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ICSE Class 10 Chemistry 2019 Paper

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Question : 81 of 86
Marks: +1, -0
Find the empirical formula and the molecular formula of an organic compound from the data given below :
C=75.92%,H=6.32% and N=17.76%C=75.92\%, H=6.32\% \text{ and } N=17.76\%
The vapour density of the compound is 39.5 .
[C=12,H=1,N=14][C=12, H=1, N=14]
Solution:  
CHN75.926.3217.7675.92126.32117.76146.326.321.266.321.266.321.261.261.265.015.011551\begin{array}{c|c|c} C & H & N \\ \hline 75.92 & 6.32 & 17.76 \\ \hline \frac{75.92}{12} & \frac{6.32}{1} & \frac{17.76}{14} \\ \hline 6.32 & 6.32 & 1.26 \\ \hline \frac{6.32}{1.26} & \frac{6.32}{1.26} & \frac{1.26}{1.26} \\ \hline 5.01 & 5.01 & 1 \\ \hline 5 & 5 & 1 \end{array}
Empirical formula : C5H5N\mathrm{C}_5\mathrm{H}_5\mathrm{N}
Empirical formula Mass
=C5H5N= \mathrm{C}_5\mathrm{H}_5\mathrm{N}
=12×5+1×5+14= 12 \times 5 + 1 \times 5 + 14
=60+5+14=79= 60 + 5 + 14 = 79
V.D.=39â‹…5\text{V.D.} = 39 \cdot 5
Mol. Mass=2×V.D.\text{Mol. Mass} = 2 \times \text{V.D.}
=2×39.5=79= 2 \times 39.5 = 79
n=Mol. MassEmp. formula Massn = \frac{\text{Mol. Mass}}{\text{Emp. formula Mass}}
=7979=1= \frac{79}{79} = 1
Molecular formula
=(Empirical formula)n= (\text{Empirical formula})_n
=C5H5N×1= \mathrm{C}_5\mathrm{H}_5\mathrm{N} \times 1
=C5H5N= \mathrm{C}_5\mathrm{H}_5\mathrm{N}
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