Question

Question (a)Calculate the empirical formula of a compound that has the following composition by mass. 29.0% Na, 40.5% S, O=30.4%

Answer 1

The first step to answer this question is to assume that the given percents are masses.

Then, convert these masses to moles using the corresponding molar mass:

$\begin{gathered} 29.0gNa\cdot\frac{1molNa}{22.98gNa}=1.26molNa \\ 40.5gS\cdot\frac{1molS}{32.065gS}=1.26molS \\ 30.4gO\cdot\frac{1molO}{15.999gO}=1.9molO \end{gathered}$

Divide each result by the least result (1.26mol):

$\begin{gathered} \frac{1.26mol}{1.26mol}=1 \\ \frac{1.26mol}{1.26mol}=1 \\ \frac{1.9mol}{1.26mol}=1.5 \end{gathered}$

Since not all of these are whole numbers, we have to multiply them times 2:

$\begin{gathered} 1\cdot2=2 \\ 1\cdot2=2 \\ 1.5\cdot2=3 \end{gathered}$

These numbers are the subscripts of each element in the empirical formula.

It means that the empirical formula of this compound is:

$Na_2S_2O_3$