What is the empirical formula of the compound?

[匿名]

2016-10-18 1:06 am

Combustion of a 0.9835-g sample of a compound containing only carbon, hydrogen, and oxygen produced 1.900g of CO2 and 1.236g 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 C4H11O2 ....but i do not know how to get there.

回答 (1)

Roger the Mole

2016-10-18 1:41 am

✔ 最佳答案

(1.900 g CO2) / (44.00964 g CO2/mol) × (1 mol C / 1 mol CO2) = 0.0431724 mol C
(0.0431724 mol C) × (12.01078 g C/mol) = 0.518534 g C

(1.236 g H2O) / (18.01532 g H2O/mol) × (2 mol H / 1 mol H2O) = 0.137217 mol H
(0.137217 mol H) × (1.007947 g H/mol) = 0.138307 g H

(0.9835 g total) - (0.518534 g C) - (0.138307 g H) = 0.326659 g O
(0.326659 g O) / (15.99943 g O/mol) = 0.0204169 mol O

Divide by the smallest number of moles:
(0.0431724 mol C) / 0.0204169 mol = 2.115
(0.137217 mol H) / 0.0204169 mol = 6.721
(0.0204169 mol O) / 0.0204169 mol = 1.000

[There is no way this can be turned into C4H11O2. Are you sure you copied all the numbers correctly? In particular I doubt "1.236g of H2O".]

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