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$6.46 \ g$ of $BaSO_4$ is obtained from $4.81 \ g$ of a compound in the estimation of sulphur by the Carius method. Calculate the percentage of sulphur in the compound. (Atomic masses: $Ba = 137, S = 32, O = 16$)
(A) The molar mass of $BaSO_4 = 137 + 32 + (4 \times 16) = 233 \ g/mol$.
The mass of sulphur in $233 \ g$ of $BaSO_4$ is $32 \ g$.
The mass of sulphur in $6.46 \ g$ of $BaSO_4 = (32 / 233) \times 6.46 \approx 0.887 \ g$.
The percentage of sulphur = $(\text{mass of sulphur} / \text{mass of compound}) \times 100$.
Percentage of sulphur = $(0.887 / 4.81) \times 100 \approx 18.44 \%$.
Rounding to one decimal place,the result is $18.4 \%$.
The mass of sulphur in $233 \ g$ of $BaSO_4$ is $32 \ g$.
The mass of sulphur in $6.46 \ g$ of $BaSO_4 = (32 / 233) \times 6.46 \approx 0.887 \ g$.
The percentage of sulphur = $(\text{mass of sulphur} / \text{mass of compound}) \times 100$.
Percentage of sulphur = $(0.887 / 4.81) \times 100 \approx 18.44 \%$.
Rounding to one decimal place,the result is $18.4 \%$.
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