Question

calculate the percentage composition of various element in sodium carbonate (Na2CO3) [Na=23 C=12 O=16]​

Answer 1

The first step is to calculate the Molecular Weight (or the Formula Weight) of the chemical compound by adding the atomic weights of the atoms (elements) that constitute the compound.

Molecular weight of Na2CO3

= (2 x Atomic weight of Na) + Atomic weight of C + (3 x Atomic weight of O)

= (2 x 23) + 12 + (3 x 16)

= 46 + 12 + 48 = 106

Now, 106 grams of Sodium Carbonate [Na2CO3] contain 46 grams of Sodium [Na], 12 grams of Carbon [C], and 48 grams of Oxygen [O].

So,

Percentage of Sodium [Na] in Sodium Carbonate [Na2CO3] = 46/106 x 100 = 43.40%.

Percentage of Carbon [C] in Sodium Carbonate [Na2CO3] = 12/106 x 100 = 11.32%.

Percentage of Oxygen [O] in Sodium Carbonate [Na2CO3] = 48/106 x 100 = 45.28%.

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Answer 2

Solution:

Relative molecular mass of Na2CO3

= 23x2 + 12 + 16x3

= 46 + 12 + 48 =106 amu

Since 106g of Na2CO3 contains 46g of sodium,

∴ 100g of Na2CO3 contains 46*100/106 of sodium

= 4600/106 = 43.4 g of sodium

Similarly, 106 g Na2CO3 contains 12g of carbon.

∴ 100g of Na2Co3 contains 12*100/106

=1200/106 = 11.3 g of Carbon

Again, 106 g of Na2CO3 contains 48g of Oxygen.

∴ 100 g of Na2CO3 contains 48*100/106

= 4800/106 = 45.3 g of Oxygen

Answer:

In Na2CO3 :

Sodium = 43.4 %

Carbon = 11.3% and

Oxygen = 45.3%