An organic compound was found to contain $40.0 \% \mathrm{C}$ and $6.66 \% \mathrm{H}$. Find it's molecular…
An organic compound was found to contain $40.0 \% \mathrm{C}$ and $6.66 \% \mathrm{H}$. Find it's molecular formula (molar mass $=180$ )
- $\mathrm{C}_{22} \mathrm{H}_{24} \mathrm{O}_{11}$
- $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$
- $\mathrm{CH}_{2} \mathrm{O}$
- $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$
Solution
For this we make a table as:
\begin{array}{|c|c|c|c|c|}
\hline Elemen & \%Compositio & Atomic & Mole & Simple Ratio \\
\hline \mathrm{t} & \mathrm{n} & Mass & Ratio & \\
\hline \mathrm{C} & 40.00 \% & 12 & (40 / 12) & (3 \cdot 33 / 3 \cdot 33)=1 \\
\hline \mathrm{H} & 6.67 \% & 1 & (6.67 / 1) & (6.667 / 3 \cdot 33)=2 \\
\hline \mathrm{O} & 53.33 \% & 16 & (53 \cdot 33 / 16 & (3 \cdot 33 / 3 \cdot 33)=1 \\
\hline
\end{array}
$\therefore$ Formula of compound is $=\mathrm{CH}_{2} \mathrm{O}$
Molecular formula mass $=180$
Empirical formula mass $=12+2+16=30$
$\mathrm{n}=180 / 30=6$
Therefore, molecular formula = empirical formula $\times 6=\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$
Asked in: MHT CET 2020 (15 Oct Shift 2)
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