Chemistry Partner

Chemical Equation Balancing Tool

2019-06-06T23:49:00.002-07:00

Use this highly interactive tool to practice or teach balancing chemical equations.

The compound is entered in the long box and its numerical coefficient in the short box.

The Reactants column is for the compounds on the left side of the equation; the Products column is for the compounds on the right side of the equation.

Click, or tap, the Get Equation button to get a chemical equation to be balanced.

Note on Balancing Chemical Equations ∨

Practice Balancing Your Chemical Equations Here

Reactants

Products

Empirical Formula

2019-01-22T13:49:00.011-08:00

Learning Objectives

When you finish reading this text, you will be able to:

define what empirical formula is

determine a compound's empirical formula

determine a compound's chemical formula using its empirical formula and its formula mass.

Definition

The empirical formula of a compound indicates the smallest integer ratio of the compound's constituent atoms.

It simply tells us the mole-to-mole ratio of the atoms present in a compound.

Calculation of Empirical Formula

To be able to calculate the empirical formula of a compound, you need to:

Step 1	Determine the individual masses of the compound's constituent atoms.

Step 2	Convert these masses into moles.

Step 3	Divide each of these amounts in moles by the one with the smallest value.

Step 4	If the results are not whole numbers, then multiply these values by an integer that will yield the smallest whole numbers.

Example 1 and Example 2 show the step-by-step calculation of empirical formula from the known masses of atoms, and from the percent composition, of a compound, respectively.

Example 3 shows how to calculate the empirical formula of a hydrate.

Example 1 Determining the Empirical Formula of an Unknown Compound From the Masses of Atoms

Example 1

A sample of a certain compound contains the following: 0.747 g of Na, 0.195 g of C, and 0.780 g of O.

Using the step by step guide outlined above, we will determine the empirical formula of the given compound.

Step 1 Masses of the compound's constituent atoms are given in the problem.

\begin{align}Na\ \ \ 0.747\ g\\C\ \ \ 0.195\ g\\O\ \ \ 0.780\ g\end{align}

Step 2 Convert these masses into moles by dividing each mass by the molar mass of its corresponding atom.

`mo\l\e Na:`

`(0.747 g)/(22.98977 g//mo\l\e) = 0.0325 mo\l\e`

`mo\l\e C:`

`(0.195 g)/(12.011 g//mo\l\e)= 0.0162 mo\l\e`

`mo\l\e O:`

`(0.780 g)/(15.9994 g//mo\l\e)= 0.0488 mo\l\e`

Step 3 Divide these amounts in mole by the one with the smallest value.

`Na:`

`(0.0325 mo\l\e)/(0.0162 mo\l\e) = 2.0`

`C:`

`(0.0162 mo\l\e)/(0.0162 mo\l\e) = 1.0`

`O:`

`(0.0488 mo\l\e)/(0.0162 mo\l\e) = 3.0`

Step 4 Since the results are whole numbers, our calculation is done.

The empirical formula of the compound is

Na₂CO₃.

Example 2 Determining the Empirical Formula of an Unknown Compound From Percent Composition

When the individual masses of the compound's atoms are not given explicitly, they can still be obtained using percent composition.

Example 2

An organic halide has a percent composition of 12.82% C, 56.76% Cl and 30.42% F.

Let us find its empirical formula.

Step 1 We can determine the masses of the constituent atoms by multiplying their weight percentages by any amount we desire.

For convenience, we will use 100-g sample. In this way, the weights in grams are numerically equal to the weight percentages:

\begin{align}0.1282\ x\ 100\ g\ =\ 12.82\ g\ \ C\ \\0.5676\ x\ 100\ g\ =\ 56.76\ g\ \ Cl\\0.3042\ x\ 100\ g\ =\ 30.42\ g\ \ F\ \end{align}

Step 2 Convert these masses into moles by dividing each mass by the molar mass of its corresponding atom.

`mo\l\e C:`

`(12.82 g)/(12.011 g//mo\l\e) = 1.0674 mo\l\e`

`mo\l\e Cl:`

`(56.76 g)/(35.453 g//mo\l\e) = 1.6010 mo\l\e`

`mo\l\e F:`

`(30.42 g)/(18.9984 g//mo\l\e) = 1.6012 mo\l\e`

Step 3 Divide these amounts in mole by the one with the smallest value.

`C:`

`(1.0674 mo\l\e)/(1.0674 mo\l\e) = 1.0`

`Cl:`

`(1.6010 mo\l\e)/(1.0674 mo\l\e) = 1.5`

`F:`

`(1.6012 mo\l\e)/(1.0674 mo\l\e) = 1.5`

Step 4 Here, we obtain non-integer results:

C₁Cl_1.5F_1.5

So we multiply them with an integer that will give the smallest whole number. That number is 2.

C_{1 x 2}Cl_{1.5 x 2}F_{1.5 x 2}

Hence, the empirical formula is

C₂Cl₃F₃.

Example 3 Determining the Empirical Formula of a Hydrate

A hydrate is a compound with water molecules adsorbed on its surface. The problem is about finding the number of water molecules bound to the compound.

Example 3

A 5.00-g sample of the hydrated CoCl₂ was analyzed and found to contain 2.27 g of H₂O.

We are going to determine the compound's empirical formula.

Step 1 Instead of using the masses of the atoms, we use the masses of CoCl₂ and H₂O.

`H_2O\ \ \ \ \ \ 2.27\ g`

`CoCl_2\ \ \ 5.00\ g - 2.27\ g = 2.73\ g`

Step 2 To convert these masses into moles, we divide them by their respective formula masses instead.

`mo\l\e\ \ H_2O:`

`(2.27\ g)/(18.0152\ g//mo\l\e)=0.1260\ mo\l\e`

`mo\l\e\ \ CoCl_2:`

`(2.73\ g)/(129.8392\ g//mo\l\e)=0.0210\ mo\l\e`

Step 3 Divide these amounts in mole by the one with the smaller value.

`H_2O:`

`(0.1260\ mo\l\e )/(0.0210\ mo\l\e) = 6.0`

`CoCl_2:`

`(0.0210\ mo\l\e )/(0.0210\ mo\l\e) = 1.0`

Step 4 Since we obtain whole numbers, the empirical formula is

(CoCl₂)₁(H₂O)₆

which can be rewritten as

CoCl₂·6H₂O

The compound is cobalt(II) chloride hexahydrate.

Use of Empirical Formula

Empirical formula is used in determining the chemical formula of a compound.

A compound's chemical formula is a multiple of its empirical formula.

Depending on the compound, the chemical formula may consist of one or more empirical formula units. (See table below.)

Table of Some Compounds With Their Chemical and Empirical Formulas

Show Table ∨

Compound	Formula	Empirical Formula
benzene	C₆H₆	CH
acetylene	C₂H₂	CH
mercury(I) nitrate	Hg₂(NO₃)₂	HgNO₃
ammonium oxalate	(NH₄)₂C₂O₄	NH₄CO₂

For a compound whose chemical formula consists of more than one empirical formula unit, deriving its correct chemical formula requires a knowledge of its empirical formula and its formula mass.

The formula mass is divided by the empirical formula mass to obtain the number of empirical formula units (efu):

`efu=(f\o\r\m\u\la\ mass)/(em\p\irical\ \f\o\r\m\u\la\ mass)`

where:

formula mass, usually determined by experiment, is the sum of the atomic masses of the compound's constituent atoms

empirical formula mass is the sum of the atomic masses of the atoms in the empirical formula

NOTE: Formula mass is a generic term for the molar mass of molecular and ionic compounds.

The subscripts of the empirical formula are multiplied by this calculated number of efu in order to obtain the correct chemical formula of the compound.

Example 4 Determining the Molecular Formula of an Unknown Molecular Compound

Show Example 4 ∨

Percent Composition Calculator

2018-10-03T23:21:00.003-07:00

In order to calculate the percent composition of a compound, you need to determine the weights of its constituent elements. For details and examples, see my previous post Percent Composition.

After determining the weight percentages of a compound's constituent elements, you can now calculate the individual weights of the elements contained in any given weight of that compound.

Percent Composition Calculator does just that. It provides you a quick check on your calculations whether you're solving or composing percent composition problems.

Just give the correct formula, weight and percent purity (if any) of the compound:

Formula. For the most commonly encountered binary inorganic compounds, a message is provided below the formula entry line if you enter their incorrect formula.

Weight. You can use any unit for the weight, meaning if your unit of weight is in grams then the computed weights of the compound's components will also be in grams.

% Purity. It is optional. If you don't enter a value, the application assumes that the compound is pure.

You need to round off the results to the required number of significant digits.

Percent Composition Calculator

Enter the following information about the compound:

Formula:

Weight:

% Purity:

(optional)

Percent Composition

2018-07-12T15:30:00.003-07:00

Percent composition of a compound refers to the mass percentages (also expressed as % w\w) of the constituent elements of the compound. It indicates the individual weights of the elements in any given weight of the compound.

For a pure compound, percent composition gives its elemental analysis; for an impure compound, percent composition gives the percent purity of the compound.

Self-Assessment Test

You may take this test before and/or after reading this text.

Calculate the percent by weight of ?

Calculation

If the individual weights of the constituent elements of a compound are known, then calculation of percent composition is straightforward.

Just divide the weight of each of the constituent elements by the weight of the compound and then multiply by 100:

`((weight of e\l\ement)/(weight of compound)) x 100 = % weight`

Example 1

A certain oxide of copper is found by experiment to contain 2.157 g of Cu and 0.543 g of O. Calculate its percent composition.

Step 1 To find the total weight of the compound, add the weights of the individual elements.

\begin{align}Cu\ \ \ 2.157\ g\\O\ \ \ 0.543\ g\\\_\_\_\_\_\_\_\_\_\\2.700\ g\end{align}

Step 2 In order to calculate the percent weights of the constituent elements, divide each weight of the elements by the weight of the compound and then multiply by 100.

