Question

1 g of a compound contains 0.433 g of sodium, 0.453 g oxygen by mass, and the rest is carbon. What is the molecular formula of the compound if its molecular weight is 106?​

Answer 1

Explanation:

Correct option is

A

Na

2

CO

3

Dividing their percentages by their atomic weights.

Sodium=

23

43.4

=1.88

Carbon=

12

11.3

=0.941

Oxygen=

16

45.3

=2.83

Dividing all these numbers with simple numbers

Sodium=

0.94

1.88

=2

Carbon=

0.94

0.94

=1

Oxygen=

0.94

2.83

=3

Their ratio 2:1:3

Empirical formula = Na

2

CO

3

Answer 2

Given: Mass of the compound = 1g

Given mass of sodium = 0.433 grams

Given mass of oxygen = 0.453 grams

Molecular weight of compound = 106

To Find: Molecular formula of the compound

Solution:

• Firstly, calculate the mass of the carbon = Mass of the compound - (mass of sodium + mass of oxygen)

         = 1 - (0.433 + 0.453)

         = 0.114 grams.

• Now, divide the percentages of given mass of sodium, carbon and oxygen with their molar mass respectively.
• The molar mass of sodium, carbon and oxygen are 23, 12 and 16 respectively.
• By dividing the percentage of given mass of sodium with molar mass of sodium =
• $\frac{0.433 \times 100}{23} = 1.88$
• By dividing the percentage of given mass of carbon with molar mass of carbon =
• $\frac{0.114 \times 100}{12} = 0.94$
• By dividing the percentage of given mass of oxygen with molar mass of oxygen =
• $\frac{0.453 \times 100}{16} = 2.83$
• To find the ratio of sodium, carbon and oxygen, divide all these numbers with a common number like 0.94.
• By dividing sodium with 0.94 we get,
• $\frac{1.88}{0.94} = 2$
• By dividing sodium with 0.94 we get,
• $\frac{0.94}{0.94} = 1$
• By dividing sodium with 0.94 we get,
• $\frac{2.83}{0.94} = 3$
• Hence, the final ratio of sodium, carbon and oxygen = 3:1:2 and thus we get the molecular formula of the compound as $Na_{2} CO_{3}$.
• To find that $Na_{2} CO_{3}$ is the correct formula calculate its molecular weight and equalize it with the given molecular weight of the compound i.e 106.
• $Na_{2} CO_{3}$ = (23×2) + 12 + (16×3) = 106

Hence, the molecular formula of the compound is $Na_{2} CO_{3}$.