An organic compound has 42.1% carbon, 6.4% hydrogen and remainder is oxygen. If its molecular weight is 342, then its molecular formula is:

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Q.
An organic compound has 42.1% carbon, 6.4% hydrogen and remainder is oxygen. If its molecular weight is 342, then its molecular formula is: [2024]

1 $C_{12} H_{22} O_{11}$

2 $C_{11} H_{18} O_{12}$

3 $C_{12} H_{20} O_{12}$

4 $C_{14} H_{20} O_{10}$

Ans.
(1)

Element Mass percent Number of atoms = mass percent/atomic mass Divide by smallest no. Simple ratio

C 42.1%
$\frac{42.1}{12} = 3.5$
$\frac{3.5}{3.22} = 1.09$ $1.09 \times 11 = 12$

H 6.4%
$\frac{6.4}{1} = 6.4$
$\frac{6.4}{3.22} = 1.99$ $1.99 \times 11 = 22$

O 100 - (42.1 + 6.4) = 51.5
$\frac{51.5}{16} = 3.22$
$\frac{3.22}{3.22} = 1$ $1 \times 11 = 11$

Empirical formula = $C_{12} H_{22} O_{11}$

Empirical formula mass = $(12 \times 12 + 22 \times 1 + 11 \times 16) u = 342 u$

As empirical formula mass is same as molecular mass, molecular formula is also $C_{12} H_{22} O_{11}$ .

Solution

Q.
An organic compound has 42.1% carbon, 6.4% hydrogen and remainder is oxygen. If its molecular weight is 342, then its molecular formula is: [2024]

1 $C_{12} H_{22} O_{11}$

2 $C_{11} H_{18} O_{12}$

3 $C_{12} H_{20} O_{12}$

4 $C_{14} H_{20} O_{10}$

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Element	Mass percent	Number of atoms = mass percent/atomic mass	Divide by smallest no.	Simple ratio
C	42.1%	$\frac{42.1}{12} = 3.5$	$\frac{3.5}{3.22} = 1.09$	$1.09 \times 11 = 12$
H	6.4%	$\frac{6.4}{1} = 6.4$	$\frac{6.4}{3.22} = 1.99$	$1.99 \times 11 = 22$
O	100 - (42.1 + 6.4) = 51.5	$\frac{51.5}{16} = 3.22$	$\frac{3.22}{3.22} = 1$	$1 \times 11 = 11$

Solution

Q. An organic compound has 42.1% carbon, 6.4% hydrogen and remainder is oxygen. If its molecular weight is 342, then its molecular formula is: [2024]

1 C12H22O11

2 C11H18O12

3 C12H20O12

4 C14H20O10

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Q.
An organic compound has 42.1% carbon, 6.4% hydrogen and remainder is oxygen. If its molecular weight is 342, then its molecular formula is: [2024]

1 $C_{12} H_{22} O_{11}$

2 $C_{11} H_{18} O_{12}$

3 $C_{12} H_{20} O_{12}$

4 $C_{14} H_{20} O_{10}$