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Firstly divide percentages by molar mass of that element

Na

43.4/23=1.9

C

11.3/12.01=0.94

O

45.3/16=2.83

Then divide the result of the first step by the smallest answer

Na

1.9/.94=2

C

.94/.94=1

O

2.83/.94=3

Empirical Formula is Na2CO3

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Q: Can you Calculate the empirical formula of the compound made from 43.4 percent Na 11.3 percent C and 45.3 percent O by mass enter the empirical formula?
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