Question

An organic compound on analysis gave the following % composition C- 54.24%, H-9.05% and the rest is oxygen. The vapour density of compound was 44. Find empirical formula and molecular formula of the compound. DON'T SPAM .

Answer 1

Let,

• The amount of Organic compound = 100 g

Given,

• Amount of Carbon = 54.24 g

• Amount of Hydrogen = 9.05 g

• Amount of Oxygen = 36.71 g

We know,

• Mass of Carbon = 12 g

• Mass of Hydrogen = 1 g

• Mass of Oxygen = 16 g

Here,

• The number of mole of Carbon = $\sf\dfrac{54.24}{12}$ =  4.52

• The number of mole of Hydrogen = $\sf\dfrac{9.05}{1}$ = 9.05

• The number of mole of Oxygen = $\sf\dfrac{36.71}{16}$ = 2.29

Taking the ratio,

= 4.52 : 9.05 : 2.29

≈ 2 : 4 : 1

Therefore,

The empirical formula of organic compound = C₂H₄O

Here,

Molecular weight = (2×12) + (4×1) + (1×16)

⇒ Molecular Weight = 24 + 4 + 16

⇒ Molecular Weight = 44 g

We know,

$\huge\boxed{\sf Molecular \:Weight=2\times Vapour\:Density}$

Here,

• Vapour Density = 44 g

Applying the values to the equation,

Molecular Weight = 2 × 44

⇒ Molecular Weight = 88 g

______________________________

► Empirical Formula = C₂H₄O ◄

► Molecular Weight = 88 g ◄

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Answer 2

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