Question

A substance on analysis gave
the following percent composition
Na 43.4% (= 11.3% and a 453%
Calculate empirical formula
Ha (23W C = 120 0 = 16U​

Answer 1

Answer:

Na = 43.4%

C = 11.3

O = 43.3%

relative number of moles

of Na = 43.4/23 = 1.88

of C = 11.3/12 = 0.94

of O = 43.3/16 = 2.71

simple ratio of moles  

of Na = 1.88/0.94 = 2

of C = 0.94/0.94 = 1

of O = 2.71/0.94 = 2.87 ~ 3

empirical formula = Na2CO3

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Explanation: