Chemical Equations

Molar quantities

Balancing equations

Objectives

Perform calculations using moles,

Avogadro’s number and molar masses

Determine percent composition from formula

Determine empirical formula from percent

composition

Determine molecular formula

Write and balance chemical equations

Counting crows:

How many H atoms are in the molecule

C6H4(CH3)(C2H5)?

1. 12

2. 4

3. 7

4. 9

How many H atoms are in the molecule

C6H3(CH3)2(C2H5)?

1. 5

2. 8

3. 11

4. 14

How many O atoms in

CuSO4.5H2O?

1. 4

2. 5

3. 9

4. 14

Molecules or moles

The numbers (coefficients) in chemical equation

can refer to molecules

But for practical applications, we need a more

useful number: we cannot count molecules

The Mole

The mole is a unit of quantity used in

chemistry to measure the number of atoms

or molecules

DEFINITION:

The number of atoms in exactly 12 g of 12C

A mole of anything always has the same

number of particles: atoms, molecules or

potatoes – 6.02 x 1023 – Avogadro’s number

Mole conversions

Particle – mole conversions

23 1No particles = Moles (mol) x 6.02 x 10 ( )

mol

23

No particles Moles (mol) =

16.02 x 10 ( )

mol

Atomic and molecular weights

Two scales:

Atomic mass unit scale

The mass of an individual atom or molecule in

atomic mass units (amu)

Molar mass scale

The mass of a mole of atoms or molecules in

grams

Confusing?...

The Good News

The mass of a single atom or molecule in amu has

same numerical value as molar mass in grams

The atomic mass of carbon is 12 amu

The molar mass of carbon is 12 g/mol

The same is true for molecules and compounds

The formula mass of H2O is 18 amu (1+1+16)

The molar mass of H2O is 18 g/mol

Calculations with molar mass

Moles =

How many moles are in 13.88 g of lithium if

the atomic mass of Li is 6.94 amu?

2.00

massmolar

mass

How many H atoms in 1 mol of CH4? 1 mol = 6.02 x 1023 particles

How many moles of O2 in 64 g of oxygen?

Atomic mass of O = 16 AMU

Calculate the formula mass of Na2SO4 in AMU.

Use atomic mass Na = 23 AMU, O = 16 AMU, S = 32 AMU

Significance of formula unit

Ionic compounds do not contain molecules. Simplest

formula is the formula unit

Covalent compounds, the molecular formula is the

formula unit

Percent composition and empirical

formula

Chemical analysis

gives the mass % of

each element in the

compound

Molar masses give the

number of moles

Obtain mole ratios

Determine empirical

formula

Determining percent composition

Percent composition is obtained from the actual masses.

Example: Sample contained 0.4205 g of C and 0.0795 g of H.

Total mass = 0.5000 g (0.4205 + 0.0795)

Therefore: in 100 g there are:

(84.10 %)

(15.90 %)

Percent composition: 84.10 % C, 15.90 % H

100 g0.4205 = 84.10 g C

.5000 gx

100 g0.0795 = 15.90 g H

.5000 gx

Percent composition from formula

What is percent composition of C5H10O2? 1 mol C5H10O2 contains 5 mol C, 10 mol H and 2 mol O

atoms

Mass of each element

Total mass = 102.13 g

12.01 g C5 mol C = 60.05 g C

1 mol C1.008 g H

10 mol H = 10.08 g H1 mol H

16.00 g O2 mol O = 32.00 g O

1 mol O

Convert masses into percents

Percent composition:

58.8 % C, 9.870 % H, 31.33 % OCheck result:

58.80 % C + 9.870 % H + 31.33 % O = 100.00%

5 10 2

60.05 g C% C = x100 = 58.80 % C

102.13 g C H O

5 10 2

10.08 g H% H = x100 = 9.870 % H

102.13 g C H O

5 10 2

32.00 g O% O = x100 = 31.33 % O

102.13 g C H O

Empirical formula from percent

composition: 84.1 % C, 15.9 % H

1. Convert percents into moles84.10 g of C ≡ 7.00 mol C

15.9 g of H ≡ 15.8 mol H

2. Determine mole ratioMole ratio H:C =

Simplest formula (decimal form): C1H2.26

Make smallest integers by multiplying

C4H9

May require rounding. Errors in real data cause problems

Do percent composition and empirical formula exercises

84.10 g C

12.00 g/mol

15.9 g H

1.008 g/mol

15.8 mol H2.26 :1

7.00 mol C

What is percent C content of C2H6? Molar mass C = 12 g/mol; molar mass H = 1 g/mol

Empirical formula with more than two elements

Percent composition of vitamin C is: 40.9 % C, 4.58 % H, 54.5 % O

1. Convert into moles

2. Determine mole ratios

3. Find lowest whole numbers

Rounding or not: the role of chemical

intuition

Formulae are always written with integers

Experimental ratios are always fractions

Two choices:

Round to nearest whole number

Multiply top and bottom to find ratio of whole

numbers with same value

Choice depends on the type of substance

Hydrocarbons: A case for not

rounding

There are millions of different hydrocarbons

What if H:C is 2.20 rather than 2.26? An error of only 3 %

Formula becomes C5H11 rather than C4H9

What if H:C is 2.30 rather than 2.26? An error of only 2 %

Formula becomes C3H7

All formulae are reasonable

So how do I know what the composition is?

