Question

A substance on analysis gave the following percentage composition: $Na=43.4\%,C=11.3\%$ and $O=45.3\%$. Calculate the empirical weight (Atomic weight in amu of $Na=23,C=12,O=16$).

Answer 1

$\begin{array}{cccc}\text{Element}& \text{Percentage composition}& \text{Atomic ratio}& \text{Least ratio}\\ \text{Sodium}& 43.4& \frac{43.4}{23}=1.89& \frac{1.89}{0.94}=2\\ \text{Carbon}& 11.3& \frac{11.3}{12}=0.94& \frac{0.94}{0.94}=1\\ \text{Oxygen}& 45.3& \frac{45.3}{16}=2.83& \frac{2.83}{0.94}=3\end{array}$

Hence, the empirical formula is $N{a}_{2}C{O}_3$ and the empirical weight is 106.