Question

A salt contains only strontium and one of the halide ions. A 0.0910-g sample of the salt was dissolved in water, and an excess of sulfuric acid was added to form strontium sulfate (SrSO₄), which was filtered, dried, and weighed. Its mass was found to be 0.133 g. What is the formula of the strontium halide? (Omit states-of-matter from your answer.)

Answer 1

The balanced equation between SrX2 and H2SO4 can be written as follows.

SrX2 + H2SO4 $\rightarrow$ SrSO4 + 2HX

Here, the no. of moles of SrSO4 formed = 0.091 g/(183.676 g/mol) = 7.241×10^-4 mol

According to the equation shown above, the mole ratio of SrX2: SrSO4 = 1: 1

Therefore, the no. moles of SrX2 = 7.241×10^-4 mol

Hence, the molar mass of SrX2 = 0.091 g/(7.241×10^-4 mol) = 125.6 g/mol = molar mass of SrF2

i.e. The formula of strontium halide is SrF2