Chemistry · General chemistry I · Worked example

Find an empirical and a molecular formula

A compound is 40.0% carbon, 6.71% hydrogen and 53.3% oxygen by mass, and its molar mass is 180.16 g/mol. Find its empirical and molecular formulas.

40.0 % C, 6.71 % H, 53.3 % O

Take a 100 g sample

In 100 g of the compound, the percentages become masses: 40.0 g of carbon, 6.71 g of hydrogen and 53.3 g of oxygen.

Convert each mass to moles

Divide by each atomic mass. Carbon and oxygen come out equal, 3.33 mol, and hydrogen is twice as much.

\begin{aligned} n_{C} & = \frac{40.0 g}{12.01 g/mol} = 3.33 mol \\ n_{H} & = \frac{6.71 g}{1.008 g/mol} = 6.66 mol \\ n_{O} & = \frac{53.3 g}{16.00 g/mol} = 3.33 mol \end{aligned}

Divide by the smallest amount

Dividing each amount by 3.33 mol gives 1 : 2.00 : 1.00. The ratios are already whole numbers, so the empirical formula is CH₂O.

Mole ratios in the sample
Element	Moles	÷ 3.33	Ratio
C	3.33	1.00	1
H	6.66	2.00	2
O	3.33	1.00	1

Find the empirical formula mass

CH₂O has a mass of 12.01 + 2(1.008) + 16.00 = 30.03 g/mol.

M ({CH}_{2} O) = 12.01 + 2 (1.008) + 16.00 = 30.03 g/mol

\begin{aligned} M ({CH}_{2} O) & = 12.01 + 2 (1.008) \\ + 16.00 = 30.03 g/mol \end{aligned}

Scale up to the molecular formula

Divide the molar mass by the empirical formula mass: 180.16/30.03 = 6.00. Multiply every subscript in CH₂O by 6.

\frac{180.16 g/mol}{30.03 g/mol} = 6.00

Result

The empirical formula is CH₂O, and the molecular formula is C₆H₁₂O₆.

Your turn

A hydrocarbon is 85.6% carbon and 14.4% hydrogen by mass. What is its empirical formula?

Show the answer and explanation

CH₂.

In 100 g: 85.6 g of carbon ÷ 12.01 = 7.13 mol and 14.4 g of hydrogen ÷ 1.008 = 14.3 mol. Dividing both by 7.13 gives 1 : 2.01, which is 1 : 2 within rounding, so the empirical formula is CH₂.

Keep exploring

Open the data in Composition & formulas, then change the molar mass to 90.08 g/mol and see the molecular formula halve to C₃H₆O₃.

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