid enough, it will turn into a gas (evaporate). As it gets an increase in temperature, its getting an increase in kinetic energy (and speed of particles). If you get high enough, the particles will be moving fast enough and be going crazy, and becomes a gas.
Note that temperature is a measure of the average kinetic energy of all the particles. Not all particles will be traveling at the same speed. But the average of the energies is related to the temperature.
There is no upper limit on temperature. You can always give something more and more energy, and the particles will just get faster and faster. There is a lower bound tho. Since the slowest a particle can move is 0 mph, temperature will be at its lowest when the particles are absolutely still. The temperature at which this theoretically happens is called absolute zero. This is very very very cold, much more cold than the coldest day in the North Pole. It's -459.67 ^oF, or -273.15 ^oC. You know, that's cold. This is the temperature at which all motion of particles will stop. This temperature has never been reached by anyone anywhere, but people have come close to it (within .00000001 ^oC or something). Wouldn't it make sense if we started a temperature scale that had this temperature as 0? So that we wouldn't need to deal with negative numbers? The Kelvin scale is the scale we need. The symbol is K (without the stupid circle; not ^oK.) 0 K is absolute zero. Therefore there are no negative Kelvins. The good thing is, 1 K = 1 ^oC. The Kelvin and the degree Celsius are the same unit, equal. The only thing different is where they start. So, converting is easy. Let's assume that we can round -273.15 to -273. This is the number commonly used to convert. All you do to the Celsius temperature is add 273 to it, and that's how many Kelvins you have. A quick few examples:

0 ^oC = 273 K.
10 ^oC = 283 K.
-30 ^oC = 243 K.
100 ^oC = 373 K.

You get the idea. And to go the other way; if you wanted Celsius from Kelvin, just subtract 273. You generally don't do that, unless a question asked for units of Celsius or something.
Why go through all this? What was wrong with the Celsius system, it never failed us! But it did, because the numbers weren't proportional. Something at 20 degrees Celsius did not have twice as much kinetic energy as something at 10 degrees Celsius. Back when they invented the Celsius and Fahrenheit scales, they had no idea that there was a bottom of the scale. They thought it went infinite in both directions. So they just picked a temperature and called it zero. (For celsius, it was the freezing point of water.) But with the Kelvin scale, you can truly say that something at 20 K has twice as much kinetic energy as something at 10 K. All of the equations that involve temperature in this section will need to be in Kelvins. So remember to convert!
And one last note: I always said that temperature was related to the average kinetic energy, but not exactly how. Here is the equation (you'll probably never use it, but if you're interested... knock yourself out.)

KE is the average kinetic energy, T is temperature (in Kelvins of course!) and R is...well...let's talk about that next.
R, the Gas Constant
R is a gas constant. It has a constant value. It's used in many equations. There are two forms of this constant (meaning different units). There's a version that involes energy, and the version that involves volume and pressure. Learn to use which one in each equation! (For example, since the equation above in the last section had energy, but not pressure nor volume, you would use the energy form.)

Pressure-Volume form:

Energy form:

In case you don't have a graphics browser, the pressure-volume form is .08206 L atm / (K mol), and the energy form is 8.3145 J / K mol. These two values are exactly the same, they are just used in different equations.
A Few Gas Laws
There are laws all over in science. There just so happens that gases must follow some laws of their own. Let us go over some relationships between some properties of gases.
Boyle's Law
This law says: The pressure and the volume of a gas are inversly proportional to each other, temperature kept constant. What this means, is that if you keep the temperature constant, then if you increase the pressure on the gas, the volume will decrease, and if you lessen the pressure on a gas, the volume will increase.
If you crush a gas, it's gonna shrink.
Real world example: that balloon thing again. If you squeeze a balloon enough, it pushes back, until the gas is too pressurized and the balloon pops. Bigger balloons are easier to squeeze. (Note: I don't know anything about squeezing balloons. I have never done it. I'm just assuming this stuff, because it's the only example I could think of. If I am misinformed about balloon squeezing, please tell me and I will fix it, hehe.)
Boyle's Law can be expressed like this: for every gas, assuming temperature is constant, there exists a constant, say, b, and this equation will hold true for all values of P and V obtained simultaneously: b = P / V.
Believe it or not, you can do some example problems with this!

Example:

Please solve the following problems.

The value of P / V for a certain gas is 3.42 atm / L, at a temperature of 300 K. How much space is needed to contain the gas at 5 atmospheres at 300 K?
At 273 K, 10 liters of a gas was pressurized at 3.00 atmospheres. How pressurized would that gas be if it only took up 5 liters of space at 273 K?

Answers

This is middle-school stuff. Sorry! So P / V = 3.42 atm/L. You know V = 5 L, so just solve for P. P = (3.42 atm/L) x V
P = (3.42 atm/L)(5 L)
P = 17.1 atm.
Here is a way that my Chem I teacher taught how to do this stuff. Boyle's Law can also be written as:
V_oP_o = V_NP_N
This is pronounced "Vopo equals Vinpin." V stands for volume, P for pressure, and subscripts 'o' and 'N' denote whether it's the volume/pressure of the old state or the new one. In this equation, the temperature must remain constant. Since the temperature stays at 273 K in this equation, let's try it.
What are we looking for? We have two states: one at 10L/3 atm, and a new one, at which V = 5L, but pressure is unknown. So we are looking for pressure of new state, "Pin", or P_N.
Solving for P_N in that equation, we have:

P_N = V_oP_o / V_N
Substitute:
P_N = (10 L)(3.00 atm) / (5 L) = 6.00 atm.
The new pressure will be 6.00 atm. That makes sense; since we reduced volume (crushed it), it should be more pressurized. Furthermore, since we halved the volume, we should expect the pressure to double.

Charles's Law
This law says: The temperature and the volume of a gas are directly proportional, assuming pressure is kept constant. That means as the temperature goes up, the volume will go up as well.
This is easy to picture. If the temperature goes up, the particles get faster, and they want to move outward more. The gas expands, and the space it takes up increases. The opposite also happens; if temperature is lowered, volume will be less.
This is easily observed in a hot air balloon. To make a balloon go up, they light a fire inside of it. This will heat up the gas inside the balloon. So the volume will go up. Since mass stays the same, the density of the air inside will become less. The reason a balloon floats is because the air inside is less dense than the air outside. Wow...
Just as we could write Boyle's Law with some sort of equation, you can write Charles's Law too.
For every gas there exists a constant, say, d, at which V = dT, pressure being held constant.
So here are some examples! (Actually, just one.)

Example:

Solve: A sample of gas is to be kept at a constant 3.00 atm. The gas is at 265 K, and takes up 2.45 L of space. If the temperature is to be increased 30 degrees, what should the new volume of the gas be, to keep the pressure equal?
Answer
Just like the above shortcut, there is a nice little equation that will compare two states of the gas. The way to say it is "Votin equals Vinto." Equationwise, it looks like this:

V_oT_N = V_NT_o

For an explanation of subscripts and such, see previous example.
Solving for the new volume, or Vn, you should have:

V_N = V_oT_N / T_o

Plugging in, your answer should be:
V_N = (2.45 L)(265 K + 30 K) / (265 K)
V_N = 2.73 L

So Boyle's Law and Charles's Law have to do with relationships between different properties of gases, how they vary with each other. There is one more, and this one's easy: Avogadro's Law.
Avogadro's Law
This one is easy: the moles of gas is directly proportional to the volume. Easy; if you have twice as many moles of gas, it will take up twice as much space, assuming you keep that other crap constant (such as pressure and temperature).
This also leads to something interesting. Before we go into this "interesting" thing, I introduce to you STP. Standard Temperature and Pressure. When a gas is at STP, it is understood that it is at O ^oC, and at 1 atm.
Now for the interesting thing. At STP, one mole of gas, any gas, will occupy exactly 22.4 liters. Doesn't matter how big the particles are, and such.
If you must need an example on this law...well...hrm...you suck!
Combining Boyle's and Charles's Laws
This one is combining volume, pressure and temperature into one easy equation to remember. It is pronounced, "Vopotin equals Vinpinto", and you can find one of them, given the other five. The equation is written as follows.

