Question

1. The percentage composition of various elements in
a compound is as follows

C =62%
H=10.4%
O=27.6%

what is the empirical formula?

​

Answer 1

Answer:

C3H6O

Explanation:

relative no. of atoms = percentage of element / it's atomic mass

C= 62/12 = 5.16

H= 10.4/1= 10.4

O = 27.6 / 16 = 1.72

now, divide all the three resultant values by the least value amongst the three.

C=5.16/1.72 = 3

H = 10.4/1.72 = 6

O = 1.72/1.72 = 1

compound - C3H6O

Note: you must round of the numbers to nearest whole number ratio in the last step.

Answer 2

Empirical formula of the compound is C₃H₆O

GIVEN

Percentage composition of Carbon = 62

Percentage composition of Hydrogen = 10.4

Percentage composition of Oxygen = 27.6

TO FIND

The empirical formula of the compound.

SOLUTION

We can simply solve the above problem as follows

Let us consider the total weight of the compound = 100 grams.

So,

Weight of carbon = 62% of 100 = 62grams.

Weight of Hydrogen = 10.4 % of 100 = 10.4 grams.

Weight of Oxygen = 27.6% of 100 = 27.6 grams.

Now,

we will calculate the mole equivalent of Carbon, Hydrogen, and Oxygen.

We know that,

Moles = Given weight/molar weight

Moles of carbon =Given Weight of carbon /molar mass of carbon.

= 62/12

= 5.16 moles.

Similarly,

Moles of Hydrogen = 10.4/1 = 10.4 moles

And,

Moles of Oxygen = 27.6/16 = 1.725 moles

The ratio of moles of Carbon : Hydrogen: Oxygen

= 5.16 : 10.4: 1.725

= 2.99: 6.03: 1

Rounding off to nearest whole number -

3 : 6 : 1

Empirical formula of the compound is C₃H₆O

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