# molecular formula in chemistry

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Thread starter 8 months ago
#1
I need to do this question and its due tonight any help would be appreciated x
an unknown compound was analysed and the following information was determined:
1-when 0.350g was heated to 100 degrees it had a gaseous volume of 83.95cm3 (molar gas volume at 100 degrees is 30.7dm3mol-1
2 it has a percentage composition of 65.63% C 9.37% H 25% O
determine molecular formula
0
8 months ago
#2
(Original post by isabellexx)
I need to do this question and its due tonight any help would be appreciated x
an unknown compound was analysed and the following information was determined:
1-when 0.350g was heated to 100 degrees it had a gaseous volume of 83.95cm3 (molar gas volume at 100 degrees is 30.7dm3mol-1
2 it has a percentage composition of 65.63% C 9.37% H 25% O
determine molecular formula
I'd start with finding the Mr from the first bit of information
1
8 months ago
#3
(Original post by philogrobized)
I'd start with finding the Mr from the first bit of information
I agree
1. Find Mr- find moles of gas in the 0.35g sample, moles=volume/molar volume

2. Then find empirical formula i.e.simplest ratio of moles of C; H : O in the compound (convert the masses of each element in 100g of compound into no of moles by dividing by the atomic mass)

3. Combine 1 and 2 together to find the molecular formula
e.g the empirical formula may have been CH3 but if the Mr was 30, then the molecular fomula is not CH3 but double that i.e.C2H6

If you look at the attachment I have explained it more clearly.
0
Thread starter 8 months ago
#4
(Original post by Davies Chemistry)
I agree
1. Find Mr- find moles of gas in the 0.35g sample, moles=volume/molar volume

2. Then find empirical formula i.e.simplest ratio of moles of C; H : O in the compound (convert the masses of each element in 100g of compound into no of moles by dividing by the atomic mass)

3. Combine 1 and 2 together to find the molecular formula
e.g the empirical formula may have been CH3 but if the Mr was 30, then the molecular fomula is not CH3 but double that i.e.C2H6

If you look at the attachment I have explained it more clearly.
this is really helpful thank you
0
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