What is the empirical formula of a substance that is 53.5% C, 15.5% H, and 31.19% N by weight?

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What is the empirical formula of a substance that is 53.5% C, 15.5% H, and 31.19% N by weight?

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Answer 1 · 2016-02-25T15:39:51

$C_{2} H_{7} N$ $C_2H_7N$ is the simplest whole number ratio defining constituent atoms in a species, and is thus the empirical formula.

Explanation:

As always with these problems, we assume that there are $100$ $100$ $g$ $g$ of substance, and calculate the elemental composition (we divide thru by the atomic masses of each component):

$% C$ $%C$ $=$ $=$ $\frac{53.5 \cdot g}{12.011 \cdot g \cdot m o l^{- 1}}$ $(53.5*g)/(12.011*g*mol^-1)$ $=$ $=$ $4.45$ $4.45$ $m o l$ $mol$ $C$ $C$ .

$% H$ $%H$ $=$ $=$ $\frac{15.5 \cdot g}{1.00794 \cdot g \cdot m o l^{- 1}}$ $(15.5*g)/(1.00794*g*mol^-1)$ $=$ $=$ $15.38$ $15.38$ $m o l$ $mol$ $H$ $H$ .

$% N$ $%N$ $=$ $=$ $\frac{31.19 \cdot g}{14.01 \cdot g \cdot m o l^{- 1}}$ $(31.19*g)/(14.01*g*mol^-1)$ $=$ $=$ $2.23$ $2.23$ $m o l$ $mol$ $N$ $N$ .

We divide thru by the smallest quotient, that of $N$ $N$ to give an EMPIRICAL formula of $C_{2} H_{7} N$ $C_2H_7N$ (i.e. we divide each molar quantity by $2.23$ $2.23$ !). Without details of molecular weight, this is as far as we can go.

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What is the empirical formula of a substance that is 53.5% C, 15.5% H, and 31.19% N by weight?

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