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A compound is found to contain 63.65% nitrogen and 36.35% oxygen by mass. What is the empirical formula for this compound

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You will change the percent to gram

N = 63.65 % => 63.65 g

O = 36.35 % => 36.35 g

The formula for this question is mole = mass / Molar mass

N = 63.65 g / 14.01 g/mol = 4.54 mol

O = 36.35 g / 16.00 g/mol = 2.27 mol

Take the smallest number in 2 element and divide them.

N = 4.54 / 2.27 = 2

O = 2.27 / 2.27 = 1

The empirical formula is N2O

Hope this help!

User Saladi
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