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JAMB Chemistry 2024 Paper

Question : 10

Total: 138

An organic compound contains 53.1% Carbon, 6.2% Hydrogen, 12.4% Nitrogen, and 28.3% Oxygen by mass. What is the molecular formula of the compound if its vapour density is 56.5? [ C =12, H = 1, N = 14, O = 16].

C $_{5}$ H $_{7}$ NO $_{2}$
C $_{5}$ H $_{6}$ NO $_{2}$
C $_{3}$ H $_{6}$ NO $_{2}$
C $_{3}$ H $_{7}$ NO $_{2}$

Validate

Solution:

Concept:

To find the molecular formula, first determine the empirical formula from the percentage composition, then use the vapour density to calculate the molecular mass and the value of

n

Explanation:

Convert the mass percentages of each element into moles by dividing by their relative atomic masses.

Carbon:

\frac{53.1}{12} = 4.425

mol

Hydrogen:

\frac{6.2}{1} = 6.2

mol

Nitrogen:

\frac{12.4}{14} = 0.886

mol

Oxygen:

\frac{28.3}{16} = 1.769

mol

Divide each result by the smallest number of moles, which is

0.886

\frac{4.425}{0.886} \approx 5

\frac{6.2}{0.886} \approx 7

\frac{0.886}{0.886} = 1

\frac{1.769}{0.886} \approx 2

So the empirical formula is

\mathrm{C_5H_7NO_2}

Calculate the empirical formula mass:

(12 \times 5) + (1 \times 7) + 14 + (16 \times 2) = 60 + 7 + 14 + 32 = 113

Molecular mass

= 2 \times

vapour density

= 2 \times 56.5 = 113

Using

(\text{empirical formula})_n = \text{molecular formula}

113n = 113

, therefore

n = 1

Answer:

The molecular formula is

\mathrm{C_5H_7NO_2}

, which corresponds to option A.

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