Asked by Jeff
Empirical formula 12.67 aluminum, 19.73 nitrogen, 67.60 oxygen
Answers
Answered by
DrBob222
Do you mean these are percents?
Take a 100 grams sample which will give you
12.67 g Al
19.73 g N
67.6 g O
Convert each to mol (mol = g/atomic mass), then find the ratio of the elements to each other with the smallest number being 1.00. The easy way to do that is to divide the smallest number by itself. Round to whole numbers but don't round too much; i.e., 3.5 rounds to 7 (2*3.5).
Take a 100 grams sample which will give you
12.67 g Al
19.73 g N
67.6 g O
Convert each to mol (mol = g/atomic mass), then find the ratio of the elements to each other with the smallest number being 1.00. The easy way to do that is to divide the smallest number by itself. Round to whole numbers but don't round too much; i.e., 3.5 rounds to 7 (2*3.5).
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