`% Cu:`

`((2.157 g)/(2.700 g)) x 100 = 79.89%`

`% O:`

`((0.543 g)/(2.700 g)) x 100 = 20.11%`

Check that the total percentage is equal to 100.

If the weights are unknown, then percent composition can still be calculated from the known chemical formula of a compound.

It is made possible by considering the amounts of a compound and its components in terms of moles.

Example 2

Let's calculate the percent composition of ammonium sulfate, (NH₄)₂SO₄:

Step 1 Determine the number of moles of the constituent elements contained in 1 mole of the compound.

1 mole of (NH₄)₂SO₄ contains:

2 moles N
8 moles H
1 mole S
4 moles O

Step 2 Convert these number of moles to weight in grams by multiplying them by the molar weights of the elements.

`weight of N:`

`2 mo\l\es x (14.0067 g)/(mo\l\e) = 28.0134 g`

`weight of H:`

`8 mo\l\es x ( 1.0079 g)/(mo\l\e) = 8.0632 g`

`weight of S:`

`1 mo\l\e x (32.0666 g)/(mo\l\e) = 32.0666 g`

`weight of O:`

`4 mo\l\es x (15.9994 g)/(mo\l\e) = 63.9976 g`

Step 3 Add all the weights of the elements. This total is the weight of 1 mole of the compound. It is also known as the formula weight of the compound.

\begin{align}28.0134\ g\\8.0632\ g\\32.0666\ g\\63.9976\ g\\\_\_\_\_\_\_\_\_\_\\132.1408\ g\end{align}

Step 4 Divide each weight of the elements by the formula weight of the compound.

`% N:`

`((28.0134 g)/(132.1408 g)) x 100 = 21.20%`

`% H:`

`((8.0632 g)/(132.1408 g)) x 100 = 6.10%`

`% S:`

`((32.0666 g)/(132.1408 g)) x 100 = 24.27%`

`% O:`

`((63.9976 g)/(132.1408 g)) x 100 = 48.43%`

Their total percentage must be equal to 100.

Uses

Percent composition is used to calculate the weight of a particular element in any given weight of a compound.

Example 3

Let's calculate the individual weights of the constituent elements present in 5.0 g of (NH₄)₂SO₄.

Step 1 Find the percent composition of the compound.

As calculated in Example 2 , the percent composition of (NH₄)₂SO₄ is:

\begin{align}N\ \ \ \ 21.20\ \%\\H\ \ \ \ \ \ 6.10\ \%\\S\ \ \ \ \ 24.27\ \%\\O\ \ \ \ \ 48.43\ \%\end{align}

Step 2 To determine the individual weights of the elements, multiply the mass percentage of each element by the given weight of the compound.

`weight of N:`

`5.0 g x 0.2120 = 1.1 g`

`weight of H:`

`5.0 g x 0.0610 = 0.3 g`

`weight of S:`

`5.0 g x 0.2427 = 1.2 g`

`weight of O:`

`5.0 g x 0.4843 = 2.4 g`

Step 3 Check that the total of the weights of the individual elements is equal to the given weight of the compound.

\begin{align}N\ \ \ \ 1.1\ \ g\\H\ \ \ \ 0.3\ \ g\\S\ \ \ \ \ 1.2\ \ g\\O\ \ \ \ \ 2.4\ \ g\\\_\_\_\_\_\_\_\_\_\\5.0\ \ g\end{align}

Percent composition is also used in the determination of the empirical formula of a compound. This is another topic that will be covered in a subsequent post.

Formula Weight Calculation

2017-11-17T14:23:00.019-08:00

This text teaches you how to calculate the formula weight of a substance.

Before you begin, please take note that you should have a basic understanding of the following concepts:

ions and molecules
mole
hydrated compounds

Self-Assessment Test

You may take this test before and/or after reading this text.

Click START button to begin the test comprising of four questions.

Definition of Formula Weight

Formula weight is the sum of the atomic weights of all the atoms indicated in the chemical formula of a substance. It is the weight in grams of 1 mole of that substance, usually expressed as grams per mole. Hence, it is also sometimes called the molar weight.

The quantity of the atoms in a given formula is indicated by a numerical subscript after each atom. If no subscript is present after an atom, it is assumed to have a numerical subscript of one.

Formula Weight or Molecular Weight?

Formula weight is a generic term that may be used for both substances that exist as ions and molecules. Whereas, molecular weight is applicable only to substances that exist as molecules.

Uses of Formula Weight

Formula weight is used in the following:

stoichiometry problems
mole calculation
percent composition calculation
empirical formula calculation

Note: The following calculations are based on atomic weights with three significant digits.

Example Calculation 1

Formula weight of water, H2O :

1.008

15.999

2.016

15.999

18.015

Example Calculation 2

Formula weight of ammonium carbonate, (NH4)2CO3 :

2 x 1

2 x 4

14.007

1.008

12.011

15.999

28.014

8.064

12.011

47.997

96.086

Example Calculation 3

Formula weight of calcium chloride hexahydrate, CaCl2.6H2O :

6 x 2

6 x 1

40.078

35.453

1.008

15.999

40.078

70.906

12.096

95.994

219.074

Compounds With Planar Shape

2013-03-12T15:04:00.001-07:00

Featured in this post are compounds having planar shape other than trigonal planar.

The following compounds are all planar, with the last two having more than one central atom.

xenon tetrafluoride, XeF₄

benzene molecule, C₆H₆

CH₂=NH

Geometry of Xenon Tetrafluoride, XeF₄

The Lewis structure of xenon tetrafluoride, XeF4, below shows that the compound's central atom has four bonding electron pairs and two lone electron pairs.

According to the VSEPR theory, the arrangement that affords the maximum distance between each of the mentioned electron pairs is a square bipyramidal.

The axial positions of the two lone electron pairs offer the least net repulsion than it would have been if one or both of these non-bonding electron pairs were in the equatorial positions.

In the illustration shown above, a square bipyramidal outline is superimposed on the xenon tetrafluoride molecule.

shape of xenon tetrafluoride: square planar
F-Xe-F bond angle: 90°

Geometry of Benzene, C₆H₆

The resonance structures of benzene are shown below; the third structure is a suggested representation of the delocalization state of the pi electrons.

Each of the six carbon atoms can be taken as a central atom about which the geometry is trigonal planar.

The result is a hexagonal planar molecule with one carbon atom and one hydrogen atom jutting out at each of the six vertices.

Geometry of CH₂=NH

The CH2=NH molecule has two central atoms, carbon and nitrogen (see the Lewis structure below).

About each center, the geometry is trigonal planar.

However, the shape about the nitrogen atom is angular due to the presence of a lone electron pair.

All of the atoms lie in one plane.

Compounds With Bent Shape

2013-01-11T06:48:00.001-08:00

The following chemical species have bent geometry:

water molecule, H₂O

nitrogen dioxide molecule, NO₂

nitrite ion, NO₂^-

sulfur dioxide molecule, SO₂

The illustrations and discussion below show how the presence of non-bonding electron pair(s) on a molecule's central atom affects the geometry of the molecule and the bond angle of its atoms.

Geometry of Water, H₂O

As shown in its Lewis structure below, the central atom of water has four pairs of electrons: two bonding pairs and two non-bonding or lone pairs.

These pairs of electrons are presumed to take a tetrahedral arrangement in space.

Here, the water molecule is depicted as being superimposed in a tetrahedral outline.

The lone pairs of electrons are relatively closer to the nucleus of the central atom and they tend to crowd the two bonding pairs together so that the H-O-H bond angle is less than the ideal tetrahedral angle of 109.5°.

shape of water: bent or angular
H-O-H bond angle: 104.5°

Geometry of Nitrogen Dioxide, NO₂

The central atom of nitrogen dioxide, nitrogen, has two sigma bond electron pairs and an unpaired non-bonding electron as shown in its Lewis electron dot structure below.

In order for this group of electrons to be farthest from each other, it is predicted that they take a trigonal planar arrangement.

Shown in the next illustration is the nitrogen dioxide molecule superimposed on a trigonal planar outline.

The observed O-N-O bond angle of 134.1° in nitrogen dioxide molecule, being closer to trigonal angle of 120°, confirms this prediction.

The opening out of the O-N-O bond angle is due to the less crowding affected by the half filled non-bonding orbital of the central atom.

This half-filled orbital accounts for the nitrogen dioxide (NO₂) being paramagnetic.

shape of nitrogen dioxide: bent or angular
O-N-O bond angle: 134.1°

Geometry of Nitrite Ion, NO₂^-

An addition of an electron to the central atom of nitrogen dioxide (NO₂) creates the nitrite anion, NO₂^-.

In the above diagram, the resonance structures of nitrite ion is shown. The oxygen atoms in red partial shades are atoms with negative formal charge.

As is the case with the nitrogen dioxide, the arrangement of the two sigma bond electron pairs and the lone electron pair of nitrite ion's central atom lies on a trigonal plane.

The only difference is the observed O-N-O bond angle of 115° for the nitrite ion. This bond angle now being much closer but less than the trigonal angle of 120° is attributed to the filling up of the non-bonding orbital of the nitrite ion's central atom.

The lone electron pair is relatively closer to the central atom and occupies more space than the adjacent bonding electron pairs do.

shape of nitrite ion: bent or angular
O-N-O bond angle: 115°

Geometry of Sulfur Dioxide, SO₂

Referring to the contributing Lewis structures of sulfur dioxide below, there is a -1 formal charge on one of the oxygen atoms and +1 formal charge on the sulfur atom which cancel each other out, leaving the molecule with zero net charge.

Given the two sigma bond electron pairs and one lone electron pair on the central atom of sulfur dioxide, as shown in the figure above, SO₂ is predicted to have a trigonal planar geometry.

A trigonal planar outline is shown below superimposed on the sulfur dioxide molecule.

Given this geometry, it is expected that SO₂ molecule has an angular shape. This is supported by the fact that sulfur dioxide has a dipole moment.

