# A level Chemistry

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

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?

can anyone explain how to answer this?

0

reply

(Original post by

Is it a multiple choice?

**GabiAbi84**)Is it a multiple choice?

0

reply

Report

#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 hydrocarbonxygen 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.

Using 15:60 you can work out that the ratio for hydrocarbonxygen 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.

1

reply

(Original post by

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 hydrocarbonxygen 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.

**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 hydrocarbonxygen 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.

0

reply

Report

#6

(Original post by

bless your soul !! may i ask why it os 1/2y h20? is this a standard ratio equation or did you work it out?

**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?

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??

0

reply

(Original post by

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

**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

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

0

reply

Report

#8

(Original post by

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

**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

But yes the same principle applies.

Do you understand why it’s 1/2y now?

0

reply

(Original post by

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?

**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?

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

reply

Report

#10

(Original post by

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?

**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?

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.

0

reply

(Original post by

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.

**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.

0

reply

Report

#12

(Original post by

thanks so much once again!!

**summer044**)thanks so much once again!!

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top