Question

What is the empirical formula of a compound with a percent composition of 18.66% N, 6.71% H, 32.00% C, and 42.63% O?

Answer 1

C- 32.00%

H- 6.71%

N- 18.66%

O- 42.63 %

Assume that you have 100 g of the compound, so you have 32.00 g C, 6.71 g H, 18.66 g N, and 42.63 % O.

moles C = 32.00 g(1 mole C/12.011 g C) = 2.664 mol C

moles H = 6.71 g(1 mole H/1.00794 g H) = 6.657 mol H

moles N = 18.66 g(1 mole N/14.0067 g N) = 1.332 mol N

moles O = 42.63 g(1 mole O/15.9994 g O) = 2.664 mol C

The smallest number of moles is 1.332 mol N, so divide all of the other numbers of moles to find the relative numbers of each element in the compound.

2.664/1.332 = 2.00

6.657/1.332 = 5.00

1.332/1.332 = 1.00

2.664/1.332 = 2.00

empirical formula = C2H5NO2 - this is in agreement with your statement of the answer.

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