Question

The percent composition by mass of an unknown compound witha molecular mass of 180.156 amu is 40.002% C, 6.7135% H, and 53.284% O. Determine the compound's empirical and molecular formulas. 1st attempt (1 point) See Periodic Table Part 1 Empirical formula: х х. Не

Answer 1

Assuming that you have 100 g of compound, the moles of elements are calculated:

n C = g / MM = 40/12 = 3.33 mol

n H = 6.71 / 1 = 6.71 mol

n O = 53.28 / 16 = 3.33 mol

They are divided by the smallest number:

C: 3.33 / 3.33 = 1

H: 6.71 / 3.33 = 2

Or: 3.33 / 3.33 = 1

It has the empirical formula CH2O, with a molar weight of 30 amu. The F factor is calculated:

F = MM molecular formula / MM empirical formula = 180/30 = 6

The empirical formula is multiplied by the F factor and the molecular one is obtained:

C6H12O6

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