2.0 text{ g} of a bromo hydrocarbon (X) was s | vedclass

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(D) $1$. Calculate moles of $Br$: $	ext{Moles of } AgBr = \frac{3.36 	ext{ g}}{188 	ext{ g/mol}} \approx 0.01787 	ext{ mol}$. Since $1 	ext{ mol

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(D) $1$. Calculate moles of $Br$: $	ext{Moles of } AgBr = \frac{3.36 	ext{ g}}{188 	ext{ g/mol}} \approx 0.01787 	ext{ mol}$. Since $1 	ext{ mol } AgBr$ contains $1 	ext{ mol } Br$, moles of $Br = 0.01787 	ext{ mol}$.$2$. Calculate mass of $Br$: $	ext{Mass of } Br = 0.01787 	ext{ mol} 	imes 80 	ext{ g/mol} \approx 1.43 	ext{ g}$.$3$. Calculate mass of $C$: $	ext{Mass of } C = 26.7\% 	ext{ of } 2.0 	ext{ g} = 0.534 	ext{ g}$.$4$. Calculate moles of $C$: $	ext{Moles of } C = \frac{0.534 	ext{ g}}{12 	ext{ g/mol}} = 0.0445 	ext{ mol}$.$5$. Determine empirical formula ratio: $	ext{Ratio } C:Br = 0.0445 : 0.01787 \approx 2.5 : 1$. To convert to the simplest integer ratio, multiply by $2$, giving $C:Br = 5:2$. Thus, the empirical formula is $C_5Br_2$. The number of carbon atoms is $5$.

$2.0 \text{ g}$ of a bromo hydrocarbon $(X)$ was subjected to Carius analysis, which gave $3.36 \text{ g}$ of $AgBr$. The percentage of carbon in the compound $(X)$ is $26.7\%$. The total number of carbon atoms in the empirical formula for compound $(X)$ is . . . . . . .

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