A sample of a compound contains 41.33 g of carbon and 8.67 g of hydrogen. The molar mass of the compound is 87.18 g/mol. What is the molecular formula for the compound?
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C6H15

Explanation:

First, we calculate the empirical formula as follows:

C = 41.33 g

H = 8.67 g

We convert each mass value to mole by dividing each element by its molar mass (C = 12g/mol, H = 1g/mol)

C = 41.33g ÷ 12g/mol = 3.44mol

H = 8.67g ÷ 1g/mol = 8.67mol

Next, we divide each mole value by the smallest (3.44mol)

C = 3.44mol ÷ 3.44mol = 1

H = 8.67mol ÷ 3.44mol = 2.52

We multiply this ratio by 2 to get a simple whole number ratio

C = 1 × 2 = 2

H = 2.52 × 2 = 5.04

Based on this, the whole number ratio of C and H is 2:5, hence, the empirical formula is C2H5.

The molecular mass of the compound is given as 87.18 g/mol, hence, the molecular formula is calculated as follows:

(C2H5)n = 87.18

[12(2) + 1(5)]n = 87.18

[24 + 5]n = 87.18

(29)n = 87.18

n = 87.18 ÷ 29

n = 3.006

Approximately to whole number, n = 3

Hence, the molecular formula of the compound is [C2H5]3

= C6H15

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