KCET · Chemistry · Some Basic Concepts of Chemistry
Empirical formula of a compound is \(\mathrm{CH}_{2} \mathrm{O}\) and its molecular mass is 90 , the molecular formula of the compound is
- A \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\)
- B \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\)
- C \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)
- D \(\mathrm{CH}_{2} \mathrm{O}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\)
Step-by-step Solution
Detailed explanation
The empirical formula of compound \(=\mathrm{CH}_{2} \mathrm{O}\)
Empirical formula mass
\(=(1 \times 12)+(2 \times 1)+(1 \times 16)=30\)
Molecular mass \(=90\)
\(\therefore n=\frac{\text { Molecular mass }}{\text { Empirical mass }}=\frac{90}{30}=3\)
and molecular formula \(=3 \times \mathrm{CH}_{2} \mathrm{O}\)
\(=\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\)
Empirical formula mass
\(=(1 \times 12)+(2 \times 1)+(1 \times 16)=30\)
Molecular mass \(=90\)
\(\therefore n=\frac{\text { Molecular mass }}{\text { Empirical mass }}=\frac{90}{30}=3\)
and molecular formula \(=3 \times \mathrm{CH}_{2} \mathrm{O}\)
\(=\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\)
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