Question

A compound containing chromium and sulfur contains 76.4 mass percent chromium.

Determine its empirical formula.

Answer 1

Divisiblity factor for Cr $= \dfrac{weight percent}{molar mass } =\dfrac{76.4}{52}=1.47$

Divisiblity factor for S $=$ $\dfrac{weight percent}{molar mass } =\dfrac{23.6}{32}=0.73$

Now divide the above both terms with the smallest number between the two so and that will be the new divisiblity factor of the both.

So new factor for Cr = 2

So new factor for S = 1

Hence the emperical formula will became = $Cr_2 S$