The elemental composition of a compound is 54.2% C, 9.2% H and 36.6% O. If the molar mass of the compound is 132 g the molecular formula of the compound is: [Given : The relative atomic mass of C : H : O = 12 : 1 : 16] [2025]

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Q.
The elemental composition of a compound is 54.2% C, 9.2% H and 36.6% O.

If the molar mass of the compound is 132 g ${mol}^{- 1}$ the molecular formula of the compound is:

[Given : The relative atomic mass of C : H : O = 12 : 1 : 16] [2025]

1 $C_{6} H_{12} O_{3}$

2 $C_{4} H_{8} O_{2}$

3 $C_{4} H_{9} O_{3}$

4 $C_{6} H_{12} O_{6}$

Ans.
(1)

Element Mass % Number of moles of atom in 100 g compound = Mass% / molar mass of atom Relative number Simple ratio

C 54.2 54.2 / 12 = 4.517 4.517 / 2.2875 = 1.97 2

H 9.2 9.2 / 1 = 9.2 9.2 / 2.2875 = 4.02 4

O 36.6 36.6 / 16 = 2.2875 2.2875 / 2.2875 = 1 1

$Empirical formula = C_{2} H_{4} O$

$Empirical formula mass = (2×12+4×1+16) g=44 g$

$n = \frac{Molecular formula mass}{Empirical formula mass} = \frac{132}{44} = 3$

$Molecular formula = {(Empirical formula)}_{n} = {(C_{2} H_{4} O)}_{3} = C_{6} H_{12} O_{3}$

Solution

Q.
The elemental composition of a compound is 54.2% C, 9.2% H and 36.6% O.

If the molar mass of the compound is 132 g ${mol}^{- 1}$ the molecular formula of the compound is:

[Given : The relative atomic mass of C : H : O = 12 : 1 : 16] [2025]

1 $C_{6} H_{12} O_{3}$

2 $C_{4} H_{8} O_{2}$

3 $C_{4} H_{9} O_{3}$

4 $C_{6} H_{12} O_{6}$

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Element	Mass %	Number of moles of atom in 100 g compound = Mass% / molar mass of atom	Relative number	Simple ratio
C	54.2	54.2 / 12 = 4.517	4.517 / 2.2875 = 1.97	2
H	9.2	9.2 / 1 = 9.2	9.2 / 2.2875 = 4.02	4
O	36.6	36.6 / 16 = 2.2875	2.2875 / 2.2875 = 1	1

Solution

Q. The elemental composition of a compound is 54.2% C, 9.2% H and 36.6% O. If the molar mass of the compound is 132 g mol-1 the molecular formula of the compound is: [Given : The relative atomic mass of C : H : O = 12 : 1 : 16] [2025]

1 C6H12O3

2 C4H8O2

3 C4H9O3

4 C6H12O6

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Q.
The elemental composition of a compound is 54.2% C, 9.2% H and 36.6% O.

If the molar mass of the compound is 132 g ${mol}^{- 1}$ the molecular formula of the compound is:

[Given : The relative atomic mass of C : H : O = 12 : 1 : 16] [2025]

1 $C_{6} H_{12} O_{3}$

2 $C_{4} H_{8} O_{2}$

3 $C_{4} H_{9} O_{3}$

4 $C_{6} H_{12} O_{6}$