V_oP_oT_N = V_NP_NT_o

This is one step below the almighty Ideal Gas Law. Try a problem out with this:

Example:

A very cautious scientist (perhaps Mieze) has measured very precise values for a gas he has. The pressure was found to be 2.349202 atms, the volume to be 2423.53929 mL, and the temperature was 24.23411 ^oC. Just as he finished, a very clumsy student walked in (I won't say who he is, but his initials are J.A.P.), and screwed it up, breaking the scientist's thermometer in the process. The scientist frantically tries to measure the new information (finding the pressure has changed to 3.423103 atms, and the volume at 2.34131248 L) but can't use his broken thermometer. What did he calculate the temperature to be? Answer
Sorry for the extra-long question, but I have to keep you interested! We see five numbers, so obviously we can use the VoPoTn=VnPnTo equation to get the sixth. We are looking for the new temperature, so solving for Tn, you get:

T_N = V_NP_NT_o / ( V_oP_o )

Just plug-'n'-chug. Make sure you convert to proper units.
T_N = (2.34131248 L)(3.423103 atm)(24.23411 + 273 K) / (2.42353929 L)(2.349202 atms)
T_N = 418.415 K = 145.415 ^oC = 293.747 ^oF.
That J.A.P. really screwed up.

Dalton's Law of Partial Pressures
This is the easiest law. The pressures of gases are additive. If you put a gas at A atms into a container, and then put another gas of B atms into it, then the total pressure in that container is A + B. Now it's time for the big daddy of all Gas Laws...the Ideal Gas Law.
Ideal Gas Law
Well, here it comes, the law that will tie it all together: the Ideal Gas Law. It ties in all the quantities together we have discussed so far, and a simple and easy-to-memorize equation is what you get.

(Sorry for overdoing the pic!)
This equation is called "Piv-Nert". That is how you pronounce it. You will be using this for many many things, so get used to it! Basically, this equation will let you find the property that you don't know, given that you know the other three.
P - pressure (in atms).
V - volume (in liters).
n - quantity of gas (in moles).
R - the gas constant discussed earlier. Hopefully, you recognize this R as being the pressure-volume form, and is therefore .08206 L atm / K mol.
T - temperature, IN KELVINS!!! The biggest mistake people make early on is forgetting to convert to Kelvins. So do it!
And that is it. This equation is probably most useful for finding the moles of gas, since all the other quantities can be measured easily. Pressure is found by some kind of gauge, volume is simply the volume of the container, and temperature can be found with a thermometer (I hope you knew that.)
I won't give any examples, because you can do plug-'n'-chug, and because any other problem that might require some tinkering with the Ideal Gas Law can be easily solved with the VoPoTn = VnPnTo equation.
Note that this is called the "Ideal" Gas Law. Real gases don't exactly follow this law, mostly because the pressure and volume aren't as they seem to be. So what is an ideal gas? There isn't such a thing in the real world, it's simply a model that tries to mimic experiment data as much as possible, and still be simple. Let's go back and see what's wrong with this model.
Pressure. The Ideal Gas Law assumes that there are no interactions between molecules. They just bounce around, with some kinetic energy. But in reality, the molecules are attracted to each other. Therefore, they wouldn't be hitting the walls of the container as hard as are expected. Therefore the actual pressure will be greater than the observed pressure.
Volume. The Ideal Gas Law assumes that the molecules are points that don't take any actual space themselves. But in reality...molecules take up space!!! (Don't act surprised.) So the actual volume that is there is less than what is observed, since the molecules take some space up which the gas should be taking up according to the Ideal Gas Law.
These factors will be taken into consideration when coming up with a more realistic gas law.
Back to Stoich
I'm so sorry, but we do have to return back to the world of moles and LRs and formulas. It was a good break from it tho. This section will not teach you anything new, just help you connect stoich and this new junk you had to learn. The key is...moles!!! Stoich is always about the moles. So if you get something in grams... convert to moles! If you get a volume, pressure, and temperature... use Pivnert to get moles!
Here's a typical problem you might have to do.

Example:

Solve the following, oh pretty please? A 100 mL of CH₃OH (Density = .850 g/mL) is mixed with some oxygen, 32.21 liters at 3.00 atmospheres and 28 ^oC. The products of the reaction are carbon dioxide and water. Find the number of moles of water formed.
Answers
Write the reaction first:

CH₃OH + O₂ ---> CO₂ + H₂O

Check to see if it is balanced. It isn't, so balance it!

2CH₃OH + 3O₂ ---> 2CO₂ + 4H₂O

Now you must use all those numbers above to get moles for some of these things. Let's begin with CH₃OH. There is 100 mL, and density is .850 g/mL, so we can easily find the mass, 85.0 grams (no calculator there, folks!) Now to get moles, simply calculate the molar mass (12 + 3 + 16 + 1 = 32 g/mol), and divide.
85.0 / (32 g/mol) = 2.65625 moles CH₃OH.
The second one is the part we learned in this section. You can just use the Ideal Gas Law to find the moles needed.
PV = nRT
n = PV / RT
n = (3.00 atm)(32.21 L) / [(.08206 L atm / K mol)(28 + 273 K)]
n = 3.91214 moles.
Now we must find the limiting reactant. Let's assume oxygen is the limiting reactant. Then let's see if we have enough of that other stuff:
3.91214 moles O₂ x [2 moles CH₃OH / 3 moles O₂] = 2.60809 moles of other stuff.
2.60809 moles of stuff is what is required. We have more than that (2.65625), so oxygen is indeed what will run out first. From here on out you know what to do.
2.60809 moles CH₃OH x [ 4 moles H₂O / 2 moles CH₃OH ] = 5.22 moles H₂O.

Here's a special formula. Nothing out of the ordinary, just derived from the Ideal Gas Law. It's how to find the molar mass of the gas, given pressure, temperature, and density.

d is density, in grams per liter, instead of milliliter, because R has liters in it.
There's some things to notice in this. If the temperature goes up of a gas, then density goes down. That's how a hot air balloon works, by decreasing the density of the gas inside by increasing temperature.
Mole fractions. I thought I mentioned them before, looks like I didn't. This is how much a mixture is made up of a particular gas. The symbol for it is a greek 'x' thing, like X. It's very easy. Let's say you have x moles of gas A, and y moles of gas B put together. The mole fraction of gas A will be:

x moles
-------------------------
Total moles (x + y)

It's like a percentage, but you don't multiply by 100. Mole fractions have no units, since mol cancels out.
Quikie: You have 2 moles of helium mixed with 5 moles of xenon. The mole fraction of helium is 2 moles / (2 moles + 5 moles) = 2/7, or 0.286. The mole fraction of xenon is .714. Note that all the mole fractions of all the gases making up a mixture must add up to 1. (Just like all percentages add up to 100%). You can think of the mole fraction as the probability of picking up one random particle and getting that gas.
Now, the cool thing is that pressure and mole fraction are related. When two or more gases are put together in the same container, their mole fractions are proportional to the pressures, and vice-versa. Why? Because when gases are put together, their volumes are the same, and so is their temperatures. The only thing that is different is their pressures and mole fractions, and so they are proportional.
What does this mean for you? It means you can do stupid little problems like... (Two gases, with .432 and .568 for mole fractions, have a total combined pressure of 3.00 atms. What is the pressure of each gas?)
Note that the pressures of the individual gases added together will be the total pressure observed, because of Mr. Dalton's law.
Let's talk about vapor pressure, since it is needed in some problems.
Vapor Pressure
Let's say you want to collect the gas of a reaction. One way of doing this is like so:

You let the gas bubble out to the top, and there's a little space above the water where the gas will reside. One very important thing to realize is that the pressure of the 'stuff' above there isn't the pressure of the gas being collected. There is also water vapor in that space, which is contributing to the pressure.
You say, 'Where is this vapor coming from?' From the water, of course! This means the pressure observed is greater than the actual pressure of the gas being collected, since the pressure measured includes both the water vapor pressure and the gas pressure.
So what do you do? You will get the pressure of water vapor as a given, with the problem. Then just take the pressure you measured, and subtract the water vapor pressure. And there's your true pressure of the gas.

I'm very sorry for the lack of examples in this section, since that is what it was intended for; to help you solve problems. But the truth is, it's very hard to make up a problem without copying it from the book, and I don't like to do that. Plus, it's hard on the fingers. Maybe when this whole page is done, I will come back and do these. Of course, by then, you will have graduated college and wouldn't care, would you?

Kinetic Molecular Theory of Gases

The Kinetic Molecular Theory of Gases (henceforth the "KMT") is the model we've been using, in these Laws, including Ideal Gas Law. It's a model, which tries to mimic real-world observations as much as possible, but also keeping it simple. As a result, you get a really simple equation to work with, but it makes some assumptions that aren't really true. Here is what this model is assuming:

The molecules of gas don't take up any volume. They are simply points of mass.
The molecules are always moving, and the average of the energy is related to temperature.
The molecules don't have any forces exerted upon each other.

