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Question

1. A compound contains 46.7% nitrogen and 53.3% oxygen. If the molecular mass of the compound is 60.0 g/mol, what is the molecular formula?

2. A compound is found to contain 24.8% carbon, 2.0% hydrogen, and 73.2% chlorine with a molecular mass of 96.9 g/mol. What is the molecular formula?

3. An oxide of chromium is found to have the following percent composition: 68.4% Cr and 31.6% O. Determine the empirical formula.

4. A compound contains 87.4% nitrogen and 12.6% hydrogen. If the molecular mass of the compound is 32.05 g/mol, what is the molecular formula?

Answer 1

1). 60.0 g/mol x 0.467 N = 28.02 g N

60.0 g/mol x 0.533 O = 31.98

moles N = 28.02 g x 1 mol/14 g = 2 moles N

moles O = 31.98 g x 1 mol/16 g = 2 moles O

molecular formula = N₂O₂

2). Do it the same way as #1

3). For empirical formula, given % composition, assume 100 g of material so that % is now equal to grams.

68.4 g Cr x 1 mol Cr/51.99 g = 1.32 moles Cr

31.6 g O x 1 mol/16 g = 1.98 moles O

Divide by lowest values to get whole numbers:

1.32/1.32 = 1 mol Cr

1.98/1.32 = 1.5 mol O

Multiply both by 2 to get whole numbers:

2 mol Cr

3 mol O

Empirical formula = Cr₂O₃

4). Same as #1 and #2