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Help empirical formula question

How do I work this out? Thanks
You can work out the moles of CO2 and H2O from the information given using mol=mass/Mr.

Then you can see the ratio of how much Carbon and Hydrogen was burned which is x and y
Reply 2
I got the ratio of CO2 to H20 to be 2:1.5 after that I'm stuck, what am I supposed to do with the mass of he hydrocarbon which we are given?

Original post by Mingzhe Feng
You can work out the moles of CO2 and H2O from the information given using mol=mass/Mr.

Then you can see the ratio of how much Carbon and Hydrogen was burned which is x and y
Reply 3
Original post by sarah99630
How do I work this out? Thanks


Hey I did that question yesterday!
Reply 4
Original post by Reshyna
Hey I did that question yesterday!


Really! Can you please send your working out?(or explain, whichever's easier for you) thanks!
Original post by sarah99630
I got the ratio of CO2 to H20 to be 2:1.5 after that I'm stuck, what am I supposed to do with the mass of he hydrocarbon which we are given?


Because 1 Carbon is used for CO2 and 2 Hydrogen for H2O, you can then look back to the compound to figure out what ratios x and y are in.
Reply 6
Original post by sarah99630
Really! Can you please send your working out?(or explain, whichever's easier for you) thanks!


oh hey. Just saw the notification but I see that you've already got a help.

Original post by Mingzhe Feng
Because 1 Carbon is used for CO2 and 2 Hydrogen for H2O, you can then look back to the compound to figure out what ratios x and y are in.


This was exactly what I did. I got the ratios 2:1.5 and I looked for the options that matched the ratio.
Original post by Reshyna
oh hey. Just saw the notification but I see that you've already got a help.



This was exactly what I did. I got the ratios 2:1.5 and I looked for the options that matched the ratio.


Yes but you have to bear in mind that there's 2Hs for a formation of a H2O, so 1.5 would need to be doubled. Then after you get the answer you can double check by using dividing the compounds Mr from it's mass. I'm guessing you did something along those lines?
Reply 8
Original post by Mingzhe Feng
Yes but you have to bear in mind that there's 2Hs for a formation of a H2O, so 1.5 would need to be doubled. Then after you get the answer you can double check by using dividing the compounds Mr from it's mass. I'm guessing you did something along those lines?


ya then I got 4:3 and balanced the equation. I sort of did this by myself. There was no one to check my working but my final answer matched.
Original post by Reshyna
ya then I got 4:3 and balanced the equation. I sort of did this by myself. There was no one to check my working but my final answer matched.


You get 2:3 no?
Original post by Mingzhe Feng
You get 2:3 no?


My answer was C. Yeah I got the ratio.
Original post by Reshyna
My answer was C. Yeah I got the ratio.


Ah okay, nice
thank you!
Original post by Reshyna
oh hey. Just saw the notification but I see that you've already got a help.



This was exactly what I did. I got the ratios 2:1.5 and I looked for the options that matched the ratio.
Original post by Mingzhe Feng
Because 1 Carbon is used for CO2 and 2 Hydrogen for H2O, you can then look back to the compound to figure out what ratios x and y are in.

I finallyyyy get it!!thank you! I appreciate your help:smile:

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