The O-S-O bond angle is expected to be somewhat less than the trigonal angle of 120° due to the presence of a lone electron pair on the central atom.

shape of sulfur dioxide: bent or angular
O-S-O bond angle: less than 120°

Molecules With Linear Shape

2012-09-03T15:02:00.000-07:00

Presented in this post are the illustrations of the geometry of the following linear molecules:

BeH₂

CO₂

HC≡CH

HC≡N

BeF₂

BeCl₂

Geometry of BeH₂ ( beryllium dihydride )

shape of molecule: linear
bond angle: 180°

The Lewis structures of BeH₂ and CO₂

Geometry of CO₂ ( carbon dioxide )

shape of molecule: linear
bond angle: 180°

Geometry of C₂H₂ ( ethyne or acetylene )

shape of molecule: linear
bond angle: 180°

The Lewis structures of C₂H₂ and HCN

Geometry of HC≡N ( hydrogen cyanide )

shape of molecule: linear
bond angle: 180°

Geometry of BeF₂ ( beryllium difluoride )

shape of molecule: linear
bond angle: 180°

The Lewis structures of BeF₂ and BeCl₂

Geometry of BeCl₂ ( beryllium dichloride )

shape of molecule: linear
bond angle: 180°

Compounds With Octahedral Shape

2011-12-08T10:57:00.001-08:00

For compounds whose central atoms possess six bonding electron pairs, their bonded atoms are oriented towards the corners of an octahedron.

The best examples are the PF₆^- ( phosphorus hexafluoride or hexafluorophosphate) ion and the SF₆ ( sulfur hexafluoride ) molecule whose Lewis structures are shown below.

The ground state electronic configuration of the two central atoms:

P	1s² 2s² 2p⁶ 3s² 3p_x¹ 3p_y¹ 3p_z¹
S	1s² 2s² 2p⁶ 3s² 3p_x² 3p_y¹ 3p_z¹

In the excited state of the two central atoms above, their valence electrons are assumed to be distributed this way:

1s² 2s² 2p⁶ 3s¹ 3p_x¹ 3p_y¹ 3p_z¹ 3d_z²¹ 3d_{x² - y²}¹

Since six equal orbitals are required, it is assumed that sp³d² hybridization is used to form six hybrid orbitals:

1s² 2s² 2p⁶ (sp³d²)¹ (sp³d²)¹ (sp³d²)¹ (sp³d²)¹ (sp³d²)¹ (sp³d²)¹

Phosphorus acquires an extra electron for its sixth orbital.

Geometry of PF₆^-

shape of ion: octahedral
bond angle of axial atoms: 180°
bond angle of equatorial atoms: 90°
bond angle between an equatorial atom and an axial atom: 90°

Axial atoms are in white circles; equatorial atoms are in black circles.

An octahedron is superimposed on the PF₆^- ion.

Geometry of SF₆

shape of molecule: octahedral
bond angle of axial atoms: 180°
bond angle of equatorial atoms: 90°
bond angle between an equatorial atom and an axial atom: 90°

Molecules With Trigonal Bipyramidal Shape

2011-11-28T06:28:00.001-08:00

Molecules like PCl₅ (phosphorus pentachloride) and PBr₂Cl₃ (phosphorus dibromo-trichloride) possess five covalent bonds whose electron pairs are arranged in a trigonal bipyramidal geometry.

The central atom of these molecules, phosphorus, has only three unpaired electrons out of its five valence electrons as can be seen from its electronic configuration below:

1s² 2s² 2p⁶ 3s² 3p_x¹ 3p_y¹ 3p_z¹

ground state configuration

But the Lewis structures of the given molecules, as shown below, indicate that the phosphorus atom uses five unshared electrons to form five covalent bonds.

It is, therefore, assumed that phosphorus uses the empty d orbital to "promote" 1 electron from 3s orbital to d orbital,

1s² 2s² 2p⁶ 3s¹ 3p_x¹ 3p_y¹ 3p_z¹ 3d_z²¹

excited configuration

and, through hybridization, form five sp³d hybrid orbitals which are then used to bond with chlorine and bromine atoms.

1s² 2s² 2p⁶ (sp³d)¹ (sp³d)¹ (sp³d)¹ (sp³d)¹ (sp³d)¹

excited configuration

Geometry of PCl₅

shape of molecule: trigonal bipyramidal
bond angle of axial atoms: 180°
bond angle of equatorial atoms: 120°
bond angle between an equatorial atom and an axial atom: 90°

Axial atoms are in white circles; equatorial atoms are in black circles.

A trigonal bipyramid is superimposed on the PCl₅ molecule.

Geometry of PBr₂Cl₃

Due to the different axial and equatorial positions the two bromine atoms can assume, PBr₂Cl₃ (phosphorus dibromo-trichloride) has three geometric isomers as shown below.

Except for the first isomer in which the bromine atoms are in axial positions, the other two geometric isomers of PBr₂Cl₃ are polar.

Molecular Geometry: Molecules With Trigonal Planar Shapes

2011-11-23T12:00:00.001-08:00

Molecules with trigonal planar shape are characterized by their central atoms having 3 bonding electron pairs.

These electron pairs refer to those ones involved in sigma bonds only.

Electron pairs from pi bonds of double and triple bonds do not count.

In this particular coordination geometry, each electron pair is at an angle of 120° from any of the other two electron pairs.

The illustrations of the geometry and the Lewis structures given below are for the following molecules:

BF₃

BCl₃

SO₃

H₂C=O

Geometry of BF₃

shape of molecule: trigonal planar
F-B-F bond angle: 120°

A trigonal plane is superimposed on the BF₃ molecule.

Lewis structures of BF₃ and BCl₃

Due to its 3 valence electrons, boron does not follow the octet rule and thus forms electron-deficient compounds such as BF₃ ( boron trifluoride ) and BCl₃ ( boron trichloride ).

The characteristic reaction of the above compounds is to accept/share an electron pair from other atoms or compounds.

The reaction is called Lewis acid-base reaction.

A Lewis acid accepts a pair of electrons; a Lewis base donates a pair of electrons.

Geometry of BCl₃

Geometry of SO₃

Lewis structures of SO₃ and H₂C=O

Shown above is the Lewis structures of SO₃ ( sulfur trioxide ) and H₂C=O ( formaldehyde ) molecules.

Geometry of H₂C=O

Molecular Geometry: Compounds With Tetrahedral Shapes

2011-11-14T05:14:00.000-08:00

A compound with four electron pairs at its central atom has a tetrahedral shape.

If one of the four electron pairs is unshared or non-bonding, then the compound becomes pyramidal in shape.

Although triangular pyramid is also a tetrahedron, there is a distinction between compounds with tetrahedral and pyramidal shapes: the location of their central atoms.

A compound with triangular pyramidal or simply pyramidal shape has its central atom located at the apex of the pyramid whereas a compound with tetrahedral shape has its central atom located at the center of tetrahedron.

Here are the lists of compounds included in this post:

compounds with tetrahedral shapes

CH₄

F₃SN

CCl₄

ClO₄^-

NH₄⁺

BF₄^-

SO₄⁼

compounds with pyramidal shapes

NH₃

SO₃⁼

ClO₃^-

NF₃

XeO₃

F₂SeO

Along with the illustration of a compound's shape, its Lewis structure is also included together with a brief discussion regarding the effect of lone electron pairs or double/triple bonds on the bond angles and/or the use of hybrid orbitals by certain atoms for covalent bonding.

The first three compounds discussed in details below serve to represent the rest of the compounds.

Methane, CH₄

shape of molecule: tetrahedral
H-C-H bond angle: 109.5°

The Lewis structure and one of the four equal H-C-H bond angles of CH₄.

A tetrahedron superimposed on a CH₄ molecule. The four hydrogen atoms are oriented towards the corners of the tetrahedron.

Ammonia, NH₃

The ammonia molecule has a non-bonding electron pair at the top of the tetrahedron as depicted below.

The Lewis structure of NH₃ shows it has a lone pair of electrons which occupies more space than any of the bonding electron pairs resulting to H-N-H bond angles being less than the ideal tetrahedral angle of 109.5°.

Since molecular shapes involve atoms only, the shape of ammonia will be minus the lone pair of electrons. It is a triangular pyramid with the nitrogen atom at the apex.

shape of molecule: triangular pyramidal
H-N-H bond angle: 106.6°

A triangular pyramid superimposed on the ammonia molecule.

Trifluorothionitrile, F₃S≡N

shape of molecule: tetrahedral
F-S-F bond angle: less than 109.5°
N-S-F bond angle: greater than 109.5°

Referring to the Lewis structure above, sulfur atom exceeds the octet configuration or the eight valence electrons required for bonding by utilizing empty d orbitals and creating hybrid orbitals (for a discussion of constructing Lewis structure of compounds, see Drawing Lewis Electron Dot Structure or Formula).

F₃SN has a tetrahedral shape which means it has four electron pairs directed towards the four corners of a tetrahedron but its Lewis structure shows 6 bonding electron pairs (see my previous post for a discussion of compounds' different coordination geometries and the corresponding number of electron pairs of their central atoms).

In determining the shapes of compounds, only electron pairs involved in sigma bonds are considered.

As for the bond angles, greater repulsion exerted by electron pairs from double and triple bonds cause the bond angles containing them to be greater than those bond angles containing single bonds only.

Compounds With Tetrahedral Shape

The Geometry of CCl₄

The Lewis structures of CCl₄ and ClO₄^-.

The chlorine atom in ClO₄^- or perchlorate ion acquires one electron to complete its eight valence electrons and utilizes empty d orbitals to bond with four oxygen atoms by sharing all of its eight valence electrons.

Referring to the illustration above, atoms in partial grey and red shades are atoms with formal charges. To learn more about the calculation of formal charges, read Drawing Lewis Electron Dot Structure or Formula.

The Geometry of ClO₄^-

The Geometry of NH₄⁺

The Lewis structures of NH₄⁺ and BF₄^-.

