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Moles and gas volumes HELP!!!! watch

1. (Original post by brookleahy)
So would x=5 to make C5H12 Pentane? I worked this out with my own knowledge rather than the method please could advise me which numbers to use to get 5?
You'd be correct, but if you just used the general formula CnH2n+2, you can't rely on that. It works in this case, but other hydrocarbons exist (for example alkenes) with other formulae that don't follow the alkane pattern.

Method:
50cm^3 = 0.05dm^3. Moles of CO2 = 0.05/24 = 0.00208 moles.
From earlier method to find y, there are 0.000417 moles of hydrocarbon
0.00208 moles of carbon dioxide contains 0.00208 moles of carbon atoms.
x = 0.00208/0.000417 = 5
The hydrocarbon is C5H12 or pentane.
2. (Original post by TheMindGarage)
You'd be correct, but if you just used the general formula CnH2n+2, you can't rely on that. It works in this case, but other hydrocarbons exist (for example alkenes) with other formulae that don't follow the alkane pattern.

Method:
50cm^3 = 0.05dm^3. Moles of CO2 = 0.05/24 = 0.00208 moles.
From earlier method to find y, there are 0.000417 moles of hydrocarbon
0.00208 moles of carbon dioxide contains 0.00208 moles of carbon atoms.
x = 0.00208/0.000417 = 5
The hydrocarbon is C5H12 or pentane.
I understand now! You've been such a great help- thankyou
3. While mathematically, the final answer would be the same, I would suggest caution when using moles of gas = volume/24.

That relation applies only under room conditions (25 degrees and 1 atm), which was not specified in the question.

It is conceptually more sound to use the ratios of the VOLUMES of the hydrocarbon to carbon dioxide formed. Ratio of the volume of gases = ratio of the moles of gases

This relationship is valid long as the conditions (temp, pressure) are the same during measurement and unlike moles = volume/24, is not only limited to room conditions (25 degrees and 1 atm)

Therefore,
hydrocarbon : CO2
10 cm^3 : 50 cm^3
1 mole : 5 moles

Therefore 1 mole of hydrocarbon consists of 5 moles of carbon

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