Question

Determine the empirical formula for the compound C=42.86% H=2.4% N=5.88% Na=9.65% O=25.75% S=13.46%

Answer 1

To determine empirical formula,first divide the percentage of elements by their molar mass to find their relative number of moles

Relative number of moles of $C=\frac{42.86}{12}=3.57167$

Relative number of moles of $H=\frac{2.4}{1}=2.4$

Relative number of moles of $N=\frac{5.88}{14}=0.42$

Relative number of moles of $Na=\frac{9.65}{23}=0.41956\approx0.42$

Relative number of moles of $O=\frac{25.75}{16}=1.60$

Relative number of moles of $S=\frac{13.46}{32}=0.42$

Then divide the relative number of moles by the smallest one.

For $C\equiv \frac{3.57167}{0.42}=8.5$

For $H\equiv \frac{2.4}{0.42}=5.71$

For $N\equiv \frac{0.42}{0.42}=1$

For $Na\equiv \frac{0.42}{0.42}=1$

For $O\equiv \frac{1.60}{0.42}=3.81$

For $S\equiv \frac{0.42}{0.42}=1$

The equivalent ratio is

$C:H:O:Na:S:N\equiv$

$8.5:5.7:3.8:1:1:1\equiv$

$1.9(4.5:3:2):1:1:1\approx$

$2(4.5:3:2):1:1:1\equiv$

$9:6:4:1:1:1$

Hence the empirical formula will be $C_9H_6O_4NaSN$