The NH₄⁺ or ammonium ion is formed by NH₃ acting as a Bronsted base and accepting a proton.

On the other hand, BF₄^- or boron tetrafluoride ion is formed by BF₃ acting as a Lewis acid and bonding with a fluoride ion acting as a Lewis base.

For a discussion of Bronsted acid-base reactions, see Solving Weak Acid/Base Dissociation Problems.

The Geometry of BF₄^-

The Geometry of SO₄⁼

The Lewis structures of SO₄⁼

Given the geometry of SO₄⁼ or sulfate ion as shown above, there are two possible Lewis structures for it. The second Lewis structure is considered to be the correct one because it is more stable and consistent with the experimental data.

The sulfur atom in the second Lewis structure exceeds the octet configuration or the 8 valence electrons required for bonding by acquiring two extra electrons and utilizing empty d orbitals.

Compounds With Pyramidal Shape

The Geometry of SO₃⁼

The Lewis structures of SO₃⁼ and ClO₃^-

The sulfur and chlorine atoms in the above Lewis structures acquire extra electrons to fill up its valence electron shells and are assumed to use hybrid orbitals involving empty d orbitals to form covalent bonds with oxygen atoms.

The Geometry of ClO₃^-

The Geometry of NF₃

The Lewis structures of NF₃ and XeO₃

The Geometry of XeO₃

The Geometry of F₂SeO

The Lewis structure of F₂SeO

Geometries of Molecules and Ions

2011-11-02T20:01:00.000-07:00

The shapes of compounds, either molecules or polyatomic ions, are very important in helping us understand better their reactions.

Fortunately, the geometries of most molecules and ions can be predicted quite reliably even by considering only their electron-electron pair interactions.

The idea is that the repulsive forces that exist between bonding and non-bonding pairs of electrons of a molecule or an ion cause those pairs of electrons to adapt certain spatial arrangement that allows minimum repulsion.

The spatial arrangement of the electron pairs of a molecule depends on its number of atoms and the number of valence electrons of its central atom.

So, in order to predict the geometry of a molecule or an ion, one needs to know its number of bonding and non-bonding pairs of electrons by determining the following:

total number of atoms in the molecule or ion
number of valence electrons of the molecule's central atom
the Lewis structure of the molecule or ion

In the following illustrations, all of the possible coordination geometries for different compounds are depicted using ball-and-stick models.

In each illustration, information such as number of electron pairs of the compound and the number of its atoms are given.

Linear Geometry

number of electron pairs: 2
coordination geometry: linear
number of atoms: 3 ( 1 central atom, 2 bonding atoms)

Trigonal Planar Geometry

number of electron pairs: 3
coordination geometry: trigonal planar
number of atoms: 4 ( 1 central atom, 3 bonding atoms)

Tetrahedral Geometry

number of electron pairs: 4
coordination geometry: tetrahedral
number of atoms: 4-5 ( 1 central atom, 3-4 bonding atoms)

Trigonal Bipyramidal Geometry

number of electron pairs: 5
coordination geometry: trigonal bipyramidal
number of atoms: 6 ( 1 central atom, 5 bonding atoms)

Octahedral Geometry

number of electron pairs: 6
coordination geometry: octahedral
number of atoms: 7 ( 1 central atom, 6 bonding atoms)

Pentagonal Bipyramidal Geometry

number of electron pairs: 7
coordination geometry: pentagonal bipyramidal
number of atoms: 8 ( 1 central atom, 7 bonding atoms)

Hydrogen Bonding

2011-01-12T12:01:00.000-08:00

When a hydrogen atom forms a covalent bond with a highly electronegative atom (such as oxygen or nitrogen) in a molecule, a polar bond is formed between them.

The highly electronegative atom draws the shared electron pair to itself most of the time and thus attains a partially negative charge.

Consequently, the hydrogen atom attains a partially positive charge, being virtually electron deficient most of the time.

As shown from the figure above, the electron cloud is asymetrically located between the two bonding atoms. The arrow simply indicates the net dipole moment of the molecule and the direction of increasing electronegativity.

This frequent uneven sharing of electrons between the two atoms imparts a polarity to the molecule containing these bonded atoms.

The resulting polarity creates a dipole-dipole attraction between molecules called hydrogen bonding.

Hydrogen bonding accounts for the following phenomena:

associated compounds have higher boiling point than non-associated compounds with nearly the same molecular weights;

greater solubility of associated compounds in water;

variations in the boiling points and solubility in water of the isomers of certain phenolic compounds.

Hydrogen bonding can occur between:

the same molecules;

different molecules (intermolecular hydrogen bonding);

within the molecule (intramolecular hydrogen bonding).

Intermolecular Hydrogen Bonding

In the following figures below, the conceptualized intermolecular hydrogen bonding of several organic molecules with themselves and with water molecules is shown.

The hydrogen bonding of the ethyl alcohol molecules above explains its higher boiling point than those of propane's and dimethyl ether's.

The wide difference in the boiling points of the primary, secondary and tertiary amines can be attibuted to hydrogen bonding. Primary amines, like methyl amine molecules shown in hydrogen bonding above, and secondary amines form hydrogen bonds while tertiary amines can not do so due to lack of hydrogen atom bonded to nitrogen atom.

Phenol has higher boiling point than toluene because of hydrogen bonding of phenol molecules as illustrated above.

The association of ethyl alcohol molecules (above) and phenol molecules (below) with water molecules through hydrogen bonding explains their relatively greater solubility in water as compared to those compounds with nearly the same molecular weights.

Intramolecular Hydrogen Bonding

The isomers of some phenolic compounds exhibit great difference in their boiling points and water solubility because of intramolecular hydrogen bonding (that is, association within the molecules) that occurs in the ortho isomers. This intramolecular hydrogen bonding reduces the chances of intermolecular hydrogen bonding of the said isomers resulting to their relatively lower boiling points and lower water solubility.

Lewis Structures of Some Common Compounds and Ions

2010-11-14T11:11:00.002-08:00

It's been quite a while since my last post....

...and there's so much catching up to do!

And so without further ado, Chemistry Partner resumes to make your life in chemistry a little bit easier.

This current update serves as a supplement to my October 23, 2009 post, Drawing Lewis Electron Dot Structure or Formula.

It consists of illustrations of Lewis structures showing the covalent bonding of the following:

Inorganic Compounds

NH₃ (ammonia)

HNO₃ (nitric acid)

SO₃ (sulfur trioxide)

H₂S (hydrogen sulfide)

N₂ (nitrogen gas)

O₃ (ozone)

H₂SO₄ (sulfuric acid)

H₂O (water)

Organic Compounds

HNCO (isocyanic acid)

CO (carbon monoxide)

HCN (hydrogen cyanide)

CH₃COOH (acetic acid)

H₂CO₃ (hydrogen carbonate or carbonic acid)

Polyatomic Ions

SO₄^2- (sulfate ion)

NH₄⁺ (ammonium ion)

PO₄^3- (phosphate ion)

C₂O₄^2- (oxalate ion)

NO₃^- (nitrate ion)

CO₄^2- (carbonate ion)

In order to identify each atom's respective valence electrons easily, different colors for atoms were used.

Also, red and grey shades were used to highlight the atoms with negative and positive formal charges.

Lewis Structures of Inorganic Compounds

Where necessary, I used different colors for the same kind of atom. Here in the example of O₃ (ozone) below, I used three different colors to distinguish the three different oxygen atoms.

For the structures of HNO₃ (nitric acid) and SO₃ (sulfur trioxide) above, the use of one color for the same kind of atom is enough to distinguish easily the individual atom's valence electrons.

For the rest of the illustrations below, you'll find the same methods of color assignment for the atoms.

When more than one Lewis structures for a species are possible, the resonance structures are given as in the case of ozone, isocyanic acid, nitrate and carbonate ions.

For determining the most stable resonance structure(s) of a species, click here to see my earlier post about it.

Lewis Structures of Organic Compounds

With the exception of hydrogen, you can see that an atom can share one or more of its valence electrons, with a maximum of six electrons shared between any two atoms.

Lewis Structures of Ions

Here, the charge on a species is determined by either of the following:

the sum of all of the valence electrons of the atoms of the species minus the total number of electrons in the Lewis electron-dot structure of the species

the algebraic sum of all the formal charges of the species

For a discussion of the determination of the formal charges of atoms of a species, follow the link given above.

For some ions, just as the case is with the ammonium ion, the total number of electrons in its electron-dot structure is less than the total number of valence electrons of its atoms.

Referring to the ammonium ion above, it should have a total of 9 valence electrons (5 from N + 4x1 from H), but its electron-dot structure has only eight electrons.

Ammonia acts as Brönsted base, though a weak one, by accepting a proton thus forming ammonium ion. (See my article Solving Weak Acid/Base Dissociation Problems for a discussion of Brönsted acid-base reaction.)

And for some ions, the total number of electrons in its electron-dot structure is more than the total number of valence electrons of its atoms.

Observe that some oxygen atoms of the phosphate, oxalate, nitrate and carbonate ions have acquired electrons from atoms other than those its own atoms.

2009-12-13T22:41:00.000-08:00

Solving Weak Acid/Base Dissociation Problems

Dissociation problems are about calculations of hydronium ion concentration, [H₃O⁺], and/or pH of aqueous solutions containing any of the following:

a weak acid or a weak base

a strong acid or a strong base

a salt of a weak acid or a salt of a weak base

a weak acid and its salt (weak acid/conjugate base composition) or a weak base and its salt (weak base/conjugate acid composition)

Weak acid and weak base dissociate (or ionize) partially in aqueous solution. Their dissociation products combine to form again the initial weak acid and base.These two reactions occur simultaneously and form an equilibrium. This is in contrast to the strong acids and bases that undergo complete dissociation or ionization in aqueous solution.

Basically, these reactions are about giving up and gaining protons. Those that donate protons are called Bronsted-Lowry acids; and those that accept protons are called Bronsted-Lowry bases.

