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Empirical Formula Question Help with Molecules

The questions says: Caffeine contains C,H,O and N. Combustion of 1.000 mg of Caffeine produces 1.813 mg CO2, 0.4649 mg H20 cand 0.2885 mg N2. Estimate the molar mass of caffeine which lies between 150 and 200 g/mol.

First of all what are the calculations because I'm lost. Also, wtf is Molar Mass and how do you even calculate the empirical formula when it has molecules.
Original post by Rtdsv
The questions says: Caffeine contains C,H,O and N. Combustion of 1.000 mg of Caffeine produces 1.813 mg CO2, 0.4649 mg H20 cand 0.2885 mg N2. Estimate the molar mass of caffeine which lies between 150 and 200 g/mol.

First of all what are the calculations because I'm lost. Also, wtf is Molar Mass and how do you even calculate the empirical formula when it has molecules.


Use the values given to find the actual mass of C,H,O and N in the compound sample.

Then divide each value by the relative atomic mass of each element (i.e. the experimental mass of carbon by 12.011 etc)

Then manipulate the values mathematically to get the simplest integral ratio.
Reply 2
1
Original post by charco
Use the values given to find the actual mass of C,H,O and N in the compound sample.

Then divide each value by the relative atomic mass of each element (i.e. the experimental mass of carbon by 12.011 etc)

Then manipulate the values mathematically to get the simplest integral ratio.


Thanks but im having issues. For example, I'd do:
C=1.813/12=0.151

But since there's O2, would I divide by 16 or 32?
This is the working out I've done, please correct me since I think I'm wrong
C=1.813/12=0.15
O=1.813/16=0.113
H=0.4639/1=0.4639
O=0.4639/16=0.29
N=0.2885/14=0.2

Then I'd add both O's up and get 0.4 and divide all those answers by 0.15 and I get 1, 1.333, 2.666, 3 respectively.

Then multiply by 3 and get C3, O6, H8, and N9.
But I know that's wrong so I must've done something incorrect. The forumla is actually C8H10N402
Original post by Rtdsv
1

Thanks but im having issues. For example, I'd do:
C=1.813/12=0.151

But since there's O2, would I divide by 16 or 32?
This is the working out I've done, please correct me since I think I'm wrong
C=1.813/12=0.15
O=1.813/16=0.113
H=0.4639/1=0.4639
O=0.4639/16=0.29
N=0.2885/14=0.2

Then I'd add both O's up and get 0.4 and divide all those answers by 0.15 and I get 1, 1.333, 2.666, 3 respectively.

Then multiply by 3 and get C3, O6, H8, and N9.
But I know that's wrong so I must've done something incorrect. The forumla is actually C8H10N402


You are interested in the relative proportion of the elements so you divide by the relative ATOMIC mass.
Reply 4
Original post by charco
You are interested in the relative proportion of the elements so you divide by the relative ATOMIC mass.


Yeah thanks a lot. I did try that but still the numbers do not quite add up. Surely I've done something else wrong?
You're burning it in oxygen; that's why burning 1 mg of caffeine leads to ~ 2 mg of product - there's extra oxygen in the products.

So as a start, to work out the amount of carbon in the sample you'd do:

1.813 * 12/(16*2 + 12) = 0.494 mg carbon

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