Question a7489

Oct 3, 2016

Percentage Composition of constituent elements in a compound

C->74.0%

H->7.4%

N->8.6%

O->10.0%#

Considering their atomic masses as

$C \to 12 \text{ g/mol}$

$H \to 1 \text{ g/mol}$

$N \to 14 \text{ g/mol}$

$O \to 16 \text{ g/mol}$

Now dividing each percentage composition by the respective atomic mass and then taking the ratio we get the ratio of number of atoms of constituent elements in the compound as

$C : H : N : O = \frac{74}{12} : \frac{7.4}{1} : \frac{8.6}{14} : \frac{10}{16}$

$= 6.16 : 7.4 : 0.614 : 0.625$

$= \frac{6.16}{0.614} : \frac{7.4}{0.614} : \frac{0.614}{0.614} : \frac{0.625}{0.614}$

$= 10 : 12 : 1 : 1$

Hence the empirical formula is

${C}_{10} {H}_{12} N O$