Question

A compound has a percent composition of 38.7% carbon, 9.76% hydrogen, 51.5% oxygen.
Laboratory data shows that the compound's molar mass is 62.0 g/mol. What is the
molecular formula of the compound?

Answer 1

The molecular formula of the compound that has a percent composition of 38.7% carbon, 9.76% hydrogen, 51.5% oxygen is C2H6O2.

How to calculate molecular formula?

The molecular formula can be calculated from the empirical formula. The empirical formula of the compound is calculated as follows:

• C = 38.7% = 38.7g
• H = 9.76% = 9.76g
• O = 51.5% = 51.5g

Next, we convert the mass to moles by dividing by their atomic mass:

• C = 38.7 ÷ 12 = 3.23mol
• H = 9.76 ÷ 1 = 9.76mol
• O = 51.5÷ 16 = 3.22mol

Next, we divide by the smallest (3.22)

• C = 1
• H = 3
• O = 1

Hence, the empirical formula of the compound is CH3O

If the molar mass of the compound is 62g/mol;

(CH3O)n = 62

31n = 62

n = 2

(CH3O)2 = C2H6O2

Therefore, the molecular formula of the compound that has a percent composition of 38.7% carbon, 9.76% hydrogen, 51.5% oxygen is C2H6O2.

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