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Here we have 2 atoms of Na, 1 of C, and 3 of O, which is why we multiply the atomic weights by those numbers.
Na = 22.989770 * 2 = 45.97954 g/mol
C = 12.0107 * 1 = 12.0107 g/mol
O = 15.9994 * 3 = 47.9982 g/mol


If we sum these three products, we get a total of 105.98844 g/mol. This is the molecular weight of Na2CO3. Then, we divide:
45.97954 / 105.98844 = .4338 = 43.38%
12.0107 / 105.98844 = .1133 = 11.33%
47.9982 / 105.98844 = .4529 = 45.29%

Hope that helps.

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14y ago
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10y ago

Sodium: 43.382%

Carbon: 11.332%

Oxygen: 45.286%

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11y ago

Sodium: 32,370 %

Sulfur: 22,575 %

Oxygen: 45,055 %

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Q: What is the percent composition of Na2SO4?
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