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Percentage composition and Molecular formula Questions in English

Class 11 Chemistry · Some Basic Concepts of Chemistry · Percentage composition and Molecular formula

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$2.0 \text{ g}$ 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 . . . . . . .
A
$1$
B
$2$
C
$3$
D
$5$

Solution

(D) $1$. Calculate moles of $Br$: $\text{Moles of } AgBr = \frac{3.36 \text{ g}}{188 \text{ g/mol}} \approx 0.01787 \text{ mol}$. Since $1 \text{ mol } AgBr$ contains $1 \text{ mol } Br$,moles of $Br = 0.01787 \text{ mol}$.
$2$. Calculate mass of $Br$: $\text{Mass of } Br = 0.01787 \text{ mol} \times 80 \text{ g/mol} \approx 1.43 \text{ g}$.
$3$. Calculate mass of $C$: $\text{Mass of } C = 26.7\% \text{ of } 2.0 \text{ g} = 0.534 \text{ g}$.
$4$. Calculate moles of $C$: $\text{Moles of } C = \frac{0.534 \text{ g}}{12 \text{ g/mol}} = 0.0445 \text{ mol}$.
$5$. Determine empirical formula ratio: $\text{Ratio } C:Br = 0.0445 : 0.01787 \approx 2.5 : 1$. 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 $5$.

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