Question

A 19.13 gram sample of chromium is heated in the presence of excess bromine. A metal bromide is formed with a mass of 77.92 g. Determine the empirical formula of the metal bromide.

Answer 1

Answer:

The empirical formula of the compound is $CrBr_2$.

Explanation:

Mass of chromium = 19.13 g

Mass metal bromide formed = 77.92 g

Mass of bromine in metal bromide = 77.92 g - 19.13 g = 58.79 g

Moles of chromium metal :

$=\frac{19.13 g}{52 g/mol}=0.3679 mol$

Moles of bromine:

$=\frac{58.79 g}{80 g/mol}=0.7349 mol$

For empirical formula divide the smallest number of moles of element from the all the moles of elements:

chromium : $\frac{0.3679 mol}{0.3679}=1$

bromine : $\frac{0.7349 mol}{0.3679}=2$

The empirical formula of the compound is $CrBr_2$.