1.2 Formulas

1.2.5 Define the terms empirical formula and molecular formula1.2.6 Determine the empirical formula and/or the molecular formula of a given compound

• empirical formula from the % composition or from experimental data• % composition from the formula of a compound• molecular formula when given both the empirical formula and the molar mass

Terms:

The law of definite proportions describes that elements in a given compound are always present in the same proportions by mass.

The Percentage composition of a compound refers to the relative mass of each element in the compound.

The term molecular mass describes the mass of a molecule.

The term formula mass describes the mass of the smallest repeating unit of an ionic compound.

Finding Percentage composition from a chemical formula:

1. Find the mass of the compound, then find the mass of the individual parts (atoms).

2. Divide the atoms by the compound and multiply by 100%

For example: H2O

H2 = 2 x 1.01 = 2.02 +O = +16

Total molecular mass = 18.02 Percentage by mass of H in water = 2.02/18.02 =.11 x 100% = 11% Percentage by mass of O in water = 16/18.02

=.888 x 100% = 89%

Percentage Composition from massPercent Composition by mass

Take the mass of the atom and divide by the total mass of the compound, then multiply by 100%

Example: Mass of compound is 48.72g, it contains 32.69g of zinc and 16.03g of sulfur

%Zn = (32.69g/ 48.72g) x 100% = 67.10%

%S = (16.03g/ 48.72g) x 100% = 32.90%

Practice:

1. Calculate the mass percentage of nitrogen in each compound:

a) N2O

b) Sr(NO3)2

c) NH4NO3

d) HNO3

e) NH4

f) KNO3

Empirical Formulas

The empirical formula is simply the simplest whole number ratio of the elements in a compound.

The molecular formula is the actual number of atoms that makes up the molecule

For example, the empirical formula of ethene (C2H4) is CH2

the empirical formula of butene (C4H8) is CH2

Comparing

Name Molecular Empirical

Low ratio

Hydrogen peroxide H2O2 HO 1:1

Glucose C6H12O6 CH2O 1:2:1

Benzene C6H6 CH 1:1

Ethene C2H4 CH2 1:2

Butene C4H8 CH2 1:2

Aniline C6H7N C6H7N 6:7:1

Water H2O H2O 2:1

Finding Empirical formula:

Calculate empirical formula of a compound that is 85.6% carbon and 14.4% hydrogen.

First, assume that you have 100g of the substance, so you really have 85.6g of C and 14.4g of H.

Next, convert it to moles using molar mass.

Finally, find the lowest whole ratio between the two.

1. 85.6g C and 14.4g H

2. mol C = 85.6g x mol = 7.13mol C

12.01g

mol H = 14.4g x mol = 14.3 mol H

1.01g

3. Lowest mole ratio:

C = 7.13 /7.13 = 1

H = 14.3/ 7.13 = 2.01 , so 2

ANSWER: CH2

Practice: Find the empirical formula for the following:

1. 17.5% hydrogen and 82.5% nitrogen

2. 46.3% lithium and 53.7% oxygen

3. 15.9% boron and 84.1% fluorine

4. 52.52% chlorine and 47.48% sulfur

Finding Molecular FormulaThe empirical formula is CH2O, and its molar

mass is 150g/mol. What is molecular formula?

Find empirical molar mass of CH2O: 12.01 + 2.02 + 15.99 g/mol =30.02 g/mol

Divide real molar mass by empirical molar mass: 150/30.02 = 5

Multiply subscripts of empirical formula through by ratio (5): C5H10O5