What is the empirical formula of the compound?

[匿名]

2016-10-18 1:03 am

A compound is composed of only C,H, and O. The combustion of 0.519-g sample of the sample of the compound yields 1.24g of CO2 and 0.255g of H2O. What is the empirical formula of the compound?

I would like to know how to solve this step by step. I am lost on how to begin.
The answer is C3H3O but i do not know how to get there.

回答 (1)

Roger the Mole

2016-10-18 1:23 am

✔ 最佳答案

(1.24 g CO2) / (44.00964 g CO2/mol) × (1 mol C / 1 mol CO2) = 0.028176 mol C
(0.028176 mol C) × (12.01078 g C/mol) = 0.33842 g C

(0.255 g H2O) / (18.01532 g H2O/mol) × (2 mol H / 1 mol H2O) = 0.028309 mol H
(0.028309 mol H) × (1.007947 g H/mol) = 0.028534 g H

(0.519 g total) - (0.33842 g C) - (0.028534 g H) = 0.152046 g O
(0.152046 g O) / (15.99943 g O/mol) = 0.0095032 mol O

Divide by the smallest number of moles:
(0.028176 mol C) / 0.0095032 mol = 2.965
(0.028309 mol H) / 0.0095032 mol = 2.979
(0.0095032 mol O) / 0.0095032 mol = 1.000

Round to the nearest whole numbers to find the empirical formula:
C3H3O

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