Their dissociation equilibria given below show that a weak acid loses a proton and forms conjugate base while a weak base gains a proton and forms conjugate acid.

The extent of dissociation (or ionization) is determined by their so called dissociation constants ( K_a for acid and K_b for base ), measured at 25° C.

The dissociation constant is a measure of the tendency of a weak acid to give up a proton and tendency of a weak base to gain a proton. A very small dissociation constant value indicates a very low degree of dissociation.

Values in [ ] denote concentration in moles/liter.

Since the [H₂O] in each case is considered constant, it is incorporated into K_a and K_b.

Take note here that water acts as a weak base in weak acid dissociation and acts as a weak acid in weak base dissociation. A solvent that can act either as an acid or as a base depending upon the solute is called amphiprotic ( or amphoteric ) solvent.

The following discussion is about the calculations of hydronium ion concentration of the aqueous solutions containing (a) a weak acid or a weak base, (b) a strong acid or a strong base, (c) a salt of a weak acid or a salt of a weak base, and (d) a weak acid and its salt or a weak base and its salt.

A. Aqueous Solution of a Weak Acid or a Weak Base

The [H₃O⁺] and/or pH of this solution can be calculated using equations 1 and 2.

Since the amount (or concentration) of dissociated acid or base is equal to the amount (or concentration) of dissociation products:

[dissociated acid] = [H₃O⁺] = [A^-] = x, and

[ionized base] = [BH⁺] = [OH^-] = x, then

we can set equations 1 and 2 to one unknown:

Equations 3 and 4 have the general form of ax² + bx + c = 0 which can be solved by using the quadratic equation:

If the dissociation constant value is very low such that x is negligible, the following approximation can be made:

If the dissociation constant value or % error is high, equations 3 and 4 should then be used for calculation.

B. Aqueous Solution of a Strong Acid or a Strong Base

At 25° C, water has the following dissociation equilibrium:

Since [H₂O] is considered constant, it is incorporated into K_w.

K_w is called the ion-product constant of water at 25° C. This means that pure water has this much amount dissociated at this temperature, that is, [H₃O⁺] = [OH^-] = 1 x 10^-7 M. The [H₃O⁺] or [OH^-] of an aqueous solution can be calculated when one of them is known.

Addition of a strong acid or base in water will shift the equilibrium to the left, as predicted by Le Chatelier principle, lowering the [H₃O⁺] and [OH^-] by depressing the dissociation of water.

Since a strong acid or base dissociates completely in water, then:

[acid] = [H₃O⁺]

[base] = [OH^-]

So,

Setting equation 7 to one unknown:

[dissociated water] = [H₃O⁺] = [OH^-] = x

1 x 10^-14 = ( [acid or base] + x )( x )

which can be solved by using quadratic equation.

But x is very small, we can assume that:

The sample problem below shows the validity of our assumption that x is negligible.

C. Aqueous Solution of a Salt of a Weak Acid or a Salt of a Weak Base

Salt of a weak acid dissociates completely in water as Na⁺ and A^-.

A^-, as a conjugate base of HA, reacts with water to form the following equilibrium:

Multiplying K_a (equation 1) by K_b, we get:

Equation 8 indicates the relationship that the strength of a weak acid or base tends to increase as the strength of its conjugate pair decreases, or vice versa.

This equation enables us to compute the dissociation constant value of the conjugate base of a weak acid and the conjugate acid of a weak base, as well as the [H₃O⁺] of this solution.

D. Aqueous Solution of a Weak Acid And Its Salt or a Weak Base And Its Salt

An aqueous solution containing a weak acid and its salt (or a weak base and its salt) is called a buffered solution or simply buffer solution. This is so because a buffer solution's hydronium ion concentration changes very little upon its dilution or upon addition of a strong acid/base.
The changes in pH value are so small such that the pH of the solution remains practically constant.

Calculation of the buffer solution's pH involves the use of equation 1 or 2:

By taking the negative logarithm of each term of the rearranged equation 1, we get equation 9 which is known as the Henderson-Hasselbach equation.

This equation is commonly used for pH calculation in biochemistry problems.

Problems

The best way to improve your problem solving skill in chemistry is to practice solving as many chemistry problems as possible. Try solving the following problems.

Problems are solved using approximation and the quadratic equation. Values in parenthesis are obtained using the quadratic equation.

For more problems on this topic and other chemistry topics, go to www.tutorpartner.blogspot.com.

1.	Calculate the dissociation constant of the conjugate base of HC₂H₂ClO₂. K_a = 1.36 x 10^-3.
	Answer: 7.35 x 10^-12

2.	A one liter aqueous solution contains 9.54 g KOH. What is the pH of this solution? The dissociation constant for water, K_w, is 1 x 10^-14. ( KOH = 56.1056 g )
	Answer: pH = 13.23

3.	An aqueous solution has a strength of 0.05 M HOCN and 0.07 M NaOCN. Calculate the pH of this solution. ( K_a = 3.3 x 10^-4)
	Answer: pH = 3.63 ( 3.63 )

4.	Calculate the pH of a 0.3 M NaC₆H₅O solution. (K_a for phenol acid is 1.05 x 10^-10; K_w = 1 x 10^-14)
	Answer: pH = 11.73 ( 11.72 )

5.	110 mL of 0.01 M NaOH solution is added to 110 mL of 0.11 M HNO₂ solution. Find the pH of the resulting solution. Given: K_a for nitrous acid is 4.5 x 10^-4.
	Answer: pH = 2.35 ( 2.56)

6.	A one liter aqueous solution contains 9.2 g NaOH. What is the pH of this solution? The dissociation constant for water, K_w, is 1 x 10^-14. ( NaOH = 39.99707 g )
	Answer: pH = 13.36

7.	What is the pH of an aqueous solution having a concentration of 0.07 M HOCN and 0.02 M NaOCN? ( K_a = 3.3 x 10^-4)
	Answer: pH = 2.94 ( 2.97 )

8.	If a solution of HOCN has a concentration of 0.9 M, what is its pH? K_a for cyanic acid is 3.3 x 10^-4.
	Answer: pH = 1.76 ( 1.77 )

9.	Find the pH of a 0.86 M NaC₆H₅COO solution. (K_a for benzoic acid is 6.3 x 10^-5; K_w = 1 x 10^-14)
	Answer: pH = 9.07 ( 9.07 )

10.	An aqueous solution of 0.12 M CH₃COOH and 0.12 M NaCH₃COO has a volume of 340 mL. If 1.6 g of NaOH is added to this solution, find the change in pH. Assume no change in the volume of solution. ( K_a = 1.8 x 10^-5; NaOH = 39.99707 g )
	Answer: pH = 6.75 ( 6.75)

11.	What is the pH of a solution of 0.28 M NaCH₃COO? Calculate the percent hydrolysis. (K_a = 1.8 x 10^-5; K_w = 1 x 10^-14)
	Answer: pH = 9.1; 4.45 x 10^-3 % ( 9.1; 4.45 x 10^-3 % )

12.	What is the pH of an aqueous solution of 0.09 M C₆H₅NH₃Cl? ( Given: K_b = 3.94 x 10^-10; K_w = 1 x 10^-14)
	Answer: pH = 2.82 ( 2.82 )

13.	If 3 L of 0.01 M HOC₂H₄NH₂ solution has a pH of 10.37, find the weight of HOC₂H₄NH₃Cl dissolved in the solution. ( K_b for ethanolamine is 3.18 x 10^-5; mol. wt. = 97.5443 )
	Answer: 0.4 g

14.	In a buffer solution of 0.07 M HC₂H₂ClO₂ and 0.07 M NaC₂H₂ClO₂, HCl is added to it such that its calculated concentration in the solution is 0.02 M. Assuming a constant volume of solution, find the change in pH of the solution. ( K_a = 1.36 x 10^-3)
	Answer: pH = 2.61 ( 2.64 )

15.	Find the [H⁺] and [OH^-] of a 0.24 M of HF solution. K_a = 6.7 x 10^-4; K_w = 1 x 10^-14.
	Answer: [H⁺] = 1.27 x 10^-2 M; [OH^-] = 7.89 x 10^-13 M ( 1.24 x 10^-2; 8.1 x 10^-13 )

16.	What is the dissociation constant of the conjugate acid of HOC₂H₄NH₂. K_b = 3.18 x 10^-5.
	Answer: 3.14 x 10^-10

17.	What is the pH of a 0.38 M CH₃COOH solution. ( K_a = 1.8 x 10^-5 )
	Answer: pH = 2.58; ( 2.58)

18.	What is the pH of an aqueous solution of NH₂C₂H₄NH₂ having a concentration of 0.01 M? The dissociation constant of ethylenediamine is 8.5 x 10^-5.
	Answer: pH = 10.96 ( 10.94)

2009-10-23T21:34:00.000-07:00

Drawing Lewis Electron Dot Structure or Formula

A Lewis electron dot structure shows how the atoms of a molecule or an ion share their outermost electrons or valence electrons to form covalent bonds with each other.

It indicates the number of shared and unshared valence electrons.

This sharing of electrons results in their outermost shells being filled up thus attaining electronic configuration similar to those of the noble gases'.

Observe the electronic configurations of the noble gases below.

He	1s²

Ne	1s²2s²2p⁶

Ar	1s²2s²2p⁶3s²3p⁶

Kr	1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶

Xe	1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶

Rn	1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶

You can see that their outermost shells are filled up which account for their being highly stable.

Now, compare them to the electronic configurations of several atoms that commonly form covalent bonds with other atoms.

H	1s¹

C	1s²2s²2p²

O	1s²2s²2p⁴

N	1s²2s²2p³

S	1s²2s²2p⁶3s²3p⁴

B	1s²2s²2p¹

P	1s²2s²2p⁶3s²3p³

Si	1s²2s²2p⁶3s²3p²

F	1s²2s²2p⁵

With the exception of hydrogen, they share their valence electrons with other atoms to fill up their outermost shells which contain a maximum of eight electrons. Hence, a Lewis electron dot formula must follow the octet (eight) rule.