These are wrong! But... they can be ignored, resulting in a formula that actually does a good job of predicting gases in the real world, at least in certain situations.
Now my book goes through the trouble of deriving the Ideal Gas Law, but I don't want to bother you. You can thank me later.
Perhaps the only thing you might need to know is the root mean square velocity. This equation will give you the average of the squares of the speeds of the particles. This can be easily derived, since you know the mass of a particle, and the energy (you learned the equation way back up there). Here it is:

Capital M? Doesn't this mean molarity? Nope, not here. There are 26 letters, actually 52 both UPPER and lower, and almost as many Greek letters, and it's funny why they reuse the same letters... M means kilogram per mole. So just divide the molecular weight of a particle of gas by 1000.
Effusion and Diffusion
Tired of gases? We still have two quick topics to cover until you become a full-fledged PhD in AP Gases. Effusion and diffusion wasn't really covered in my class. They basically compare the rates at which gases travel from one place to another. This is only relative rates; you'll see more later.
Effusion is the rate at which particles will travel from one compartment to another totally empty compartment (vacuum), separated by only a small hole. To help ya visualize this, of course here's a visual (that's what they're for):

The particles go in the direction of that yellow arrow, towards the empty chamber. It does so at a constant rate, until they are balanced. Different gases will do this at different speeds. Generally, the lighter the particle is, the faster it will effuse. All right, all right, I'll stop yakking and just give you the formula, that's all you ingrates want anyway.

Note that this will not give you any rates. It will tell how much faster Gas A will effuse than Gas B. Note the capital 'M' again, meaning the kilograms per mole. But in this case, it doesn't really matter, since units cancel. Just keep molecular weight units consistent with both gases.
So let's find out how fast hydrogen gas will effuse compared to oxygen. Since hydrogen is lighter, we know that it will be faster, but by how much? Let's call hydrogen Gas B, since its rate will be on top on the left part. Therefore, helium's molecular weight will go on top on the right (note that it's flip-flopped?)
Rate of H₂ / Rate of O₂ = Ö(32 / 2) = 4.
This means hydrogen effuses 4 times as fast as oxygen.
Diffusion is when a gas travels across a distance. Like in a room, when someone farts at one end of the room, and it travels to the other end, it's diffusing. Some gases travel faster than others. Generally, lighter gases travel faster, and the whole concept is exactly the same as effusion. Same formula, that's good news 4 u.
An Alternative to the Ideal Gas Law
We already said that the Ideal Gas Law could be better, accounting for the fact that molecules take up space, and have attractions to each other. Well, here it comes. We know that we must subtract from volume, and add to pressure to account for those things. Therefore, here's the "real" (or at least better) Ideal Gas Law.

There are no new variables except for this 'a' and 'b'. They are constants that depend only on the identity of the gas. This formula will be more accurate than the Ideal Gas Law. The only thing is, you can't use them unless you are given the constants a and b. Once you do, tho, you can solve for anything.

States of Matter

2011-10-29T20:47:00.000-07:00

Matter: States of Matter

by Anthony Carpi, Ph.D.

As a young boy, I remember staring in wonder at a pot of boiling water. Searching for an explanation for the bubbles that formed, I believed for a time that the motion of the hot water drew air down into the pot, which then bubbled back to the surface. Little did I know that what was happening was even more magical than I imagined - the bubbles were not air, but actually water in the form of a gas.
The different states of matter have long confused people. The ancient Greeks were the first to identify three classes (what we now call states) of matter based on their observations of water. But these same Greeks, in particular the philosopher Thales (624 - 545 BCE), incorrectly suggested that since water could exist as a solid, liquid, or even a gas under natural conditions, it must be the single principal element in the universe from which all other substances are made. We now know that water is not the fundamental substance of the universe; in fact, it is not even an element.
To understand the different states in which matter can exist, we need to understand something called the Kinetic Molecular Theory of Matter. Kinetic Molecular Theory has many parts, but we will introduce just a few here. One of the basic concepts of the theory states that atoms and molecules possess an energy of motion that we perceive as temperature. In other words, atoms and molecules are constantly moving, and we measure the energy of these movements as the temperature of the substance. The more energy a substance has, the more molecular movement there will be, and the higher the perceived temperature will be. An important point that follows this is that the amount of energy that atoms and molecules have (and thus the amount of movement) influences their interaction with each other. Unlike simple billiard balls, many atoms and molecules are attracted to each other as a result of various intermolecular forces such as hydrogen bonds, van der Waals forces, and others. Atoms and molecules that have relatively small amounts of energy (and movement) will interact strongly with each other, while those that have relatively high energy will interact only slightly, if even at all, with others.
How does this produce different states of matter? Atoms that have low energy interact strongly and tend to “lock” in place with respect to other atoms. Thus, collectively, these atoms form a hard substance, what we call a solid. Atoms that possess high energy will move past each other freely, flying about a room, and forming what we call a gas. As it turns out, there are several known states of matter; a few of them are detailed below.

©Corel Corporation

Solids are formed when the attractive forces between individual molecules are greater than the energy causing them to move apart. Individual molecules are locked in position near each other, and cannot move past one another. The atoms or molecules of solids remain in motion. However, that motion is limited to vibrational energy; individual molecules stay fixed in place and vibrate next to each other. As the temperature of a solid is increased, the amount of vibration increases, but the solid retains its shape and volume because the molecules are locked in place relative to each other. To view an example of this, click on the animation below which shows the molecular structure of ice crystals.

Liquids are formed when the energy (usually in the form of heat) of a system is increased and the rigid structure of the solid state is broken down. In liquids, molecules can move past one another and bump into other molecules; however, they remain relatively close to each other like solids. Often in liquids, intermolecular forces (such as the hydrogen bonds shown in the animation below) pull molecules together and are quickly broken. As the temperature of a liquid is increased, the amount of movement of individual molecules increases. As a result, liquids can “flow” to take the shape of their container but they cannot be easily compressed because the molecules are already close together. Thus liquids have an undefined shape, but a defined volume. In the example animation below we see that liquid water is made up of molecules that can freely move past one another, yet remain relatively close in distance to each other.

Gases are formed when the energy in the system exceeds all of the attractive forces between molecules. Thus gas molecules have little interaction with each other beyond occasionally bumping into one another. In the gas state, molecules move quickly and are free to move in any direction, spreading out long distances. As the temperature of a gas increases, the amount of movement of individual molecules increases. Gases expand to fill their containers and have low density. Because individual molecules are widely separated and can move around easily in the gas state, gases can be compressed easily and they have an undefined shape.

Solids, liquids, and gases are the most common states of matter that exist on our planet. If you would like to compare the three states to one another, click on the comparison animation below. Note the differences in molecular motion of water molecules in these three states.

Plasmas are hot, ionized gases. Plasmas are formed under conditions of extremely high energy, so high, in fact, that molecules are ripped apart and only free atoms exist. More astounding, plasmas have so much energy that the outer electrons are actually ripped off of individual atoms, thus forming a gas of highly energetic, charged ions. Because the atoms in plasma exist as charged ions, plasmas behave differently than gases, thus representing a fourth state of matter. Plasmas can be commonly seen simply by looking upward; the high energy conditions that exist in stars such as our sun force individual atoms into the plasma state.
As we have seen, increasing energy leads to more molecular motion. Conversely, decreasing energy results in less molecular motion. As a result, one prediction of Kinetic Molecular Theory is that if we continue to decrease the energy (measured as temperature) of a substance, we will reach a point at which all molecular motion stops. The temperature at which molecular motion stops is called absolute zero and has been calculated to be -273.15 degrees Celsius. While scientists have cooled substances to temperatures close to absolute zero, they have never actually reached absolute zero. The difficulty with observing a substance at absolute zero is that to “see” the substance, light is needed, and light itself transfers energy to the substance, thus raising the temperature. Despite these challenges, scientists have recently observed a fifth state of matter that only exists at temperatures very close to absolute zero.
Bose-Einstein Condensates represent a fifth state of matter only seen for the first time in 1995. The state is named after Satyendra Nath Bose and Albert Einstein who predicted its existence in the 1920’s. B-E condensates are gaseous superfluids cooled to temperatures very near absolute zero. In this weird state, all the atoms of the condensate attain the same quantum-mechanical state and can flow past one another without friction. Even more strangely, B-E condensates can actually “trap” light, releasing it when the state breaks down.
Several other less common states of matter have also either been described or actually seen. Some of these states include liquid crystals, fermionic condensates, superfluids, supersolids and the aptly named strange matter. To read more about these phases, visit the Phase page of Wikipedia, linked to below in the Further Exploration section.

