Calculating Empirical and
Molecular Formulas
Calculating empirical formula
A compound contains 79.80% carbon and 20.20%
hydrogen. What is the empirical formula.
79.80g C
1mol C
= 6.64 mol C = 1
12.01 g C
20.20g H
1mol H
6.64 mol
=
1.01 g H
Empirical Formula = CH3
20.00 mol H = 3.01
6.64 mol
Calculating Molecular Formulas
The molar mass of a compound is 92.00 g/mol. Analysis of a
sample of the compound indicates that it contains 0.61g of
nitrogen and 1.28 g of oxygen. Find its molecular formula.
Find the empirical formula first.
0.61 g N
1mol N
= 0.04 mol N = 1
14.01g N
1.28 g O 1mol O
0.04 mol
= 0.08 mol O
16 g O
Empirical formula = NO2
0.04 mol
= 2
Molecular formula continued
Using the empirical formula: NO2, find the molecular formula.
Step 1: determine the molar mass of the empirical formula
1mol N x 14.01g = 14.01 g/mol
2mol O x 16.00g = 32.00 g/mol
46.01 g/mol
Step 2: Divide the given molar mass by the molar mass
calculated in step 1. x = 92.00 g/mol = 2
x
46.01 g/mol
Step 3: Multiply the subscripts from the empirical formula by
what was calculated as x in step 2: (NO2)2 = N2O4
The molecular formula is N2O4
Assignment
Determine the Empirical and molecular formulas for each of the
following. Box your empirical formula and your molecular
formula.
1.
A sample compound with a molar mass of 34.00g/mol is found
to consist of 0.44g H and 6.92g O. Calculate both empirical
and molecular formulas.
2.
A compound has a molar mass of 456.18 g/mol and consists
of 3.05% Fe and 4.81% S. Calculate both empirical and
molecular formulas
3.
A compound consists of 36.48g of Na, 25.41g of S, and
38.11g of O. It has a molar mass of 252.10 g/mol. Calculate
both empirical and molecular formulas.