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<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>
<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>

Relative molecular mass

I was doing this question in the revision book for OCR B and it is as follows:
An experiment was carried out to find the relative molecular mass of a sample of LPG. 1dm3 of the LPG had a mass of 2.25g at RT and pressure. Calculate a value for the relative molecular mass of this sample. Assume that 1 mole of gas occupied 24dm3 at RT and pressure.

The answer states that it is 2.25 x 24 to get to 54 but I'm not sure how they got to that. If someone could explain I'd appreciate it ty.
Original post by misteltain
I was doing this question in the revision book for OCR B and it is as follows:
An experiment was carried out to find the relative molecular mass of a sample of LPG. 1dm3 of the LPG had a mass of 2.25g at RT and pressure. Calculate a value for the relative molecular mass of this sample. Assume that 1 mole of gas occupied 24dm3 at RT and pressure.

The answer states that it is 2.25 x 24 to get to 54 but I'm not sure how they got to that. If someone could explain I'd appreciate it ty.


I would divide moles by 24 (understand why? 24 dm3 = 1 moles, so 1 dm3 = 1/24 moles)
do mass (g) / n (moles) = rmm (theres a formula triangle for this you should learn which comes in useful) Not sure why the mark scheme did that...
(edited 8 years ago)
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misteltain
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Original post by jacksonmeg
I would divide moles by 24 (understand why? 24 dm3 = 1 moles, so 1 dm3 = 1/24 moles)
do mass (g) / n (moles) = rmm (theres a formula triangle for this you should learn which comes in useful) Not sure why the mark scheme did that...


So if 1/24 =0.04 moles, using m/n = Mr would be 2.25/0.04 to get 56.25? That's how I was planning to use the triangle but I'm not sure if this even makes sense lol anyone got anything else?

Struggling with simple mathematics :facepalm:
Original post by misteltain
So if 1/24 =0.04 moles, using m/n = Mr would be 2.25/0.04 to get 56.25? That's how I was planning to use the triangle but I'm not sure if this even makes sense lol anyone got anything else?

Struggling with simple mathematics :facepalm:


Yes that's what I did to work it out
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misteltain
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Original post by jacksonmeg
Yes that's what I did to work it out


y u do dis mark scheme. Someone put us out of our misery
Original post by misteltain
y u do dis mark scheme. Someone put us out of our misery


You won't be marked down because my method is correct, what level is this?
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misteltain
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Original post by jacksonmeg
You won't be marked down because my method is correct, what level is this?


AS. It's not a past paper or anything so I'll overlook it I guess. Thanks for your help!
Original post by misteltain
I was doing this question in the revision book for OCR B and it is as follows:
An experiment was carried out to find the relative molecular mass of a sample of LPG. 1dm3 of the LPG had a mass of 2.25g at RT and pressure. Calculate a value for the relative molecular mass of this sample. Assume that 1 mole of gas occupied 24dm3 at RT and pressure.

The answer states that it is 2.25 x 24 to get to 54 but I'm not sure how they got to that. If someone could explain I'd appreciate it ty






I agree with jacksonmeg.
what i did was see how many moles are there in 1 dm3:
24dm3= 1mole
1dm3 = x
then cross multiply to find x = 1/24 (don't change it to decimals)
next, use the formula relative molecular mass = mass/number of moles
which is 2.25 / 1/24 = (exactly) 54.
hope i could help! :smile:
(edited 8 years ago)
Reply 8
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misteltain
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[QUOTE="O/LSHAMINI;55556597"]
Original post by misteltain
I was doing this question in the revision book for OCR B and it is as follows:
An experiment was carried out to find the relative molecular mass of a sample of LPG. 1dm3 of the LPG had a mass of 2.25g at RT and pressure. Calculate a value for the relative molecular mass of this sample. Assume that 1 mole of gas occupied 24dm3 at RT and pressure.

The answer states that it is 2.25 x 24 to get to 54 but I'm not sure how they got to that. If someone could explain I'd appreciate it ty






I agree with jacksonmeg.
what i did was see how many moles are there in 1 dm3:
24dm3= 1mole
1dm3 = x
then cross multiply to find x = 1/24 (don't change it to decimals)
next, use the formula relative molecular mass = mass/number of moles
which is 2.25 / 1/24 = (exactly) 54.
hope i could help! :smile:


Thank you very much! Funny how changing it to decimals and then dividing makes such a difference -_- appreciate the help x
[QUOTE="misteltain;55558107"]
Original post by O/LSHAMINI


Thank you very much! Funny how changing it to decimals and then dividing makes such a difference -_- appreciate the help x

No problem! Glad to be of help! 😀

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