JEE MAIN - Chemistry (2022 - 28th June Evening Shift - No. 1)
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.
Explanation
Mole ratio of H, C and N
$$ = {{8.7} \over 1}:{{74} \over {12}}:{{17.\,3} \over {14}}$$
$$ = 8\,.7:6.167:1.23$$
$$ = {{8.7} \over {1.23}}:{{6.167} \over {1.23}}:{{1.23} \over {1.23}}$$
$$ = 7:5:1$$
$$\therefore$$ Emperical formula = $${{C_5}{H_7}N}$$
$$\therefore$$ Molecular formula $$ = {\left( {{C_5}{H_7}N} \right)_n}$$
Given molecular mass = 162
Molecular mass of $${\left( {{C_5}{H_7}N} \right)_n}$$
$$ = (5 \times 12 + 7 \times 1 + 14) \times n$$
$$ = (81) \times n$$
$$\therefore$$ $$81 \times n = 162$$
$$ \Rightarrow n = 2$$
$$\therefore$$ Molecular formula = C10H14N2
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