Take a look at the following examples.

For our purposes, we're going to use color codes in our illustrations to distinguish the individual atoms with their respective valence electrons.

Lewis Electron Dot Structure of Ethane, C₂H₆

The electronic configuration of carbon atom shows that it needs four more electrons to fill up its outer shell; the hydrogen atom, on the other hand, needs just one more electron for its lone electron shell.

Each carbon atom satisfies the octet rule by sharing 3 valence electrons with 3 hydrogen atoms and one with another carbon atom. The hydrogen atom fills up its shell by sharing one electron with carbon atom.

Take note that each pair of shared electrons constitue a single bond.

Lewis Electron Dot Structure of Ethylene (Ethene), C₂H₄

This time, the carbon atoms share 2 electrons with each other and 2 electrons with 2 hydrogen atoms.

Sharing of two pairs of electrons constitute double bond.

Lewis Electron Dot Structure of Acetylene (Ethyne), C₂H₂

Here the carbon atoms share 3 valence electrons with each other and 1 electron with 1 hydrogen atom.

Sharing of three pairs of electrons constitute triple bond.

So far, we've been looking at the Lewis electron dot structures of molecules. We're going to look at the electron dot structures of some ions.

Lewis Electron Dot Structure of Hydroxide Ion

From the electronic configurations given above, we can see that oxygen atom has 6 valence electrons; it needs 2 more electrons to fill up its outer electron shell.

In order to follow the octet rule, oxygen forms the hydroxide ion with the hydrogen atom by sharing one electron with one hydrogen atom and acquiring one more electron .

The gain of one electron results to the ion having a negative net charge of one.

The structure has one electron more than the total valence electrons of the ion (6 from O + 1 from H = 7 valence electrons).

Lewis Electron Dot Structure of Ammonium Ion

The nitrogen atom has 5 valence electrons as indicated by its electronic configuration; it needs three more electrons to fill up its last shell.

But the nitrogen atom in the ammonium ion satisfies the octet rule by sharing three of its valence electrons with each of the three hydrogen atoms.

The structure has one electron less than the total valence electrons of the ion (5 from N + 4 from 4 H = 9 valence electrons).

Determining The Net Charge of an Electron Dot Structure

From these examples, we can say that the:

positive net charge of an ion is equal to the number of electrons deficient in its total valence electrons;

negave net charge of an ion is equal to the number of electrons in excess of its total valence electrons.
The net charge of an ion is also determined by the algebraic sum of the formal charge of all the atoms of the ion.

The formal charge of an atom is the encircled + or - sign near the atom. It is calculated as follows:

formal charge of an atom = valence electrons of the atom - (number of unshared electrons of the atom + number of pairs of shared electrons by the atom)

Formal charge of hydroxide ion's atoms:

O = 6 - (6 + 1) = -1

H = 1 - (0 + 1) = 0

net charge of hydroxide ion = -1 + 0 = -1

Formal charge of ammonium ion's atoms:

N = 5 - (0 + 4) = +1

H = 1 - (0 + 1) = 0

net charge of ammonium ion = +1 + (4 x 0) = +1

Steps To Be Followed In Drawing A Lewis Electron Dot Structure

To summarize, the following steps may be followed to draw the Lewis electron dot structure of a chemical species.

1. Draw all possible structures by using single, double or triple pairs of electrons for atom-to-atom bonds.

2. Make sure that the structures contain the total valence electrons of all the atoms of the species.

3. If the species has a net charge, subtract electrons from the structure as indicated by the + charge or add electrons to the structure as indicated by the - charge.

4. Compute the formal charge of the atoms.

5. The structure that satisfies the octet rule for all atoms must be the correct Lewis structure.
Following the above steps, we're going to draw the electron dot structures of oxygen molecule (O₂) and cyanide ion (CN^-)

Drawing the Lewis Structure of a Molecule

Here are the 3 possible Lewis structures of N₂ molecule.

Structures I and II are incorrect because they violate the octet rule:

structure I has nitrogen atoms each containing only 6 electrons;

structure II has nitrogen atoms each containing only 5 electrons.
Structure III is the correct Lewis electron dot structure of N₂ molecule.

Drawing the Lewis Structure of an Ion

For the structures of cyanide ion as illustrated below, we added one electron to the total valence electrons of 9 (4 from C + 5 from N = 9) since it has a net charge of -1.

We rule out structures I and II because their atoms are electron deficient.

Structure III is the correct Lewis electron dot structure of cyanide ion.

The Resonance Theory

Sometimes, the step by step procedure we used above does not suffice because there are chemical species that have more than one possible Lewis electron dot structures.

One such species is the benzene molecule.

The Lewis structures shown below are equivalent but neither of them represents the actual structure of benzene molecule.

(We used single and double bonds here for easier viewing of the illustrations.)

Instead, benzene is represented by this structure:

This representation of a chemical species by a structure which is intermediate of 2 or more equivalent Lewis structures is called resonance.

The above structure is called resonance hybrid and structures I and II above are called resonance structures.

It should be noted here that resonance structures retain the same atomic arrangement; the only difference among these resonance structures is the arrangement of their electrons.

The Resonance Rules

There are several resonance rules that are used to determine the stability of resonance structures.

The most stable structures are the most important ones.

According to the resonance rules, the most stable structure is the one which has:

1. the greatest number of covalent bonds;

2. the least number of formal charges;

3. the - sign on the more electronegative atom and the + sign on the more electropositive atom, if formal charges are present.

We will use these rules to help us determine the most important Lewis structures of compounds having several resonance structures.

Determining The Most Stable Resonance Structure

Given below are the possible electron dot structures of CO₂.

We strike off structures I, II and III because of their electron deficient atoms.

Our choices then boil down to structures IV - VI.

We can see that structures V and VI are equivalent; we can use any of the two to compare with structure IV.

Being equivalent in the number of covalent bonds, we compare their number of formal charge.

Structure IV is more stable than structure V or VI because it has no formal charge. Hence, structure IV is the Lewis electron dot structure of CO₂.

Let's have another example: an ion this time.

The possible structures of thiocyanate ion (SCN^-) are given below.

How did we determine that C atom is the central atom?

We take the atom with the lowest valence electrons (except H) as the central atom because it has the maximum number of unpaired electrons that can be used to form covalent bonds with other atoms.

We exclude structures I - III from our choices because they don't follow the octet rule.

Since the above 3 structures have all equal number of covalent bonds, we will evaluate their stability based on the number of formal charge.

Structures IV and V are equivalent and are more stable than structure VI because they have less number of formal charge.

We choose structure IV as the Lewis structure of thiocyanate ion because the - sign is on the more electronegative atom.

Conclusion

By using the procedure and rules given above, you are now ready to draw the Lewis electron dot formula of most compounds having covalent bonds. Try the problems below.

Problems

The best way to improve your problem solving skill in chemistry is to practice solving as many chemistry problems as possible. Here are some common ions and molecules. Draw their Lewis electron dot structures.

For more problems on other chemistry topics, go to www.tutorpartner.blogspot.com.

1.	C₂H₅NH₃⁺

2.	CH₃COO^-

3.	H₂O₂

4.	HSO₃^-

5.	HCO₃-

6.	CH₃CN

7.	C₂O₄^-2

8.	HNO₂

9.	HCOCl

10.	N₂

11.	CH₃CONH₂

12.	SO₃^-2

13.	C₄H₆

14.	NO₃^-

15.	NO₂^-

How to Derive a Gas Law From the Ideal Gas Law Equation

2009-10-06T19:15:00.000-07:00

There are several equations that relate the properties of a gas. These relatioships of the pressure, the volume and the temperature for a fixed amount of a gas are given below.

The variables with subscript 1 indicate initial values; and those with subscript 2 indicate final values.

For one who is taking up chemistry for the first time, remembering all these equations is a little bit tricky.

Using the Ideal Gas Law equation, the above gas laws can be derived easily.

This way, you need to remember fewer equations.

The Ideal Gas Law Equation

PV = nRT

where:

P is the pressure in atmosphere

V is the volume in liters

n is the amount of gas in moles

R is the universal gas constant in L-atm/K-mol

T is the temperature in Kelvin

This law relates the four variables of an ideal gas (pressure, volume, amount, and temperature) such that the value of PV/nT is always 0.082057.

It means that a change in the value of any of the four variables will bring about corresponding changes in the values of the other three variables so that the value of R remains constant.

Mathematically, it is equivalent to:

Deriving a Gas Law Equation

As mentioned earlier, we are dealing here with a fixed amount of gas.

It means the amount of gas remains the same from the initial state to the final state. So,

The value of nR must be constant.

Since the values of P₁V₁/T₁ and P₂V₂/T₂ are both equal to nR, then

Thus, we have derived the Combined Gas Law.

Using the above method, we can easily derive the other gas laws.

Suppose a problem states that the temperature must remain constant.

Using the Ideal Gas Law equation for the initial values and final values,

The value of nRT must be constant. Hence,

This is the Boyle's Law.

Now, you can derive the rest of the gas laws using the method we used here.

Problems

Improve your problem solving skill in chemistry by solving as many chemistry problems as possible. Here are some problems dealing with gas laws. The answer is provided for each problem.

For more chemistry problems on gases and other chemistry topics, you can go to www.tutorpartner.blogspot.com.