Phase transitions

The transformation of one state of matter into another state is called a phase transition. The more common phase transitions even have names; for example, the terms melting and freezing describe phase transitions between the solid and liquid state, and the terms evaporation and condensation describe transitions between the liquid and gas state. Phase transitions occur at very precise points, when the energy (measured as temperature) of a substance in a given state exceeds that allowed in the state. For example, liquid water can exist at a range of temperatures. Cold drinking water may be around 4ºC. Hot shower water has more energy and thus may be around 40ºC. However, at 100°C under normal conditions, water will begin to undergo a phase transition into the gas phase. At this point, energy introduced into the liquid will not go into increasing the temperature; it will be used to send molecules of water into the gas state. Thus, no matter how high the flame is on the stove, a pot of boiling water will remain at 100ºC until all of the water has undergone transition to the gas phase. The excess energy introduced by a high flame will accelerate the liquid-to-gas transition; it will not change the temperature. The heat curve below illustrates the corresponding changes in energy (shown in calories) and temperature of water as it undergoes a phase transition between the liquid and gas states.

As can be seen in the graph above, as we move from left to right, the temperature of liquid water increases as energy (heat) is introduced. At 100ºC, water begins to undergo a phase transition and the temperature remains constant even as energy is added (the flat part of the graph). The energy that is introduced during this period goes toward breaking intermolecular forces so that individual water molecules can “escape” into the gas state. Finally, once the transition is complete, if further energy is added to the system, the heat of the gaseous water, or steam, will increase.
This same process can be seen in reverse if we simply look at the graph above starting on the right side and moving left. As steam is cooled, the movement of gaseous water molecules and thus temperature will decrease. When the gas reaches 100ºC, more energy will be lost from the system as the attractive forces between molecules reform; however the temperature remains constant during the transition (the flat part of the graph). Finally, when condensation is complete, the temperature of the liquid will begin to fall as energy is withdrawn.
Phase transitions are an important part of the world around us. For example, the energy withdrawn when perspiration evaporates from the surface of your skin allows your body to correctly regulate its temperature during hot days. Phase transitions play an important part in geology, influencing mineral formation and possibly even earthquakes. And who can ignore the phase transition that occurs at about -3ºC, when cream, perhaps with a few strawberries or chocolate chunks, begins to form solid ice cream.
Now we understand what is happening in a pot of boiling water. The energy (heat) introduced at the bottom of the pot causes a localized phase transition of liquid water to the gaseous state. Because gases are less dense than liquids, these localized phase transitions form pockets (or bubbles) of gas, which rise to the surface of the pot and burst. But nature is often more magical than our imaginations. Despite all that we know about the states of matter and phase transitions, we still cannot predict where the individual bubbles will form in a pot of boiling water.

A video tutorial

Stoichiometry and Chemical Equations

2011-10-29T07:52:00.000-07:00

Stoich is short for stoichiometry. Don't be scared off by that name; I remembered in ninth grade when I heard that I thought I wasn't going to get through the year. It's a 13-letter name for something that's very easy, yet important part to becoming successful in chemistry. This is where the stuff becomes important to learn, because it's directly on the test.

The Mole and Molar Calculations

I am surprised to hear that this is the point where some chemistry students lose all hope; trying to figure out what a mole is. I admit, I was confused too, but keeping this in mind made life a lot easier: a mole is simply the measure of how many. A similar unit is a dozen, and since everyone knows what this is, I'll use this first.
Let's say you have a dozen eggs. What does this mean? It just means you have twelve of something. It doesn't say how much it weighs, or how much space it occupies, just how much. A dozen siege tanks doesn't have the same mass nor volume to a dozen specks of dust. There's just twelve of each.
Moles are the same concept. One mole of something has 6.022 x 10²³ objects in it. It's a lot, but atoms are small, so it's a good unit in chem. 6.022 x 10²³ is N_A, called Avogadro's number. It's a pretty big number. One mole of seconds is 4 million times as long as the age of the earth. One mole of meters is over 600 times the diameter of the Milky Way. For our purposes, it's pretty good.
It's important to understand the relation of amu's and moles. One mole of something that weighs one amu is one gram. In other words, 1 g = 6.022 x 10²³ amus. So one mole of hydrogen (H2) would weigh 1.0079 x 2 = 2.0158 grams. One mole of carbon is exactly 12 grams; that's how they derived Avogadro's number.
Often it's important to find out how many moles is some given quantity of mass. You can use dimensional analysis (DA) for this. Let's say you want to know how many moles is 36 grams of water, and also how many molecules are in it. First you must find the molar mass of water (how many amus it weighs). Water is H₂O, that means 2 of hydrogen and one of oxygen. So the adding of those masses is:
1.008 x 2 + 16.00 = 18.00 amu.
Now comes the DA part.

So it's 2 moles. And since 1 mole is N_A molecules, there are 2 x N_A = 1.2 x 10²⁴ molecules.
The key to solving these problems is getting the g/mol (molar mass, or amus) of the substance, and then using that in a DA setup to convert back and forth between grams and moles. One amu is one g/mol.

Example:

How many moles are in 4.5 grams of glucose (C₆H₁₂O₆)?AnswerFirst you must find out the molar mass of glucose. Adding the weights of the individual atoms, you get:
6 x 12.00 amu + 12 x 1.008 amu + 6 + 16.00 amu = 180.1 amus = 180.1 g/mol
Then you use DA:

The answer is .025 moles. Be careful to use DA correctly, and that you will have to flip it so that the units in the top and bottom cancel out.

Example:

You have a sample of glucose (C₆H₁₂O₆), that weighs 2.5 grams. How many carbon molecules are in it?AnswerFind the number of moles of glucose (we already found molar mass of glucose in last problem, 180.1.)

In each glucose molecule there are six carbons. So if you have .014 moles of glucose, you must have 6 times as many carbons in it. .014 moles x 6 = .084 moles C. Just multiply it by Avogadro's number to get the answer.
.084 moles x 6.022 x 10²³=5.1 x 10²² carbon molecules.

Percent Mass

This is just another way to show how much of something is in a compound. It's how much of the substance is made of one particular element, the percent by mass (I bet you could figure that out by the title). So if there is 60 g of something, and in that compound there is 12 g of carbon, then you would say the percent mass is 20% (12 g / 60 g x 100%).
Note that this number can't be directly obtained from the moles. Percent moles is not the percent mass too. You must convert or die.

Example:

Find the mass percent of each element of penicillin (C₁₄H₂₀N₂SO₄).Here's a hint: if they don't give you how much there is, assume one mole.
So if there is one mole of penicillin, there must be 14 moles of carbon, 20 moles of hydrogen, and so on.
To find the mass of each element in this mole of penicillin, just do the DA using the g/mol conversion.
Mass of carbon = 14 moles C x 12.01 g / mol = 168.4 g.
Mass of hydrogen = 20 moles H x 1.008 g / mol = 20.16 g.
Mass of nitrogen = 2 moles N x 14.01 g / mol = 28.02 g.
Mass of sulfur = 1 mole S x 32.07 g / mol = 32.07 g.
Mass of oxygen = 4 moles O x 16.00 g / mol = 64.00 g.
The total mass can be found just by adding:
168.4g + 20.16g + 28.02g + 32.07g + 64.00g = 312.6 g
Percent mass is mass of one thing divided by the total mass, times hundred percent:
Percent mass carbon = 168.4 g / 312.6 g x 100% = 53.87%
Percent mass hydrogen = 20.16 g / 312.6 g x 100% = 6.449%
Percent mass nitrogen = 28.02 g / 312.6 g x 100% = 8.964%
Percent mass sulfur = 32.07 g / 312.6 g x 100% = 10.24%
Percent mass oxygen = 64.00 g / 312.6 g x 100% = 20.47%
And the percents add up to 99.993%, which is close enough to 100%.

Empirical and Molecular Formulas

The molecular formula is the type that we've always been using: H₂O, C₆H₁₂O₆, C₂H₄, and all the like. It's how many atoms of each element are present in one molecule of the substance. The empirical formula is the reduced form of this. It's the ratio of atoms. For H₂O, there's 2 hydrogens for every oxygen. So the empirical formula is the same. However in C₆H₁₂O₆, there is a ratio of 6 : 12 : 6. That can be reduced, so the ratio is 1 : 2 : 1. Therefore the empirical formula is CH₂O. Likewise, the ratio 2 : 4 in the last one can be reduced to 1 : 2. The empirical for it is is CH₂.

