Question

A compound is found to contain 50.05 % sulphur and 49.95 % oxygen by weight. What is the empirical formula for this compound? The molecular weight for this compound is 64.07 g/mol. What is its molecular formula?

Answer 1

1) Assume 100 g of the compound is present. This changes the percents to grams:
$S\Rightarrow 50.05g$
$O\Rightarrow 49.95g$
2) Convert the masses to moles:
$S\Rightarrow \frac{50.05g}{32.066}$ g/mol = 1.5608 mol
$O\Rightarrow \frac{49.95g}{16.00}$ g/mol = 3.1212 mol
3) Divide by the lowest, seeking the smallest whole-number ratio:
$S\Rightarrow \frac{1.5608}{1.5608}=1$
$O\Rightarrow \frac{3.1212}{1.5608}=2$
4) Write the empirical formula:
$S{O}_2$
5) Compute the "empirical formula weight:"
32 + 16 + 16 = 64
6) Divide the molecule weight by the "EFW:"
$\frac{64.07}{64}=1$
7) Use the scaling factor computed just above to determine the molecular formula:
$S{O}_2$ times 1 gives $S{O}_2$ for the molecular formula