Question

2.746 g of a compound gave on analysis 1.94 g of silver, 0.268 g of sulphur and 0.538 g of oxygen. Find the empirical formula of the compound.
(At. masses: $Ag=108,S=32,O=16$)

• A. $A{g}_{2}S{O}_4$
• B. $A{g}_4(S{O}_4{)}_2$
• C. $AgS{O}_4$
• D. $AgS{O}_2$

Answer 1

The correct option is A $A{g}_{2}S{O}_4$

Percentage composition of the compound is calculated as below:
$\text{Silver}=\frac{1.94}{2.746}\times 100=70.65\%$
$\text{Sulphur}=\frac{0.268}{2.746}\times 100=9.75\%$
$\text{Oxygen}=\frac{0.538}{2.746}\times 100=19.6\%$

To calculate empirical formula:
$\begin{array}{cccccc}& & & & & \\ \text{Element}& \%\text{age}& \text{At. mass}& \text{Relative number}& \text{Simplest ratio}& \text{Simplest whole}\\ & & & \text{of moles}& \text{of moles}& \text{no. of ratio}\\ Ag& 70.65& 108& \frac{70.65}{108}=0.654& \frac{0.654}{0.305}=2.14& 2\\ S& 9.75& 32& \frac{9.75}{32}=0.305& \frac{0.305}{0.305}=1& 1\\ O& 19.6& 16& \frac{19.6}{16}=1.22& \frac{1.22}{0.305}=4& 4\end{array}$

$\therefore$ Empirical formula is $A{g}_{2}S{O}_4$