91Ó°ÊÓ

The atomic ratio is a measure of the proportion of one type of atom I to another type of atom (ii) (j). The atomic percent (or at. percent) is a closely related concept that expresses the percentage of one type of atom relative to the total number of atoms. The molar fraction or molar percent are the molecular equivalents of these concepts.

Compound known to contain just C,H,N and O with 46.21wt \% \mathrm{C}, 9.02 wt%H ,13.74 wt%N and, by difference, 100 - 46.21 - 9.02 - 13.74 = 31.03%O .

First consider that $100 \mathrm{~g}$ of compound would contain:

$$\begin{gathered} 46.21\mathrm{g}\mathrm{of}\mathrm{C} \\ 9.02\mathrm{g}\mathrm{of}\mathrm{H} \\ 13.74\mathrm{g}\mathrm{of}\mathrm{N} \\ 31.03\mathrm{g}\mathrm{of}\mathrm{O} \end{gathered}$$

Calculating the atomic ratio of C : H : N : O :

$$\begin{gathered} \frac{\mathrm{m}}{\mathrm{M}}\left(\mathrm{C}\right):\frac{\mathrm{m}}{\mathrm{M}}\left(\mathrm{H}\right):\frac{\mathrm{m}}{\mathrm{M}}\left(\mathrm{N}\right):\frac{\mathrm{m}}{\mathrm{M}}\left(\mathrm{O}\right): \\ \frac{46.21\mathrm{g}}{12.0107\mathrm{g}/\mathrm{mol}}:\frac{9.02\mathrm{g}}{1.00794\mathrm{g}/\mathrm{mol}}:\frac{13.74}{14.00674\mathrm{g}/\mathrm{mol}}:\frac{31.03\mathrm{g}}{15.9994\mathrm{g}/\mathrm{mol}} \end{gathered}$$

Considering that 0.981 is the smallest factor we will divide all of the factors above with 0.981 which will then give us the ratio:

3.992 : 9.120 : 1.000 : 1.978

The empirical formula would then be ${\mathrm{C}}_4{\mathrm{H}}_9{\mathrm{NO}}_2$

Therefore the atomic ratio is 3.992 : 9.120 : 1.000 : 1.978 .