Compound A contains 8.7,% Hydrogen,74,% Carbo | vedclass

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(D) $C$: $74\,\%$$\frac{74}{12} = 6.16$; $\frac{6.16}{1.23} \approx 5$$N$: $17.3\,\%$$\frac{17.3}{14} = 1.23$; $\frac{1.23}{1.23} = 1$$H$: $8.7\,\%$$\

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$C_{10}H_{14}N_{2}$

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(D) $C$: $74\,\%$$\frac{74}{12} = 6.16$; $\frac{6.16}{1.23} \approx 5$$N$: $17.3\,\%$$\frac{17.3}{14} = 1.23$; $\frac{1.23}{1.23} = 1$$H$: $8.7\,\%$$\frac{8.7}{1} = 8.7$; $\frac{8.7}{1.23} \approx 7$Empirical formula $= C_{5}H_{7}N$Empirical formula mass $= (5 	imes 12) + (7 	imes 1) + (1 	imes 14) = 60 + 7 + 14 = 81\, g\, mol^{-1}$Multiplying factor $(n) = \frac{	ext{Molar mass}}{	ext{Empirical formula mass}} = \frac{162}{81} = 2$Molecular formula $= n 	imes (	ext{Empirical formula}) = 2 	imes (C_{5}H_{7}N) = C_{10}H_{14}N_{2}$

Compound $A$ contains $8.7\,\%$ Hydrogen,$74\,\%$ Carbon and $17.3\,\%$ Nitrogen. The molecular formula of the compound is ....
Given : Atomic masses of $C$,$H$ and $N$ are $12$,$1$ and $14\, amu$ respectively.
The molar mass of the compound $A$ is $162\, g\, mol^{-1}$.

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Compound $A$ contains $8.7\,\%$ Hydrogen,$74\,\%$ Carbon and $17.3\,\%$ Nitrogen. The molecular formula of the compound is ....Given : Atomic masses of $C$,$H$ and $N$ are $12$,$1$ and $14\, amu$ respectively.The molar mass of the compound $A$ is $162\, g\, mol^{-1}$.