Question

H. A substance, on analysis, gave
the following percent composition:
Na = 43.4 %, C = 11.3 % and
0 = 45.3 %. Calculate the empirical
formula. (At. mass Na = 23 u, C = 12
u, 0= 16 u).
solve step by step ​

Answer 1

Explanation :

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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