Chemistry question: Combustion
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# Chemistry question: Combustion

[From: ] [author: ] [Date: 11-09-25] [Hit: ]
50g of CO2 and 58.70g of H20.From this you can find the Mass of Carbon and Hydrogen.1) 95.- mass/molar mass x 95.- 12 / 44 x 95.......
Could someone help me go through this? I'm really confused...you don't have to give me the answer, I just want to know where to start...its so weird.
A certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 50.00 gram sample of the alcohol produced 95.50 grams of CO2 and 58.70 grams of H2O. What is the empirical formula of the alcohol?

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There's 50.00grams of Alcohol which produce ---> 95.50g of CO2 and 58.70g of H20.

From this you can find the Mass of Carbon and Hydrogen.

1) 95.50g of CO2
- mass/molar mass x 95.50g
- 12 / 44 x 95.50g
- 26.045g of Carbon.

2) 58.70g of H20
- mass/molar mass x 58.70g
- 2/18 x 58.70
- 6.522g of Hydrogen

3) Which means in the 50g of alcohol there are 26.045 g of carbon and 6.522 g of hydrogen
- from this statement you can find the mass of oxygen
- 50 - 26.045 - 6.522 = 17.4327g

4) then you can do the empirical formula calculation with these mass of each elements.
the answers will be C2H6O. Good luck! It's Ethanol =)

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(95.50 g CO2) / ( 44.0096 g CO2/mol) x (1/1) x (12.0108 g C/mol) = 26.06 g C
(58.70 g H2O) / (18.0153 g H2O/mol) x (2/1) x (1.0079 g H/mol) = 6.568 g H
(50.00 g) - (26.06 g C) - (6.568 g H) = 17.37 g O

(26.06 g C) / (12.0108 g C/mol) = 2.16971 mol C
(6.568 g H) / (1.0079 g H/mol) = 6.51652 mol H
(17.37 g O) / (15.9994 g O/mol = 1.08567 mol O
Divide by the smallest number of moles:
(2.16971 mol C) / 1.08567 = 1.9985
(6.51652 mol H) / 1.08567 = 6.002
(1.08567 mol O) / 1.08567 = 1.000
Round to the nearest whole number to find the empirical formula:
C2H6O
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