A substance on analysis, gave the following percentage composition, Na = 43.4%, C = 11.3%, 0 = 43.3% calculate its empirical formula [Na = 23, C = 12, O = 16].

Answers (1)
S Sumit

Na = 43.4%
C = 11.3
O = 43.3%
Number of moles of respective elements is- 
Na = 43.4/23 = 1.88
C = 11.3/12 = 0.94
O = 43.3/16 = 2.71
Simple ratio of moles is 
Na = 1.88/0.94 = 2
C = 0.94/0.94 = 1
O = 2.71/0.94 = 2.87 ~ 3
Empirical formula = Na2CO3

 

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