Boyle's Law Problems

1.	Calculate the pressure required to reduce the volume of 18.0 L of gas at a pressure of 0.333 atm to 6.0 L with the temperature remaining constant.
	Answer: 0.999 atm

2.	Under constant temperature, 7000 mL of a sample gas was allowed to expand to a final volume of 23000 mL. If the initial pressure of the gas was 1085 torr, what was the final pressure of the gas after its expansion?
	Answer: 330 torr

3.	A sample of a gas which occupies 0.93 L has a pressure of 3 atm. Determine the resulting volume of the gas if the pressure is changed to 9 atm and the temperature is maintained constant.
	Answer: 0.31 L

4.	Calculate the pressure required to reduce the volume of 3000 mL of gas at a pressure of 1260 torr to 1000 mL with the temperature remaining constant.
	Answer: 3780 torr

5.	Calculate the pressure required to reduce the volume of 24.0 L of gas at a pressure of 0.25 atm to 13.0 L with the temperature remaining constant.
	Answer: 0.462 atm

6.	At a pressure of 7 atm, a sample of gas has a volume of 8.634 L. If the pressure is reduced to 2.75 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
	Answer: 21.977 L

7.	Under constant temperature, 10000 mL of a sample gas was allowed to expand to a final volume of 17000 mL. If the initial pressure of the gas was 21489 torr, what was the final pressure of the gas after its expansion?
	Answer: 12641 torr

8.	At a pressure of 10 atm, a sample of gas has a volume of 0.497 L. If the pressure is reduced to 5 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
	Answer: 0.994 L

9.	A sample of a gas which occupies 2.8 L has a pressure of 1 atm. Determine the resulting volume of the gas if the pressure is changed to 8 atm and the temperature is maintained constant.
	Answer: 0.35 L

10.	Under constant temperature, 5.0 L of a sample gas was allowed to expand to a final volume of 19.0 L. If the initial pressure of the gas was 12.088 atm, what was the final pressure of the gas after its expansion?
	Answer: 3.181 atm

11.	Calculate the pressure required to reduce the volume of 14000 mL of gas at a pressure of 3281 torr to 3000 mL with the temperature remaining constant.
	Answer: 15311 torr

12.	At a pressure of 10.425 atm, a sample of gas has a volume of 0.214 L. If the pressure is reduced to 1.882 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
	Answer: 1.185 L

13.	Under constant temperature, 8.0 L of a sample gas was allowed to expand to a final volume of 25.0 L. If the initial pressure of the gas was 0.342 atm, what was the final pressure of the gas after its expansion?
	Answer: 0.109 atm

14.	At a pressure of 9.679 atm, a sample of gas has a volume of 0.397 L. If the pressure is reduced to 5.373 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
	Answer: 0.715 L

15.	At a pressure of 8.537 atm, a sample of gas has a volume of 0.45 L. If the pressure is reduced to 2.531 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
	Answer: 1.518 L

Charles' Law Problems

1.	Calculate the final volume of a 3.65 L gas at 92° C when the temperature is brought down to 9° C as the pressure remains unchanged.
	Answer: 2.82 L

2.	Calculate the final volume of a 7.39 L gas at 43° C when the temperature is brought down to 13° C as the pressure remains unchanged.
	Answer: 6.69 L

3.	A sample of gas with a volume of 6.0 L at -17° C was allowed to expand to 21.0 L while maintaining a constant pressure. What was the new temperature of the gas?
	Answer: 623° C

4.	18 L of gas at 294° C was compressed to 2 L at constant pressure. Determine the resulting temperature of the gas.
	Answer: -210° C

5.	Maintaining a constant pressure, a sample of gas having a volume of 5.0 L at -127° C is heated to 369° C. What is the final volume of the gas?
	Answer: 22.0 L

6.	20 L of gas at 582° C was compressed to 4 L at constant pressure. Determine the resulting temperature of the gas.
	Answer: -102° C

7.	A sample of gas with a volume of 3.0 L at -145° C was allowed to expand to 17.0 L while maintaining a constant pressure. What was the new temperature of the gas?
	Answer: 452° C

8.	Maintaining a constant pressure, a sample of gas having a volume of 9.0 L at -210° C is heated to -7° C. What is the final volume of the gas?
	Answer: 38.0 L

Gay-Lussac's Law Problems

1.	A sample of a gas sealed in a vessel was determined to have a pressure of 0.795 atm at 9° C. What would be the pressure of the gas if the vessel was heated to increase the temperature by 88° C?
	Answer: 1.043 atm

2.	A gas cylinder containing a certain gas is found to have a pressure of 18.053 atm at 57° C. If the cylinder is immersed in a cooling solution so that the temperature is lowered by 0° C, calculate the resulting pressure inside the cylinder.
	Answer: 18.053 atm

3.	A sample of a gas has a pressure of 10 atm and a temperature of 32° C. Find the corresponding change in the temperature if the pressure is (a) reduced to 5 atm, (b) tripled. Assume the volume of the gas is constant.
	Answer: (a) -120° C, (b) 642° C

4.	A sample of a gas sealed in a vessel was determined to have a pressure of 1.007 atm at 2° C. What would be the pressure of the gas if the vessel was heated to increase the temperature by 81° C?
	Answer: 1.304 atm

5.	A gas cylinder containing a certain gas is found to have a pressure of 147.177 atm at 50° C. If the cylinder is immersed in a cooling solution so that the temperature is lowered by 1° C, calculate the resulting pressure inside the cylinder.
	Answer: 146.721 atm

6.	A sample of a gas sealed in a vessel was determined to have a pressure of 0.906 atm at 3° C. What would be the pressure of the gas if the vessel was heated to increase the temperature by 84° C?
	Answer: 1.182 atm

Combined Gas Law Problems

1.	What will be the final temperature of a 51.413 L gas at 27° C if its volume is changed to 18.283 L and its pressure changed from 0.974 atm to 4.063 atm?
	Answer: 172° C

2.	784.96 L of a sample of gas has a pressure of 0.23 atm and a temperature of 19° C. After compressing the volume down to 1.37 L and increasing the temperature to 98° C, calculate the change in the pressure of the gas.
	Answer: 167.435 atm

3.	69.73 L of a sample of gas has a pressure of 0.98 atm and a temperature of 29° C. After compressing the volume down to 7.47 L and increasing the temperature to 72° C, calculate the change in the pressure of the gas.
	Answer: 10.451 atm

4.	An expandable air balloon containing 88.2 L of a sample of a gas has a pressure of 1.039 atm at 47° C. What will be the resulting volume of the gas if the pressure is changed to 7 atm and the temperature is increased by 15° C?
	Answer: 13.7 L

5.	What will be the final temperature of a 161.519 L gas at 12° C if its volume is changed to 24.129 L and its pressure changed from 0.225 atm to 11.314 atm?
	Answer: 1866° C

6.	An expandable air balloon containing 8.3 L of a sample of a gas has a pressure of 31.941 atm at 42° C. What will be the resulting volume of the gas if the pressure is changed to 9 atm and the temperature is increased by 14° C?
	Answer: 30.8 L

Ideal Gas Law Problems

1.	A 29.135 L tank containing a sample of gas has a pressure of 0.968 atm and a temperature of 70° C. Find the moles of gas contained inside the tank.
	Answer: 1.002 mol

2.	Calculate the volume of 9.385 mol of a gas confined in a sealed vessel with a pressure of 39.66 atm and a temperature of 36° C.
	Answer: 6 L

3.	Given a volume of 25.07 L of the following gases at STP : (a) SO₂, (b) N₂, (c) H₂S, find the weight in grams of each gas. (mol. wt.: SO₂=64.0588, N₂=28.0134, H₂S=34.0758)
	Answer: (a) 71.69 g, (b) 31.35 g, (c) 38.14 g

4.	Given a weight of 8.8 g of each of the following gases: (a) NH₃, (b) He, (c) NH₃, calculate the volume of each gas at STP. (mol. wt.: NH₃=17.0304, He=4.0026, NH₃=17.0304)
	Answer: (a) 11.57 L, (b) 49.25 L, (c) 11.57 L

2009-09-05T23:48:00.000-07:00

A Note on Dilution and Concentrations of Solutions

Problems dealing with dilution of solutions from one concentration to another concentration can sometimes be confusing.

In order to have a better understanding of it, we will use graphical representation of the solute-solvent components of a solution to help us visualize the problem.

A typical example of this kind of problem is given below:

Given 2 L of 2 M NaCl solution, make a solution having a concentration of 1 M NaCl solution. What must be the final volume of the solution?

We will use a graphical representation like the one below to explain this problem.

The blue box represents one liter of solution, and each white ellipse represents one mole of solute ( NaCl ).

The blue box containing the two ellipses represents the concentration of the solution ( 2 M NaCl = 2 moles NaCl/1 liter of solution ).

So our given solution of 2 L of 2 M NaCl will be represented like this:

By inspection, we can see that we have a total of 4 moles NaCl in 2 L of solution.

Mathematically, finding the total amount of solute in a solution is equivalent to:

Volume of solution (V)  x  Concentration of solution (C)  =  amount of solute

or simply,

V x C  =  amount of solute                         (1)

Since the problem requires us to make 1 M NaCl solution from the given solution by diluting it (adding water to the solution), the total amount of solute must remain the same before and after dilution.

To account for the 4 moles of NaCl, the graphical representation for our final solution will be like this:

By inspection again, we can see that the volume of the final solution is now increased to 4 liters.

Using equation number 1, we can find the final volume of the solution mathematically by solving for V:

So the problem can be represented like this:

2 L of 2 M NaCl

4 L of 1 M NaCl

amount of solute before dilution

amount of solute after dilution

Substituting equation 1 into this, and using subscript 1 and 2 to distinguish the initial solution from the final solution, gives us:

V₁  x  C₁

V₂  x  C₂                             (2)

Equation 2 is the formula for dilution problems involving different kinds of concentrations.

Solving for V₂:

Just knowing that the amount of solute does not change and employing the dimensional analysis to the problem will enable you to figure out the formula by yourself, as long as you know the definitions of different concentrations.

Here are the commonly used concentrations in chemistry problems and their definitions:

The last two expressions are used in very dilute solutions. They are used to express trace amount of solute in aqueous solutions. (1 gram = 1,000 milligrams = 1,000,000 micrograms)

Problems

The best way to improve your problem solving skill in chemistry is to solve as many chemistry problems as you can find. Here are some problems on dilutions and concentrations of solutions. The answer is provided for each problem.

