Question

Write the empirical and molecular formulae of a compound, if the molecular mass of a compound is 93 and compound containing 77.43% of C, 7.53% of H and 15.05% of N ?​

Answer 1

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$\underline{\underline{\bf \purple{Solution}}}$

Given,

• Molecular mass of compound = 93
• Percentage of C = 77.43%
• Percentage of H = 7.53%
• Percentage of N = 15.05%

To find ,

• Empirical formula
• Molecular formula

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Considering 100g of solution

$\begin{tabular}{||c|c|c|c|c||} \cline{1-5} \bf{Element} & \bf Atomic mass & \bf\% of mass & \bf no. of atoms & \bf Simplest Ratio \\\cline{1-5} \sf C & \sf 12 & \sf 77.43 & \sf {77.43}/{12}=6.45 & \sf 6 \\\cline{1-5} \sf H & 1 & 7.53 & 7.53/1 = 7.53 & 8 \\\cline{1-5} \sf N & 14 & 15.05 & 15.05/14 = 1.075 & 1\\\cline{1-5} \end{tabular}$

Now ,

Ratio of atoms

6 : 8 : 1

That means

• Carbon atoms = C6
• Hydrogen atoms = H8
• Nitrogen atoms = N

Empirical formula = $\sf C_6H_8N$

Now , finding molecular formula

We know that

$\dagger\; \sf Molecular\;formula=[Empirical\; formula]_{\frac{MFW}{EFW}}$

Where

• EFW : Empirical formula weight = 6(12) + 1(8) + 14 = 72 + 8 + 14 = 94
• MFW : Molecular formula weight = 93

Substituting the values we have

$\to \sf M.F = [C_6H_8N]_{\frac{93}{94}}$

Here ,

$\sf\dfrac{94}{93} = 0.98$ ≈ 1

$\bigstar\boxed{ \rm \green{Molecular\;formula = C_6H_8N}}$

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