What is the empirical formula of a substance with 74.0% C , 8.65% H, and 17.4% N?

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What is the empirical formula of a substance with 74.0% C , 8.65% H, and 17.4% N?

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Answer 1 · 2016-06-25T17:17:52

$C_{5} H_{7} N$ $C_5H_7N$

Explanation:

First, assume that there is 100 g of substance.

Next is to calculate the number of moles or each element:

$74.0 g C \cdot \frac{1 m o l C}{12.01 g} \approx 6.16 m o l C$ $74.0 g C * (1mol C )/ (12.01 g) ~~ 6.16 mol C$
$8.65 g H \cdot \frac{1 m o l H}{1.008 g} \approx 8.58 m o l H$ $8.65 g H * (1mol H )/ (1.008 g) ~~ 8.58 mol H$
$17.4 g N \cdot \frac{1 m o l N}{14.01 g} \approx 1.24 m o l N$ $17.4 g N * (1mol N )/ (14.01 g) ~~ 1.24 mol N$

After getting the mol of each element, you can calculate now the mole ratio. Do this by dividing each number of moles by the lowest number of moles obtained.

$C = (\frac{6.16}{1.24}) \approx 5$ $C= (6.16/1.24)~~ 5$
$H = (\frac{8.58}{1.24}) \approx 7$ $H= (8.58/1.24)~~ 7$
$N = (\frac{1.24}{1.24}) \approx 1$ $N= (1.24/1.24)~~ 1$

Therefore, the empirical formula of a substance with 74.0% C , 8.65% H, and 17.4% N is $C_{5} H_{7} N$ $C_5H_7N$

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What is the empirical formula of a substance with 74.0% C , 8.65% H, and 17.4% N?

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