What is the empirical formula of a compound is 36.84% N and 63.16 % O?
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Divide each % value by respective atomic mass
N = 36.84/14.01 = 2.63
O = 63.16/16.00 = 3.95
Divide by smaller
N = 2.63/2.63 = 1
O = 3.95/2.63 = 1.5
Multiply by 2 to remove fraction
N = 2
O = 3
Empirical formula = N2O3
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