Question

An organic acid contains C=70.59% and H=5.88%.Its vapour density was found 68.what is the molecular formulae of the compound.

Answer 1

It's formula is C8H8O2.

Answer 2

Heya !
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◼Firstly , let's know what is molecular formula and empirical formula ?

→Empirical Formula - It gives a simple whole number ratio of the atoms and elements present in one molecule of compound .

→Molecular Formula - It gives an exact estimate of the atoms and elements present in one molecule of compound .

=> For example , the empirical formula of Benzene is CH and it's molecular formula is C6H6 .

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=> Let's move to the question.

Step. 1 ◼ Write down the % composition of Carbon ( C ) and Hydrogen ( H) as given . Here we see that the percentage doesn't add up to 100. So the remaining percentage is of Oxygen.

•°• % of Oxygen = 100 - (70.59 + 5.88 )

=> % Of Oxygen = 23.53%...............[ 1 ]

Step 2 ◼ Now , we shall calculate the number of moles (n) thus, divide the % composition by the atomic mass. We have -

◾For Carbon => 70.59 / 12 = 5.88 ( 2 )

◾For Hydrogen => 5.88/1 = 5.88 ( 3 )

◾For Oxygen => 23.53 / 16 = 1.47 ( 4 )

Step 3 ◼ Divide equations (2) (3) and (4) By the smallest value among the three.
The smallest value is 1.47 → So divide all three values by 1.47

▪For Carbon => 5.88/ 1.47 = 4

▪For Hydrogen => 5.88 / 1.47 = 4

▪For Oxygen => 1.47/1.47 = 1

°•° Empirical Formula = C4H4O ✔

Step 4 ◼ Now we need to calculate the molecular formula.

=> We have the relation -

→ n × empirical formula = molecular formula

[ n = molecular mass / empirical formula mass ]

→Molecular Formula = Vapour Density × 2

→Molecular Formula = 68 × 2 = > 136 ✔

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★ Empirical Formula mass = Mass of C4H4O

=> 12 × 4 + 1 × 4 + 16 = 48 + 4 + 16

=> 68 ✔

•°• n = 136 / 68 = 2 ✔

=> Molecular formula = 2 × ( C4H4O)

◼Molecular Formula is C8H8O ◼

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