A compound is 40.0% c, 6.70% h, and 53.3% o by mass. assume that we have a 100.-g sample of this compound. the molecular formula mass of this compound is 180 amu. what are the subscripts in the actual molecular formula?

Question

Jester

Chemistry

18 July, 06:19

A compound is 40.0% c, 6.70% h, and 53.3% o by mass. assume that we have a 100.-g sample of this compound. the molecular formula mass of this compound is 180 amu. what are the subscripts in the actual molecular formula?

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Mercedes Nicholson · Answer 1

If you have 100 g of compound, then based on the percentages given, there are 40.0 g C, 6.70 g H, and 53.3 g O. Convert those to moles (g / AW). (AW = atomic weight from the periodic table).

C: 40.0 g / 12.0 = 3.33 moles C

H: 6.70 g / 1.01 = 6.63 moles H

O: 53.3 / 16.0 = 3.33 moles O

Dividing by the smallest (3.33), we get a C:H:O mole ratio of 1:2:1. The empirical formula is CH2O. That formula has a molar mass of (12.0 + 2 (1.0) + 16.0) = 30.0. How many times will that go into the actual molar mass of 150? 150/30 = 5, so multiply the emprical formula by 5.

5 x CH2O = C5H10O5, and that is the molecular formula.