Question

From the experimental data, the compositions are found as barium = $58.81\%$, sulfur = $13.73\%$, and oxygen =$27.46\%$. Predict the empirical formula of the compound.

• A. $BaS{O}_4$
• B. $Ba{S}_{2}O$
• C. $B{a}_{2}S{O}_3$
• D. $Ba{S}_2{O}_4$
• E. $B{a}_{2}S{O}_4$

Answer 1

The correct option is A $BaS{O}_4$

From the experimental data, the compositions are found as barium = 58.81%, sulfur = 13.73%, and oxygen =27.46%.
100 g of compound will have 58.81 g barium, 13.73 g sulfur and 27.46 g oxygen.
The atomic masses of barium, sulfur and oxygen are 137.3 g/mol, 32.1 g/mol and 16 g/mol respectively.
When we divide mass with molar mass, we get number of moles.
Number of moles of barium $=\frac{58.81}{137.3}=0.428$ moles
Number of moles of sulfur $=\frac{13.73}{32.1}=0.428$ moles
Number of moles of oxygen $=\frac{27.46}{16}=1.716$ moles
The ratio of number of moles $Ba:S:O=0.428:0.428:1.716=1:1:4$.
Hence, the empirical formula of the compound is $BaS{O}_4$.