Question

a compound contains 13.13% of hydrogen 52.14 % of carbon and 34.73% of oxygen its molar mass is 46.06 what are the empirical and molecular formula​

Answer 1

Answer:

H=13.13%

O= 34.73%

No. of moles of C= 52.14/12= 4.345

No.of moles of H= 13.13/1= 13.13

No.of moles of O= 34.73/16= 2.170

Therefore,the ratio of the atoms of C,H and O= 4.345 : 13.13 : 2.170 = 2 : 6 : 1

So,the empirical formula is C2H6O

Let the Molecular formula is (C2H6O)n

Then, molar mass = (2×12+1×6+16)n

= 46n

As the question, 46n = 46.07

or,n= 1

So,the molecular formula is (C2H6O)1= C2H6O i.e.the compound is ethanol or dimethyl ether

Answer 2

Concept:

The number of moles of the element can be calculated by dividing the given percentage by the molar mass of the element.

Given:

Percentage of H = 13.13%

Percentage of C = 52.14%

Percentage of O =  34.73%

The molar mass of the compound = 46.06

Find:

The empirical and molecular formula of the compound.

Solution:

The number of moles of  H = 13.13/1 = 13.13

The number of moles of C = 52.14/12 = 4.345

The number of moles of O = 34.73/16 = 2.170

The ratio of the atoms of elements C, H, and O are

4.345 : 13.13 : 2.170 = 2 : 6 : 1

So, the empirical formula can be written as C₂H₆O.

The Molecular formula can be written as (C₂H₆O)ₙ

Molar mass of the compound = [(2×12) + (1×6) +(16×1)]n = 46n

Given, 46n = 46.06

n= 1

So, (C₂H₆O)1 = C₂H₆O

Hence, the empirical formula is C₂H₆O and the molecular formula is C₂H₆O  which is ethanol or dimethyl ether.

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