Example:

The empirical formula of a compound is found to be C₂H₃O₄. The molar mass was found to be 273.1 g / mol. Find the molecular formula.Answer

Let's find the molecular mass of the empirical formula.
12.01 g/mol x 2 + 1.008 g/mol x 3 + 16.00 g/mol x 4 = 91.04 g / mol.
And since the molecular formula is just a multiple of the empirical, let's see if 273.1 g / mol is a multiple of 91.04 g / mol.
273.1 g/mol / 91.04 g/mol = 3.000.
Sure enough, the number goes divides almost evenly three times. That means the molecular must have three times the number of particles of the empirical. Multiplying each by three gives the molecular formula to be C₆H₉O₁₂.

It's important to remember that the molecular formula is a multiple of the empirical formula in both subscripts and molar masses.

Chemical Equations and Balancing Them
_________________________________________________________________________________
A chemical reaction is when one or more molecules rearrange their atoms to form new molecules. Here is an example:

CH₄ + O₂ ---> CO₂ + H₂O

You can see this visually below:

	+		->
Methane		Oxygen

		+
	Carbon Dioxide		Water

The things on the left side are called the reactants, and the things on the right are called the products. The reactants are generally transformed into the products until one of the reactants runs out, or the system reaches equilibrium (a balance).
As another example, if you wanted to say that two molecules combine to form one, just use the coefficients. Just like in algebra, you can put numbers in front of the molecule to show how many of them are in the equation. Let's say 2 of Molecule A and 3 of Molecule B are needed to make 4 of Molecule C. The equation would look like:

2A + 3B ---> 4C

Another thing that's important to right down is what state each reactant and product is, by little subscript letters. The symbols are:

Solid: (s)
Liquid: (l) (When you write it by hand, it's a lowercase cursive l.)
Gas: (g)
Aqueous (means dissolved in water): (aq)

So the first equation should have really been written as:

CH_{4 (g)} + O_{2 (g)} ---> CO_{2 (g)} + H₂O_(g)

It's funny how they teach you to do that, but on the AP test on the Equation Writing section, they will not look for those symbols. But your teacher most certainly will, so get into the habit!
That equation is better, but it still is missing a key point in chemical reactions. Atoms cannot be created nor destroyed; they can only be put in different orders. If you look above, there are two oxygens (green) on the left and three oxygens on the right. Where did that extra oxygen come from? The key is to realize that these aren't one to one reactions. That is, they all don't have coefficients of one. So you have to balance.
Let's start with hydrogen. There's four on the left and two on the right, so multiply the molecule on the right by two:

CH_{4 (g)} + O_{2 (g)} ---> CO_{2 (g)} + 2H₂O_(g)

Now the hydrogens are balanced, but the oxygens aren't. Now we have 4 on right and 2 on left. Multiply atom with oxygen on left by two:

CH_{4 (g)} + 2O_{2 (g)} ---> CO_{2 (g)} + 2H₂O_(g)

If you count them, everything adds up. There are the same number of atoms on the left as on the right. That is the correct chemical equation.

Here is The Art of Balancing Instruction Manual, by none other than Takalah:

Before doing anything, check if it all adds up.If the same number of atoms exist on both sides, then it's already balanced and you're done.If not, start your balancing with the atom that is combined with another on both sides, and the one that has the most atoms. You really do want to start with the most complex one. For instance, in C₂H₅OH + O₂ CO₂ + H₂O, you would start with H first, because it's in a molecule with other atoms on both sides, and the numbers of atoms (5 and 2) are greater than the other one that fits the first criteria (C).

Leave the atoms that stand by themselves for last. Like in the equation in the last rule, you would save O₂ for last. Since changing the coefficient for this only affects one atom, you can use it as a "finishing touch", balancing the oxygens after all the other ones are done, but without messing another atom up.
Don't change the molecular formula!!!As frustrating as some of these problems can be at first, don't be tempted to change one little number in some molecular formula just to make something balance. Remember that changing the coefficient will just change the number of particles that are involved in the reaction. Changing the subscripts will change the whole molecule into something else. The molecule is a unit. So don't do it!

Example:

Write balanced equations for the following:

Glucose (C₆H₁₂O₆) reacts with oxygen gas to form carbon dioxide and water.

Indium reacts with oxygen to make indium (III) oxide.

AgNO_{3 (aq)} + H₂SO_{4 (aq)} ---> Ag₂SO_{4 (s)} + HNO_{3 (aq)}

Answer

The unbalanced reaction is: C₆H₁₂O_{6 (s)} + O_{2 (g)} ---> CO_{2 (g)} + H₂O_(g)The most complicated atom is H, since it's in glucose and water, and that the subscripts are the largest. Let's see; there are 12 on the left and 2 on the right, so let's try multiplying right by 6:C₆H₁₂O_{6 (s)} + O_{2 (g)} ---> CO_{2 (g)} + 6H₂O_(g)
Now you move on to the next complicated atom; C, since it's in both carbon dioxide and glucose. There are six on left, and one on right; try multiplying right one by 6. You get:
C₆H₁₂O_{6 (s)} + O_{2 (g)} ---> 6CO_{2 (g)} + 6H₂O_(g)
Everything is basically done now, you got have to finish it by balancing oxygens using the O₂ molecule. There's 6 on left; and 6x2+6 = 18 on right. (Remember that you have to multiply coefficient by subscript to get number of atoms.) So there must be 12 atoms on right to balance it. Twelve oxygen atoms is 6 oxygen molecules, so the final answer is:
C₆H₁₂O_{6 (s)} + 6O_{2 (g)} ---> 6CO_{2 (g)} + 6H₂O_(g)
You have to remember the stuff you learned in the last part to do this. Indium (III) oxide is In₂O₃. So the equation is:In_(s) + O_{2 (g)} ---> In₂O_{3 (g)}Start with the most complicated one, which is oxygen. Multiplying left by three and right by two gives
In_(s) + 3O_{2 (g)} ---> 2In₂O_{3 (g)}
Now to balance In's, there are four total In's on right. So there must be 4 In's in left. Answer is:
4In_(s) + 3O_{2 (g)} ---> 2In₂O_{3 (g)}
AgNO_{3 (aq)} + H₂SO_{4 (aq)} ---> Ag₂SO_{4 (s)} + HNO_{3 (aq)}This is the hardest one yet. Oxygen is too hard to do right now, since it's in all four. Let's start with Ag, by getting two on left:2AgNO_{3 (aq)} + H₂SO_{4 (aq)} ---> Ag₂SO_{4 (s)} + HNO_{3 (aq)}
That changes the N to 2N. Looking at the right, we have only one N. Let's multiply that one by 2 too.
2AgNO_{3 (aq)} + H₂SO_{4 (aq)} ---> Ag₂SO_{4 (s)} + 2HNO_{3 (aq)}
Now we get 2 H's on the right, but that's favorable, since we have 2 H's on left. In fact, if you find out how many oxygen's are on each side, you gots ten, and that's your answer. Note that solving these can be like a chain reaction; solving for one gets you how many of another, then you balance that, and it goes on.

Stoichiometric Calculations

One thing to realize when doing any calculations is that moles and coefficients are interchangeable. Both mean number of particles, or multiples thereof. So in 2A + 3B ---> 4C, it either means 2 particles of A combine with 3 particles of B to make 4 particles of C, or 2 moles reacts with 3 moles to make 4 moles.
With that in mind, here is a simple problem.

Example:

The formation of water from hydrogen and oxygen gas is: 2H_{2 (g)} + O_{2 (g)} ---> 2H₂O_(l). What mass of water will form from 12.0 grams of hydrogen and excess oxygen (assuming the reaction goes to completion?)AnswerFirst what you must do in any of these problems is get all given masses into moles. You are given 12.0 grams of hydrogen, let's see how many moles that is:
12.0 g H₂ x 1 mol / 2.02 g = 5.94 mol H₂
So how much water is formed? According to the equation, for every 2 moles of hydrogen, 2 moles of water are produced, or in other words, a 1 to 1 ratio. So 5.94 moles of water will be formed. The question asks for what mass, so we're not quite done yet.
5.94 mol H₂O x 18.0 g / mol = 107. grams of H₂O.

