Isaac C. answered 05/30/17
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I mole of Na2CO3 contains 2 moles Na, 3 moles O, and one mole of C. In grams this is
2 *23 grams of Na, 3 * 16 grams of O and 12 grams of C
46 + 48 + 12 = 106 grams total.
percent Na = 46/106 = 43.4%
percent O = 48/106 = 45.3%
Percent C = 12/106 = 11.3%
Total 100.0%