Question

A substance on analysis gave the following percentage composition :Na = 43.4% C=11.3% O=45.3%. calculate the empirical formula Na = 23, C =12,O=16

Answer 1

Answer:

Explanation:

=> According to question, A substance on analysis gave the following percentage composition.

Na = 43.4%

C = 11.3%

O = 45.3%

That means, Mass of these compounds in 100g is,

Na =43.4 g, C = 11.3 g, O = 45.3 g

=> To convert mass into moles, lets divide it by their molar mass,

Na = 43.4/23 = 1.88

C = 11.3/12 = 0.94

O = 45.3 / 16.00 = 2.8

Ratio of the moles of the elements = Na : C : O = 1.88 : 0.94 : 2.8

=> Lets divide it by 0.94 (least number) to obtain whole number ratio.

C = 1.88/0.94 = 2

H = 0.94/0.94 = 1

O = 2.8/0.94 = 3

Thus, C : H : O = 2 : 1 : 3

Therefore,  the empirical formula is Na₂CO₃

Answer 2

Explanation:

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