The "oxygen in excess" is important. We're assuming that there's tons of oxygen, enough to supply whatever the hydrogen needs. If there was only .0000000000000000000001 moles of oxygen, 5.94 moles of water obviously can't form.
Let's try one more of this type before moving on to LRs.

Example:

The newly discovered element Takalahium (symbol Tak; molecular mass = 411 g/mol) combines with oxygen to form Takalahium Oxide. The unbalanced equation is:

Tak + O₂ ---> Tak₂O₃

How many grams of Tak Oxide are formed when burning 8.00 kilograms of Tak?
Answer
First and foremost, the balanced equation is needed. That would be:

4Tak + 3O₂ ---> 2Tak₂O₃

Then convert all given masses to moles:
8.00 kg Tak = 8000 g x 1 mol / 411 g = 19.5 moles.
Since there are 2 Tak's for every 1 Tak Oxide, there must be half as many moles of Tak Oxide, or 9.50 moles. You can also use DA to do the same mole ratios.
Before you can get grams, you must first find the molar mass of Tak Oxide, which is no problem:
Mass = 2 x 411 g + 3 x 16.0 g = 870 g/mol.
Then you find the mass:
9.50 moles x 870 g / 1 mol = 8260 grams = 8.26 kilograms.

In these last two problems, we always assumed that one of the reactants was in excess. This simplified things, but it's often not the case in many reactions. You have given amounts of each reactant. So how do you do this? Simple, just find out which out runs out first molewise. The one that does is called the limiting reactant, or just LR. One quick example:

Example:

Tak + Asdf ---> TakAsdf

You are given 45.0 grams of Takalahium (411 g/mol) and 500. grams of Asdfur (5620.5 g/mol). Find the limiting reactant in the formation of Takalahium Asdfite.
Answer
The limiting reactant isn't necessarily the one with the smallest mass. Remember that reactions look at number of particles, not at mass. So let's get both of these values into moles:
45.0 g Tak x 1 mol / 411 g = .109 moles.
500.g Asdf x 1 mol / 5620.5 g = .0890 moles.
As both are being consumed, you'll see that Asdf is used up first. Therefore it's the limiting reactant, even though there is over ten times as much of it by mass.

Note that mole ratios are important. Let's say 2 Tak's become 1 Asdf. Then it becomes a little bit more than just finding the bigger number, but not very harder. You start with one, say, Tak. Assume that it is the LR. Then, according to the mole ratio, you would need at least half as much Asdf for Tak to become the LR. Do you have a half? Well, one-half of .109 moles is about .05 something, and you do have that much. So if Asdf isn't your LR, then Tak must be. (Of course if it was exactly equal to half, then you could use either one, there would be no LR, and everything gets used up.)
What happens if you started with looking at Asdf? Well you need twice as much Asdf's. Twice .0890 is .17 something. You don't have that much Tak, therefore Tak is your LR, no matter how you look at it.
Once you find the LR, you can go ahead and finish the problem just like the others. So what's the difference? Once you find the LR, you use that amount and throw away the other amount. So like in the above example, you use the .0890 value, and forget about the .109. Of course, if they ask you to refer to the original amount, such as how much reactant is left over, then you need to use it. But not for stoich.

Example:

In the previous example, how much Takalahium Asdfide is produced by the reaction? And how much of each element is left over?AnswerSince the Asdf was the limiting reactant, with .0890 moles, you use this value for all remaining stoich problems. .109 moles is not how much Tak reacts, it will be less than that. Since it's a one-to-one ratio, .0890 moles of Tak will react with .0890 moles of Asdf to form .0890 moles of TakAsdf. The question asks what mass was produced. That's easy:
Molar mass = 411 g/mol + 5620.5 g/mol = 6031.5 g/mol,
.0890 mol TakAsdf x 6031.5 g / mol = 537 grams of TakAsdf.
As to the second prompt, you know that .109 moles Tak was initially present, and .0890 moles or Tak was consumed in reaction. A simple subtraction and conversion, and there's your answer:
Moles left over = Moles initially there - Moles used up = .109 mol - .0890 mol = .020 mol.
.020 mol Tak x 411 g / mol = 8.22 g
As to the Asdf, all of it was used up, so there's 0 g left.

Percent Yield

One little thing. You can test the above stuff in a lab, like measuring out the reactants, and see if you get the calculated amount of product. You probably won't get the exact number, but rather a number that's below it. There could be many reasons for it, such as maybe the reaction doesn't go to completion, or maybe there's impurities or something. The theoretical yield is how much you should get, according to stoich. The actual yield is how much you actually get, determined in a lab. The percent yield is the ratio of these two. The equation:

Yields are masswise. That is, how much you get is measured in kilograms, grams, mass.
Note that the percent yield will always be under 100%. The theoretical yield is the max you can get. (You're not going to get more substance than calculations predict.)
Now it's time for the big example, that ties together balancing equations, stoich, LRs, and percent yield.

Example:

Here's the combustion of ethane (not balanced):

C₂H₆ + O₂ ---> CO₂ + H₂O

32.0 g of ethane was burned with 15.0 grams of oxygen gas, and 10.8 grams of carbon dioxide was formed. Calculate the percent yield of carbon dioxide.
Answer
Before doing anything, get the equation balanced!

2C₂H₆ + 7O₂ ---> 4CO₂ + 6H₂O

(If you didn't know how to do that, review the rules of balancing. I do have a 20 megabyte limit here!)
Now to convert all given masses into moles:
32.0 g ethane x 1 mol / 30.0 g = 1.07 mol
15.0 g oxygen x 1 mol / 32.0 g = .469 mol
Which one is the LR? Let's look at ethane first. If ethane is indeed the LR, then the moles oxygen reacted would be:
1.07 mol ethane x 7 mol oxygen / 2 mol ethane = 3.74 moles oxygen needed.
Since the oxygen will have run out first, oxygen is the limiting reactant. Disregard the 1.07 mol.
Now to find the theoretical yield. 4 moles of carbon dioxide will form from 7 moles of oxygen:
.469 mol oxygen x 4 mol CO₂ / 7 mol oxygen = .268 moles.
Now to get the mass:
.268 moles x 44.0 g / mol = 11.8 grams.
That's how much you should get. But experiment shows that you get only 10.8 grams. Finding theoretical yield:
10.8 grams / 11.8 grams x 100 % = 91.5%

Atoms, Molecules and Compounds

2011-10-29T07:47:00.000-07:00

Law of Multiple Proportions

The Law of Multiple Proportions simply states: When two elements form a series of compounds, the ratios of the masses of the second element that combine with 1 gram of the first element can always be reduced to small whole numbers.
For those of you who are lost, it just means that only multiples of an amount of one element can react with a constant amount of another in a two element compound. For example, here is a table comparing three compounds of nitrogen and oxygen.

Compound	Amount of Nitrogen that reacts with 1 gram of oxygen
Compound A	1.750 g
Compound B	.8750 g
Compound C	.4375 g

See the smallest number, .4375? It divides into the other numbers with no remainder. That is, B/C is 2, and A/C is 4. This just shows that molecules are made up of finite sized units, or atoms. You can't have 2.3 atoms combine with 1 atom. Same thing here.
There is no way to know from this data alone exactly the ratios of the molecules. You could take the base unit to be any one of the numbers. Let's do two examples.

Using .4375 and ratio 1N : 1O (Compound C):
Let's call .4375 the ratios of the molecules 1N : 1O. Since Compound B contains twice as much N, you could say that the ratio there is 2N : 1O. Since Compound A contains four times as much N as Compound C, you could say that the ratio is 4N : 1O.

Using .8750 and ratio 2N : 2O (Compound B):
Now you say that .8750 is your base number, and the ratio 2N : 2O. What will the ratio of Compound C be? You know that there is half as much Compound C as in Compound B, so just reduce the N by one-half; 1N : 2O. And since you know there is 4 times as much in Compound A, it's 4N : 2O. It's that easy.
As you can guess, there is an infinite number of situations that satisfy the table.

______________________________________________________________________________________________
Dalton's Atomic Theory

These are pretty straightforward and require no explanation. Although you don't use these on tests or anything, it's good stuff to know (not to mention our whole understanding of chemistry is based on this):

Each element is made up of tiny particles called atoms.

The atoms of a given element are identical; the atoms of different elements are different in some fundamental way or ways.

Chemical compounds are formed when atoms combine with each other. A given compound always has the same relative numbers and types of atoms.

Chemical reactions involve reorganization of the atoms--changes in the way they are bound together. The atoms themselves are not changed in a chemical reaction.

