A substance contains 38.8% C,16.0% H,and 45 | vedclass

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Question

(A) Step $1$: Calculate the moles of each element by dividing the percentage by its atomic mass.$C = \frac{38.8}{12} = 3.23$$H = \frac{16.0}{1} = 16.0

Vedclass · Accepted Answer

$CH_3NH_2$

Vedclass · Answer

(A) Step $1$: Calculate the moles of each element by dividing the percentage by its atomic mass.$C = \frac{38.8}{12} = 3.23$$H = \frac{16.0}{1} = 16.0$$N = \frac{45.2}{14} = 3.23$Step $2$: Divide each value by the smallest mole value $(3.23)$ to find the simplest ratio.$C = \frac{3.23}{3.23} = 1$$H = \frac{16.0}{3.23} \approx 5$$N = \frac{3.23}{3.23} = 1$Step $3$: The ratio of $C:H:N$ is $1:5:1$. Therefore,the empirical formula is $CH_5N$,which can be written as $CH_3NH_2$.

$A$ substance contains $38.8\% \ C$,$16.0\% \ H$,and $45.2\% \ N$. What is the empirical formula of the substance?

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