Dilution Problems

1.	What is the volume of the diluted solution of 10.0 mL of 4.34 M K₂HPO₄ if its final concentration is 0.55 M K₂HPO₄?
	Answer: 78.9 mL

2.	If 15.0 mL of 1.9 M Cu(NO₃)₂ solution is diluted to 45.0 mL, calculate the new molarity of the solution.
	Answer: 0.6 M

3.	23.6 mL of 1.07 M K₂SO₄ solution was prepared from an unknown volume of 4.21 M K₂SO₄ solution. What was the initial volume of the solution?
	Answer: 6.0 mL

4.	20.00 mL of an unknown concentration of LiNO₃ solution was diluted to 64 mL to make a concentration of 0.59 M LiNO₃. Determine the initial concentration of the diluted solution.
	Answer: 1.89 M

5.	22.00 mL of an unknown concentration of NaClO₃ solution was diluted to 63 mL to make a concentration of 0.5 M NaClO₃. Determine the initial concentration of the diluted solution.
	Answer: 1.43 M

6.	If 8.0 mL of 0.8 M NiCl₂ solution is diluted to 63.0 mL, calculate the new molarity of the solution.
	Answer: 0.1 M

7.	20.4 mL of 1.44 M NH₄HCO₃ solution was prepared from an unknown volume of 1.63 M NH₄HCO₃ solution. What was the initial volume of the solution?
	Answer: 18.0 mL

8.	21.00 mL of an unknown concentration of KCH₃COO solution was diluted to 64 mL to make a concentration of 0.3 M KCH₃COO. Determine the initial concentration of the diluted solution.
	Answer: 0.9 M

9.	What is the volume of the diluted solution of 17.0 mL of 4.06 M NaCl if its final concentration is 2.84 M NaCl?
	Answer: 24.3 mL

10.	If 24.0 mL of 3.9 M K₂Cr₂O₇ solution is diluted to 38.0 mL, calculate the new molarity of the solution.
	Answer: 2.5 M

Concentration Problems

1.	Determine the weight in grams of the solute contained in a given volume and a given concentration of the following solutions: (a) 3.0 L of 1.97 M Ca(NO₃)₂.4H₂O solution, (b) 14.0 mL of 1.84 M CuS solution, (c) 23.0 mL of 0.91 M Cs₂SO₄ solution. (mol. wt.: Ca(NO₃)₂.4H₂O=236.1506, CuS=95.606, Cs₂SO₄=361.8684)
	Answer: (a) 1395.7 g, (b) 2.5 g, (c) 7.6 g

2.	Calculate the final molarity (M) of a mixed solution of 15.0 mL of 4.65 M NaCH₃COO solution and 10.0 mL of 2.96 M NaCH₃COO solution, assuming their volumes are additive.
	Answer: 4 M

3.	Calculate the resulting molarity (M) of NO₃^-1 from the mixture of 17.00 mL of 1.42 M NaNO₃ solution and 18.00 mL of 4.2 M Mg(NO₃)₂ solution, assuming their volumes are additive.
	Answer: 5.01 M

4.	Determine the weight in grams of the solute contained in the following solutions: (a) 14.0 mL of 0.78 M KOH solution, (b) 4.0 L of 1.81 M Ca(CH₃COO)₂ solution, (c) 6.0 L of 3.03 M AgNO₃ solution. (mol. wt.: KOH=56.1056, Ca(CH₃COO)₂=158.169, AgNO₃=169.8729)
	Answer: (a) 0.6 g, (b) 1145.1 g, (c) 3088.3 g

5.	What is the molarity (M) of the following solutions: (a) 96% (w/w) glacial acetic acid solution with a density of 1.06 g/mL, (b) 35% (w/w) ammonium hydroxide solution with a density of 0.88 g/mL. (mol. wt.: CH₃COOH=60.0524, NH₄OH=35.0456)
	Answer: (a) 16.95 M, (b) 8.79 M

6.	Calculate the resulting molarity (M) of Cl^-1 from the mixture of 10.00 mL of 2.25 M CoCl₃ solution and 20.00 mL of 3.2 M MgCl₂ solution, assuming their volumes are additive.
	Answer: 6.52 M

7.	Calculate the weight of the solute required in order to prepare the following solutions: (a) 18.0 mL of 5.21 M Na₄P₂O₇ solution, (b) 2.0 L of 1.7 M NH₄H₂PO₄ solution, (c) 5.0 L of 1.82 M Al₂(SO₄)₃ solution. (mol. wt.: Na₄P₂O₇=265.9024, NH₄H₂PO₄=115.02546, Al₂(SO₄)₃=342.13588)
	Answer: (a) 24.9 g, (b) 391.1 g, (c) 3113.4 g

8.	What is the molarity (M) of the following solutions: (a) 40% (w/w) hydrobromic acid solution with a density of 1.38 g/mL, (b) 97% (w/w) concentrated sulfuric acid solution with a density of 1.84 g/mL. (mol. wt.: HBr=80.9119, H₂SO₄=98.0734)
	Answer: (a) 6.82 M, (b) 18.2 M

9.	Determine the weight in grams of the solute contained in the following solutions: (a) 9.0 mL of 1.4 M NaClO₃ solution, (b) 1.0 L of 4.24 M Zn(NO₃)₂.6H₂O solution, (c) 4.0 L of 3.31 M Ca(NO₃)₂.4H₂O solution. (mol. wt.: NaClO₃=106.44097, Zn(NO₃)₂.6H₂O=297.481, Ca(NO₃)₂.4H₂O=236.1506)
	Answer: (a) 1.3 g, (b) 1261.3 g, (c) 3126.6 g

10.	What is the molarity of a 23.0 mL solution if it contains: (a) 6.23 g of CdCO₃, (b) 1.78 g of K₂S₂O₅, (c) 7.47 g of Na₂HPO₄.2H₂O? (mol. wt.: CdCO₃=172.4192, K₂S₂O₅=222.3136, Na₂HPO₄.2H₂O=177.9892)
	Answer: (a) 1.6 M, (b) 0.3 M, (c) 1.8 M

Chemistry Partner

Chemical Equation Balancing Tool

Practice Balancing Your Chemical Equations Here

Empirical Formula

Learning Objectives

Definition

Calculation of Empirical Formula

Example 1 Determining the Empirical Formula of an Unknown Compound From the Masses of Atoms

Example 2 Determining the Empirical Formula of an Unknown Compound From Percent Composition

Example 3 Determining the Empirical Formula of a Hydrate

Use of Empirical Formula

Table of Some Compounds With Their Chemical and Empirical Formulas

Example 4 Determining the Molecular Formula of an Unknown Molecular Compound

Percent Composition Calculator

Percent Composition

Self-Assessment Test

Calculation

Uses

Formula Weight Calculation

Self-Assessment Test

Definition of Formula Weight

Formula Weight or Molecular Weight?

Uses of Formula Weight

Example Calculation 1

Example Calculation 2

Example Calculation 3

Compounds With Planar Shape

Geometry of Xenon Tetrafluoride, XeF4

Geometry of Benzene, C6H6

Geometry of CH2=NH

Compounds With Bent Shape

Geometry of Water, H2O

Geometry of Nitrogen Dioxide, NO2

Geometry of Nitrite Ion, NO2-

Geometry of Sulfur Dioxide, SO2

Molecules With Linear Shape

Compounds With Octahedral Shape

Molecules With Trigonal Bipyramidal Shape

Molecular Geometry: Molecules With Trigonal Planar Shapes

Molecular Geometry: Compounds With Tetrahedral Shapes

Methane, CH4

Ammonia, NH3

Trifluorothionitrile, F3S≡N

Compounds With Tetrahedral Shape

Compounds With Pyramidal Shape

Geometries of Molecules and Ions

Linear Geometry

Trigonal Planar Geometry

Tetrahedral Geometry

Trigonal Bipyramidal Geometry

Octahedral Geometry

Pentagonal Bipyramidal Geometry

Hydrogen Bonding

Intermolecular Hydrogen Bonding

Intramolecular Hydrogen Bonding

Lewis Structures of Some Common Compounds and Ions

Lewis Structures of Inorganic Compounds

Lewis Structures of Organic Compounds

Lewis Structures of Ions

Solving Weak Acid/Base Dissociation Problems

A. Aqueous Solution of a Weak Acid or a Weak Base

B. Aqueous Solution of a Strong Acid or a Strong Base

C. Aqueous Solution of a Salt of a Weak Acid or a Salt of a Weak Base

D. Aqueous Solution of a Weak Acid And Its Salt or a Weak Base And Its Salt

Problems

Drawing Lewis Electron Dot Structure or Formula

Lewis Electron Dot Structure of Ethane, C2H6

Lewis Electron Dot Structure of Ethylene (Ethene), C2H4

Lewis Electron Dot Structure of Acetylene (Ethyne), C2H2

Lewis Electron Dot Structure of Hydroxide Ion

Lewis Electron Dot Structure of Ammonium Ion

Determining The Net Charge of an Electron Dot Structure

Steps To Be Followed In Drawing A Lewis Electron Dot Structure

Drawing the Lewis Structure of a Molecule

Drawing the Lewis Structure of an Ion

The Resonance Theory

The Resonance Rules

Determining The Most Stable Resonance Structure

Conclusion

Problems

Geometry of Xenon Tetrafluoride, XeF₄

Geometry of Benzene, C₆H₆

Geometry of CH₂=NH

Geometry of Water, H₂O

Geometry of Nitrogen Dioxide, NO₂

Geometry of Nitrite Ion, NO₂^-

Geometry of Sulfur Dioxide, SO₂

Methane, CH₄

Ammonia, NH₃

Trifluorothionitrile, F₃S≡N

Lewis Electron Dot Structure of Ethane, C₂H₆

Lewis Electron Dot Structure of Ethylene (Ethene), C₂H₄

Lewis Electron Dot Structure of Acetylene (Ethyne), C₂H₂