These ideas form the basis of what a lot of AP Chem is all about; chemical reactions. It's nothing more than breaking some bonds and making some bonds, to put the atoms in a different order to create new things.

______________________________________________________________________________________________

Intro to Atomic Structure

An atom consists of a nucleus and electrons orbiting around it. The nucleus is made up of positively charged particles called protons, and neutral particles called neutrons. The electrons are negatively charged particles.
Here is a simplified picture:

This atom has two protons (red), so it is a helium atom.
One thing that pictures like the above cannot express is the relative sizes of the nucleus and the atom. If the atom was really that big (diameter 108 pixels), then the nucleus would be 1/100th of a pixel big! Or on a bigger scale, if the nucleus was the size of a pea, then the atom would be as wide as a football field. Another thing is mass. Virtually all the mass is concentrated in the nucleus. The 'pea' would weigh 250 million tons!
There are two numbers to know about any atom. One is the atomic number, which is the number of protons (and electrons in a neutral atom), and the atomic mass number, which is how much the thing weighs. One proton's weight is close to one neutron's weight which is close to one. Electrons weigh virtually nothing when compared to the huge protons, so they don't affect the total mass. So the atomic mass number is simply the number of protons plus the number of electrons.
The notation of writing atoms is like this:

The top number represents the mass number.
The bottom number represents the atomic number.
Some things to remember while solving these types of problems:
(Note: Problems of this simplicity will not be on the AP test, I guarantee that. We are not so lucky. But you will be tested in class on this stuff.)

The bottom number is always the number of protons.
The bottom number is also usually the number of electrons, but only in a neutral element. Ions come later.
The number of neutrons is always the top number minus the bottom number. All you're doing is taking the total mass, taking away the portion that is due to the protons, and what you got left is the neutrons. Likewise, if you know number of neutrons and protons, to find this number all you got to do is add.

Example:

Sample Problem

Here is a sample atom. How many electrons, protons, and neutrons does it got?
Solution
The bottom number is always the number of protons you have. So there are 9 protons. It's a neutral atom, by the fact that there are no plus or minus signs anywhere, so you can assume there are 9 electrons as well. There are 19 total nucleons (protons and neutrons) in the nucleus, and 9 are protons. So there are 10 neutrons. Easy.

Molecules and Ions

Compounds are made up of two or more atoms held together in some way. Although I'll go way more in depth later, atoms can be bonded together by both sharing electrons or one taking electrons away from the other.
Those compounds that share form a covalent bond, and the resulting formation is a molecule. An example is water, H--O--H. Each H shares its electron with the O, so that the O has two shared electrons, completing its valence shell. Another is methane, CH₄, which has each H sharing an electron with C, and C sharing a total 4 electrons with the H's.

A molecule acts as a unit by itself. It moves together. To represent a molecule, you can use either a ball-and-stick model or a space filling model. Examples of H₂O are shown below.


Ball-and-stick model	Space filling model

As a general rule, a covalent bond results when two non-metals are put together.
Let's consider the case of NaCl. Chlorine has a higher affinity for electrons than sodium, and so it takes one away, shown by the following:

Cl + e^- -> Cl^-

This shows that the electron joins up with the chlorine, and becomes a negatively charged ion--also known as a anion. But where did this electron come from? From the sodium, of course:

Na -> Na⁺ + e^-

This shows that the normal sodium atom gives off an electron--and becomes a positively charged ion in the process--also known as a cation.
Since these charges have opposite signs, they attract. The positive Na⁺ and the negative Cl^- form a ionic solid, also a salt. (Keep in mind that a salt refers to any ionic solid, not just our example of table salt. Ions don't have to be just one atom; they can be polyatomic, or consisting of many atoms. Some examples are the anion C₂H₃O₂^- and the cation NH₄⁺. A general rule, not to be followed strictly; if a compound has a metal and non metal in it, it's ionic. One of many exceptions is NH₄Cl; I guess the best rule to follow is to look for those ions!
One of the most annoying things in chemistry is memorizing ions. Once every week, my class would have to memorize ten ions and have an Ion Quiz. Although you might forget them, it is a good idea to review them while preparing for the test. I have included an ion table in the Tables section.

Compound Naming

I hated this part in Chem I. But it's not too bad. On a test, a question might be to mix dinitrogen pentoxide with something else. No problem, but what's that made of? This section will show you how to take a chemical structure and give a name to it, and vice versa.
Binary Ionic Compounds
These are just compounds made up of two different ions, like sodium chloride. All you do is: Put the name of the cation, followed by the name of the anion. That's it.

Example:

Compound	Name	Ions
KI	Potassium Iodide	K⁺, I^-
CaS	Calcium Sulfide	Ca²⁺, S^2-
MgO	Magnesium Oxide	Mg²⁺, O^2-
Li₃N	Lithium Nitride	Li⁺, N^3-

For single one-atom ion compounds such as above, you keep the name of the cation, and change the ending of the anion to -ide. Other examples are hydrogen => hydride, carbon => carbide, Fluorine => Fluoride, you get the idea.
One more thing to consider is the charge on the metal atom. If it can have only one possible value (such as +1 for sodium and +2 for magnesium above), then the normal name is fine. If the metal can form cations with different charges, like iron and tin, then you must put the Roman Numeral of it in parenthesis right after the name of the cation. Here are some examples:

Metal Ion	First half of compound name
Fe²⁺	Iron (II)
Fe³⁺	Iron (III)
Pb²⁺	Lead (II)
Pb⁴⁺	Lead (IV)

How do you know if to use roman numerals or not? Generally, if the metal is a transition metal (belonging to the three row skinny middle part of the periodic table up to the metalloids), then you need to use it. Some exceptions are aluminum (+3), zinc (+2), silver (+2), and cadmium (+2). Even though these are transition metals, they have only one possible ion. These metals and the metals in the first two columns need not use it, in fact, don't use them.
Another thing; mercury is really weird here. The two possible ions with mercury are Hg²⁺2²⁺. The difference when writing this is that Mercury (II) for the first, and Mercury (I). Because there is a total of +2, and there are two.
Nothing to it. Before I get to the sample problems, there is one more important point to all this: Compounds don't have charge. So when you add up all the charge from the ions, they must equal zero. And if they don't, you must balance it. Note that in the first table, the first three ions, when charge is added up, give zero. The last one is different. It still must add up to zero, but note that N is -3, but Li is only +1. It's common sense that you need three Li atoms to cancel out the -3 in N, hence the formula.

Example:

Give the name of these compounds:

a.) CuCl b.) NaF c.) MnO₂ d.) FeBr₂ e.) Fe₂O₃

Answers

By the rules, this is Copper Chloride. But copper can have different charges, so you must figure what the charge on this one is. Let's see: you know chloride can have only one value, and that's a -1. So if they must add up to zero, Cu must be +1. The answer is copper (I) chloride.
This is sodium and fluorine, so the name is sodium fluoride. Sodium can only be +1, so there is no need for roman numerals.
The name is Manganese Oxide, but a roman numeral is necessary, since it's a transition metal. In this case, manganese is a transition metal, so it needs a roman numeral. Let's start with oxygen: Each oxide ion has a charge of -2, and there are 2 of them. So the total charge due to oxide is -4. The single manganese ion must compensate for this, so its charge must be +4. The name is manganese (IV) oxide.
Same here. It needs a roman numeral. Br is -1, and 2 of them, so Fe must be +2. Name is Iron (II) Bromide.
e.This one's a little bit different, but the same concepts apply. You start with the oxygens. 3 of -2 ions make it a total of -6 charge due to oxide. So the two irons must compensate. If two of them must equal +6, each must be...+3!!! It should be no surprise. So the name of this is Iron (III) Oxide.

Chemistry Basics (Part 2)

2011-10-29T07:39:00.000-07:00

Significant Figures

In real life, you can't get a perfect measurement. Expensive and fancy instruments in labs can measure things pretty accurately, but in high school you are stuck using rulers and graduated cylinders. It's important to realize that fact, and you must record your measurements accordingly. Here is an example:

You can see that you can't be sure whether it's 6.1489898 or 6.1501239223. The rule for this is; get as many numbers you can be sure of, and then make an educated guess about the last digit. So in this example, you can see it's definitely between 6 and 7, so you know for sure the first number is 6. You can also see it's between the .1 and .2 marks, so you know it's 6.1 for sure. But the next, and last digit is uncertain. It looks like the end lies just around halfway between .1 and .2, so a safe guess is .05. Your final measurement should be around 6.15, of which all numbers are significant figures.
(Note: If you need more examples, tell me. I don't know if some people need this reinforced or something, that's all I plan to include on this.)
You'll need to know how many sig figs are in a measurement, to make proper calculations. Here are some guidelines.

