Molecular Formula Practice
Example: If the empirical formula was Al2O3 and had a molar mass of 306 g, what would the molecular formula be?
Step 1: The molar mass is always a multiple of the empirical molar mass (i.e., M.W. = n × E.F.W.) To determine n, divide the given molar mass by the empirical molar mass.
The empirical molar mass is 2 × 27.0 g + 3 × 16.0 g = 102 g
The molecular molar mass is 306 g.
306 ÷ 102 = 3.
Step 2: Multiply all the subscripts in the empirical formula by the answer to the previous step. We multiply the subscripts in the empirical formula by 3 to get the molecular formula Aℓ6O9
Practice:
1. Determine the molecular formula of each compound from the empirical formula and the molar mass:
a) E.F. = NaS2O3, molar mass = 270.4 g/mol
b) E.F. = C3 H2Cl, molar mass = 147.0 g/mol
c) E.F. = C2 HCl, molar mass = 181.4 g/mol
d) E.F. = Na2SiO3, molar mass = 732.6 g/mol
e) E.F. = NaPO3, molar mass = 305.9 g/mol
f) E.F. = NO2, molar mass = 92.0 g/mol
2. What is the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4 N?
3. What is the molecular formula of CH3O if its molar mass is 62 g/mol?
4. Find the molecular formula for a compound with an empirical formula of C2 H8 N and a molecular mass of 46 grams per mole.
5. 4.04 g of nitrogen combine with 11.46 g of oxygen to produce a compound.
a) What is the empirical formula of this compound?
b) The molar mass of the compound is 108.0g/mol. Find its molecular formula
6. The molar mass of a compound is 92 g. Analysis of the sample indicates that it contains 0.606 g N and 1.390 g O. Find the compound’s molecular formula.
7. Ethene, a gas used extensively in preparing plastics and other polymers, has a composition of 85.7% carbon and 14.3% hydrogen. Its molar mass is 28 g. Find the molecular formula for ethene.