0.16 g of an organic compound containing sul | vedclass

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Question

(A) The molar mass of $BaSO_4$ is $233 \ g/mol$. The atomic mass of sulphur $(S)$ is $32 \ g/mol$. The formula for the percentage of sulphur is: $	ex

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) The molar mass of $BaSO_4$ is $233 \ g/mol$. The atomic mass of sulphur $(S)$ is $32 \ g/mol$. The formula for the percentage of sulphur is: $	ext{Percentage of } S = \frac{32 	imes 	ext{mass of } BaSO_4 	imes 100}{233 	imes 	ext{mass of organic compound}}$. Substituting the given values: $	ext{Percentage of } S = \frac{32 	imes 0.233 	imes 100}{233 	imes 0.16}$. Since $0.233 / 233 = 0.001$,the expression becomes: $\frac{32 	imes 0.001 	imes 100}{0.16} = \frac{3.2}{0.16} = 20 \%$. Thus,the percentage of sulphur is $20 \%$.

$0.16 \ g$ of an organic compound containing sulphur produces $0.233 \ g$ of $BaSO_4$. The percentage of sulphur in the compound is:

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