Guidelines for Sig FigsAny number that isn't a zero is always a sig fig.Examples: 6.15 has three sig figs, 4.324 has 4 sig figs, 34231.123 has 8 sig figs, and so on.Zeros that don't have any non-zeros before them are not sig figs. They only act as placeholders. Examples: .00342 has only 3 sig figs, .02 has one sig fig, 02.3 has two sig figs, and so on.
Zeros that are between two non-zeros (trapped) are always sig figs.
Examples: 100.02 has five sig figs, .0203 has only three (the first zero isn't, according to Rule 2), 0101.00203 has eight sig figs, and so on.
Zeros that are at the end of the number are sig figs only if the number contains a decimal point.
Examples: 123.00 has 5 sig figs, .003030 has 4 sig figs, 100. has three sig figs, (Note that 100 by itself only has one sig fig, since it has no decimal) 2.30 x 10^-16 has 3 sig figs, and so on.
Numbers in equations and conversions are considered to be exact, that is, have an infinite number of sig figs.
Examples: In A = pr², the 2 is exact. In 1 inch = 2.54 centimeters, 2.54 is exact.

When you do your calculations, you might naturally be tempted to write as many numbers down as your calculator feeds, thinking that it is more accurate. WRONG! Your accuracy of your answer can only be as precise as the numbers the problems give you! For example, if a problem says something like this:

3.0 liters of water is added to 3.14159265358979323 liters of water. How many liters of water are there?

The answer is not 6.14159265358979323. 3.0 liters means that the actual value could be anywhere from 2.9 to 3.1 liters (remember the last number is just a guess?). So with that much uncertainty, going all the way down to 3 x 10-12 liters doesn't have as much an effect as the +/-0.5 that you have. The correct answer is 6.1 liters. How? Here are a few simple rules governing the realm of sig figs:

To multiply or divide: You find the number with the least sig figs, and use that many sig figs in the answer.
To add or subtract: The last digit in the final answer must have the same place as the place of the least precise number. That is, the leftmost ending value.

That's all there is to it. And just in case you forgot how to round (something you should've learned in like fifth grade...)

Remember to finish all your addings and multiplyings, then round.
If the number to the right of where you're rounding to is 4 or less, leave it alone and kill the numbers to right of it.
If the number to the right of where you're rounding to is 5 or greater, add one to the number and kill the numbers to right of it.

Prefixes
Instead of the annoying British system, with its gallons and quarts and pints and who knows what, the SI system has an easier way of remembering how many small units go into one big unit. It's the prefix system, and it's based on powers of ten. Makes it very simple. Here is a table on some of the most commonly used prefixes:

Name of Prefix(symbol):	Multiplier of Base
Mega (M)	10⁶
Kilo (k)	10³
Hecto (h)	10²
Deka (da)	10¹
---	1
Deci (d)	10^-1
Centi (c)	10^-2
Milli (m)	10^-3
Micro (m)	10^-6
Nano (n)	10^-9

So, some quick examples are: one kilometer is a thousand meters; one dekagram is a hundred decigrams, and you can go on from there. These are only the most used prefixes; other ones aren't used much in chem. Anyway, a more extensive list of prefixes will be included later in the reference area.

Dimensional Analysis

Let's say that you have one unit, and need it in another unit. In other words, you need to convert. A quick and easy way to do this is dimensional analysis. Basically, it uses the concept of multiplying by one (where "one" is one unit divided by the equivalence in another unit). Here's how to convert 100. meters into inches.

Note a few things. First, to convert you multiply your starting point with a series of fractions to get the final answer. Each fraction is a unit over another, which are equal. In the example, 100 cm is equal to 1 m, and 1 inch is equal to 2.54 cm. Thus multiplying by one. The trick to figuring out what to put in each fraction is looking at the units in the top of the previous fraction, and putting that unit in the bottom of the next one. You look at meters in the top of the first fraction (it's 100. meters / 1), and put that in the bottom of the next one. To figure out what unit to put in top, you look at the final answer, and see if you can get a direct relation to it. Most people don't know directly how many inches are in a meter, but it is common knowledge that there are 2.54 cm in an inch. So perhaps converting into centimeters is a start...
Once you get the units right, plug in the numbers that make the top equal the bottom. You should know that 100 cm's is an m. Now you go to the next fraction. cm's go in the bottom, and you can go straight to inches, knowing the relation. You ultimately wants the units converting to in the top of the last fraction. That way, everything cancels out, and you're done.
Plugging the values in, and multiplying the fractions give you the answer.
Another couple of examples:
Converting 3.450 kg's into pounds.

Converting 434.2 atm's into torr:

______________________________________________________________________________________________

Usage of Periodic Table

You will be using the periodic table until your head explodes, so you'd better learn to use it quick and efficiently from the beginning onward. I made a cool periodic table image map that lets you click on an element, and gives info on it. Here is a section of it, and what the number means.

Atomic Number: This is how many protons are in the nucleus of one atom of this element. It also shows the number of electrons in a stable atom of this element.
Symbol : This is the shorthand way of writing the element, and you will see it like this in chemical equations.
Atomic Mass : This is the number you will be using the most during the AP test. It is how much one atom weighs, in amus. 1 amu is 1 gram divided by Avogadro's number (6.022 x 10^-23). So in other words, one mole of beryllium has a mass of 9.012 grams. Anytime you need a molecular mass, you will be adding multiples of atomic masses. Memorize some of the most common ones; it will save you some looking-up time.

______________________________________________________________________________________________
Chemistry Basics (Part 1)

Chemistry Basics

2011-10-29T06:49:00.001-07:00

Matter:

Anything that takes up space is considered to be matter. Matter is most commonly found in three states. Obviously, you can tell the differences between solids, liquids and gases. One can easily observe that solids have a fixed volume and shape, liquids have a fixed volume but not a fixed shape, and gases don't have fixed volumes or shapes. This is all common sense.
Now, matter exists as a homogeneous mixture or a heterogeneous mixture. A homogenous mixture is a mixture that cannot be identified as two different things with the naked eye. Heterogeneous mixtures can be easily told apart (for instance, a potato chip in a bag of pretzels).
Among mixtures, there are pure substances and not-so-pure substances (the latter isn't a real term, I just made it up). Pure substance is made of one thing. An example would be taking water, running it through a filter, and getting purified water. This is a pure substance. However, if you go down to your local creek, the water in the creek also contains dirt, various little insects, and occasionally someone's feces. The creek water is not-so-pure substance.
Pure substances contain compounds, which are substances with consant composition and can be broken down to their basic elements (H₂O into H₂ and O₂).
_________________________________________________________________________________

Protons, neutrons, and electrons

Elements make up compounds, and are considered the basic building blocks of matter. You cannot break down elements into smaller parts, but you can classify the different parts within the element. Within the element are protons, neutrons, and electrons. The neutron, a chargeless particle, can be found in the nucleus along with the proton, which is a particle only slightly smaller than the neutron but positively charged. Electrons are negatively charged and are found circling the nucleus (much like the sun and the planets in our solar system).
Within the atom, one can find a nucleus. The nucleus, which does not move around like electrons, contains both neutrons and protons. Both neutrons and protons have mass, and these two contribute almost 100% of the atomic mass of an element. Electrons, on the other hand, have almost no mass. Most calculations assume a mass of zero for electrons. Electrons, being so small, can move around very quickly around the nucleus. Protons and neutrons can also be broken down into quarks, but you won't learn about those petit particles in this chemistry year.
A common demonstration to show how little space the electrons and nucleus take up in the atom is with the football field analogy. If a football stadium was considered to be an atom, a feather on the 50 yard line would be the nucleus. That is how much empty space there is within an atom!
______________________________________________________________________________________________

Elements
As mentioned before, an element's mass is determined by its protons and neutrons. The electrons, having almost no mass, do no contribute to the mass of an element. An element is defined by the number of protons and electrons it has. The number of nuetrons an element has, however, can vary from atom to atom. Each possible atom is known as an isotope. (ex. Carbon-14) If you go look at a periodic table, you will notice that each element has a given mass. This mass is an average of its isotopes. This does NOT mean that all carbons will have 6 protons and 6 neutrons (don't worry about these calculations, they'll make sense in the next chapter). Just remember that an element can have different number of neutrons, which will change its weight.
______________________________________________________________________________________________

Chemistry Basics (Part 2)