# A level Chemistry

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#1
15cm3 of a gaseous hydrocarbon requires 90cm3 of oxygen for complete combustion, both volumes measured at 15 degrees Celcius and 1 atm. what is the formula of the hydrocarbon?

can anyone explain how to answer this?
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7 months ago
#2
Is it a multiple choice?
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#3
(Original post by GabiAbi84)
Is it a multiple choice?
yes the answer is c4h8 but i dont know how or why that is
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7 months ago
#4
I’m not entirely sure if this is the way you’re supposed to work it out but this is how I did it:

Using 15:60 you can work out that the ratio for hydrocarbon xygen is 1:6

Then with a standard equation you know

CxHy + 6O2 -> xCO2 + 1/2yH2O

So balancing the oxygen gives

6(2) = x(2) + 1/2y

12= 2x + 1/2y

Then use the multiple choice answers to see which satisfies that.
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#5
(Original post by GabiAbi84)
I’m not entirely sure if this is the way you’re supposed to work it out but this is how I did it:

Using 15:60 you can work out that the ratio for hydrocarbon xygen is 1:6

Then with a standard equation you know

CxHy + 6O2 -> xCO2 + 1/2yH2O

So balancing the oxygen gives

6(2) = x(2) + 1/2y

12= 2x + 1/2y

Then use the multiple choice answers to see which satisfies that.
bless your soul !! may i ask why it os 1/2y h20? is this a standard ratio equation or did you work it out?
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7 months ago
#6
(Original post by summer044)
bless your soul !! may i ask why it os 1/2y h20? is this a standard ratio equation or did you work it out?
Because of the 2 in H2O

So you know all the carbons from the hydrocarbon get turned into the CO2 so x from the CxHy becomes the coefficient of the CO2

All the hydrogen gets turned into H2O so the y from the CxHy needs to take into account that there are 2 Hs in water so it’s 1/2y

Does that make sense??
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#7
(Original post by GabiAbi84)
Because of the 2 in H2O

So you know all the carbons from the hydrocarbon get turned into the CO2 so x from the CxHy becomes the coefficient of the CO2

All the hydrogen gets turned into H2O so the y from the CxHy needs to take into account that there are 2 Hs in water so it’s 1/2y
A 50cm3 sample of gaseous hydrocarbon required exactly 250cm3 of oxygen for complete combustion. a volume of 150cm3 carbon dioxide was produced, which of the following is the correct formula for the hydrocarbon?

All measurements of volumes were made under the same conditions. Which one of the following is the correct formula of the hydrocarbon?
A) C3H4
B)C3H8
C)C5H10
D)C5H12

my calculation:

50 CH + 250 O2 --> x CO2 + 1/2y H20

--> Divide by 50 to get a ratio of 1:5

CH + 5O2 --> x CO2 + 1/2y H2O

10 Oxygen = 2X + 1/2 y

x= 3 y= 8

and according to the mark scheme this correct!!! THANK YOU SO MUCH WORDS CANNOT SHOW YOU HOW APPRECIATIVE I AM <333
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7 months ago
#8
(Original post by summer044)
A 50cm3 sample of gaseous hydrocarbon required exactly 250cm3 of oxygen for complete combustion. a volume of 150cm3 carbon dioxide was produced, which of the following is the correct formula for the hydrocarbon?

All measurements of volumes were made under the same conditions. Which one of the following is the correct formula of the hydrocarbon?
A) C3H4
B)C3H8
C)C5H10
D)C5H12

my calculation:

50 CH + 250 O2 --> x CO2 + 1/2y H20

--> Divide by 50 to get a ratio of 1:5

CH + 5O2 --> x CO2 + 1/2y H2O

10 Oxygen = 2X + 1/2 y

x= 3 y= 8

and according to the mark scheme this correct!!! THANK YOU SO MUCH WORDS CANNOT SHOW YOU HOW APPRECIATIVE I AM <333
This one is slightly easier also as you’re given the amount of CO2 made which means you already know X

But yes the same principle applies.

Do you understand why it’s 1/2y now?
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#9
(Original post by GabiAbi84)
This one is slightly easier also as you’re given the amount of CO2 made which means you already know X

But yes the same principle applies.

Do you understand why it’s 1/2y now?
yes i do!! it makes perfect sense, but what do i do in this case where the amount of oxygen isnt given

Complete combustion of 50 cm3 of a hydrocarbon vapour gave 350 cm3 of carbon dioxide, both gas volumes being measured at the same temperature and pressure. the formula of the hydrocarbon could be

A C8H18
B C7H16
C C6H14
D C5H12

My calculation

50CxHy + O2 --> 350CO2 + 1/2y H20

Ratio is now 1:7

the coefficient of CO2 is = x

x= 7

so the answer must be B. according to the mark scheme this is also correct. Have I done it the right way?
0
7 months ago
#10
(Original post by summer044)
yes i do!! it makes perfect sense, but what do i do in this case where the amount of oxygen isnt given

Complete combustion of 50 cm3 of a hydrocarbon vapour gave 350 cm3 of carbon dioxide, both gas volumes being measured at the same temperature and pressure. the formula of the hydrocarbon could be

A C8H18
B C7H16
C C6H14
D C5H12

My calculation

50CxHy + O2 --> 350CO2 + 1/2y H20

Ratio is now 1:7

the coefficient of CO2 is = x

x= 7

so the answer must be B. according to the mark scheme this is also correct. Have I done it the right way?
Yes so knowing the ratio between the hydrocarbon and carbon dioxide means you can work out x just like you did.
Without the oxygen amount you can’t work out y but since the multiple choice only has one that has x=7 you don’t need it.
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#11
(Original post by GabiAbi84)
Yes so knowing the ratio between the hydrocarbon and carbon dioxide means you can work out x just like you did.
Without the oxygen amount you can’t work out y but since the multiple choice only has one that has x=7 you don’t need it.
thanks so much once again!!
0
7 months ago
#12
(Original post by summer044)
thanks so much once again!!
No worries, glad I could help 0
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