Additional knowledge about the substance is helpful: melting point, boiling point, molar mass

Inorganic compounds: Rounding

makes sense

Inorganic compounds tend to have few

compositions

Iron forms three oxides: FeO, Fe3O4 and

Fe2O3

Experimental formula FeO1.75 would indicate

Fe2O3 not FeO2

Practice empirical formula problem A compound contains 62.1 % C, 5.21 % H, 12.1 % N and

20.7 % O. What is the empirical formula?

Empirical and molecular formula

Percent composition gives the empirical(simplest) formula. It says nothing about the molecular formula.

Molecular formula describes number of atoms in the molecule

May be much larger than the empirical formula in the case of molecular covalent compounds

For ionic compounds empirical formula = “molecular” formula

Elements and compounds can have molecular

formula different from simplest formula

Substance Empirical

formula

Molecular

formula

Substance Empirical

formula

Molecular

formula

Sulphur S S8Phosphorous P P4

Benzene CH C6H6Acetylene CH C2H2

Ethylene CH2 C2H4Cyclohexane CH2 C6H12

Determination of molecular formula

Require:

1. Empirical formula from percent composition analysis

2. Molar mass from some other source

Number of empirical formula units in molecule:

There are n (AaBbCc) in molecule:

Molecular formula is AnaBnbCnc

Molar mass

Empirical formula massn

Molecular formula of vitamin C

Empirical formula of vitamin C is C3H4O3

Molar mass vitamin C is 176.12 g/mol

Mass of empirical formula = 88.06 g/mol

(3 x 12.01 + 4 x 1.008 + 3 x 16.00)

Number of formula units per molecule =

Molecular formula = 2(C3H4O3) = C6H8O6

Molar mass vitamin C 176.122

Empirical formula mass vitamin C 88.06n

Ionic solids do not have molecular

formulas

Infinite lattices like ionic solids and

covalently bonded lattices are not molecular.

The formula used for an ionic compound is

the same as the empirical formula – with

one or two exceptions Hg2Cl2 rather than HgCl

Structural formula provides more

information

The molecular formula indicates the number of atoms in the molecule

The structural formula indicates how those atoms are arranged

C2H6O is the molecular formula for ethanol and

dimethyl ether

Structural formula for ethanol is CH3CH2OH

Structural formula for ether is CH3OCH3

In a recipe we would need to use the structural

formula to identify the correct reagent

In some cases the mole contents of the

compound are obtained indirectly

Analysis of the

hydrocarbon is performed

by combustion.

Mole ratios of the elements

are derived from the mole

ratios of the combustion

products CO2 + H2O

(1 mol H2O ≡ 2 mol H)

(1 mol CO2 ≡ 1 mol C)

The molecular or empirical formula can

be used to determine the percent

composition

Formula of aspirin is C9H8O4

Molar mass is 180 g

Mass of C = 108 g

Mass of H = 8 g

Mass of O = 64 g

% C = 108/180 x 100 %

% H = 8/180 x 100 %

% O = 64/180 x 100 %

The chemical equation

aA + bB = cC + dD

The Law of Conservation of Matter states that matter is neither created nor destroyed

All the atoms on the left must be the same as those on the right

Reactant

sideProduct

side

coefficient ELEMENT or

COMPOUND

Chemical book-keeping

The key to writing correct equations is to ask

the question, “Have I gained or lost any

atoms?”

Another thing is to put down the correct

formula for each reactant or product

Formulas cannot be changed in order to

balance the equation

In the reaction of hydrogen with oxygen to produce

water, the reactants are the elements H2 and O2,

and the product is H2O

Count the atoms: 4 H and 2 O 4 H and 2 O

The big number

multiplies every

atom after it

The subscript

only multiplies

the atom before

it

Molecular representation of the

same reaction

Balance the equations

CH4 + O2 = CO2 + H2O CH4 + 2O2 = CO2 + 2H2O

C3H8 + O2 = CO2 + H2O C3H8 + 5O2 = 3CO2 + 4H2O

N2 + H2 = NH3

N2 + 3H2 = 2NH3

Do balancing equation exercises