Question

30. A compound on analysis gave the following % composition:
Na=43.4%, C=11.4% and O=45.3%.
Determine the empirical and molecular formulae. ( molecular mass of the compound = 106u)​

Answer 1

Answer: Na2CO3.

Explanation:

Na = 43.4%

C = 11.4%

O = 45.3%

relative number of moles

of Na = 43.4/23 = 1.88

of C = 11.4/12 = 0.95

of O = 45.3/16 = 2.83

simple ratio of moles  

of Na = 1.88/0.95 = 2

of C = 0.95/0.95 = 1

of O = 2.83/0.95 = 2.97 ~ 3

empirical formula = Na2CO3.

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