Determining Chemical Formulas Experimentally % composition, empirical and molecular formula

Determining Chemical Formulas Experimentally

% composition, empirical and molecular formula

Percentage Composition The mass of each element in a compound

compared to the entire mass of the compound times 100 is the percent composition of that element.

Example: What is the % composition of H and O in 1 mole of H2O?%H = 1.01 g H x 2 x 100 = 11%

18.02 g H2O =100%%O = 16.00 g O x 100 = 89%

18.02 g H2O

Percentage CompositionExample: A sample of an unknown compound with

amass of 0.2370 g contains 0.0948 g of carbon, 0.126 g of oxygen, 0.016 g of hydrogen. What is the % composition of the compound?

%C = 0.0948g X 100 = 40%0.2370 g

%O = 0.126 g x 100 = 53 % 0.2370 g

%H = 0.016 g x 100 = 7% 0.2370

Empirical Formula

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

Steps:1. Find moles of each element in the compound.2. Divide each mole value by the smallest mole

value to find the mole ratio.3. Write the formula using the mole ratios

determined in step 2.

A Poem

Percent to gramsGrams to moles

Divide by smallestMultiply till whole

Empirical Formula

Example: A sample is analyzed and determined to contain 80 g C and 20 g H. What is the empirical formula?

1. 80 g C x 1 mole = 6.7 mol C 12.0 g

20 g H x 1 mole = 20 mol H 1.01 g

Empirical Formula

2. Mole ratios: C: 6.7 = 1 H: 20 = 2.98 ~ 3

6.7 6.73. CH3 is the empirical formula You may round the mole ratio if it is

within 0.05 of a whole number. If it is not a whole number you must multiply all the mole ratios by a factor to get a whole number.

Empirical FormulaExample: What is the empirical formula of

a compound that is 70% Fe and 30%O?1. 70% Fe = 70 g Fe x 1 mole = 1.25 mol

55.85 g30% O = 30 g O x 1 mole = 1.88 mol

16.00 g2. Fe = 1.25/1.25 = 1

O = 1.88/ 1.25 = 1.5

Empirical Formula3. Empirical formula must be in whole

number ratios.

Fe = 1 and O = 1.5, multiply both ratios by 2

Fe = 2 and O = 3, so the formula is:Fe2O3

Molecular Formula The formula of the actual molecular

compound is the molecular formula.Steps:1. Calculate the empirical mass.2. Divide the molecular mass by the

empirical mass.3. Multiply each subscript in the

empirical formula; by the number found in step 2.

Molecular Formula Example: The molecular mass of a

compound is found to be 180g/mol. If the empirical formula is CH2O, determine the molecular formula.

1. CH2O = 12.01g/mol + (1.01g/mol x 2) + 16.00 g/mol = 30.02 g/mol

2. I80g/mole = 6 30.02 g/mole

3. CH2O x 6 = C6H12O6