Question #148083

A stimulant in coffee and tea is caffeine, a substance of molar mass 194 g/mol. When 0.376 g of caffeine was burned, 0.682 g of carbon dioxide, 0.174 g of water, and 0.110 g of nitrogen were formed. Determine the empirical and molecular formulas of caffeine, and write the equation for its combustion.

Expert's answer

Firstly it should be known that caffeine consists of carbon,hydrogen,nitrogen and oxygen.So our formular must consist of the four.

Then we have to find the mass and percentage by mass of the elements since we were only given their compounds.

So for carbon in CO2 we have;

If 44g of CO2 consist of 12g of carbon

Then 0.682g of CO2 will consist of =0.682×12/44=0.186

To find the percentage by mass we have

(0.186/0.376)×100=49.5%

For hydrogen we have;

If 18g of H2O consist of 2g of H2

Then 0.174g of H2O consist of;

0.174×2/18=0.019

To find the percentage by mass we have;

(0.019/0.376)×100=5%

For nitrogen since we were only given the element so we just find the percentage of mass straight-up .So we have;

(0.11/0.376)×100=29%

To find the percentage of mass of oxygen, we can just add the percentage by mass of the first 3 compounds then subtract it from 100%

So we have: 100-(49.5+5+29)=16.5%

So now to find the empirical formula

C H N O

49.5% 5% 29% 16.5%

49.5/16 5/1 29/14 16.5/16

4 5 2 1

4 5 2 1

So to explain what we did above we first wrote out the percentage by mass of the elecments then we will divide by their molar mass respect ively.After that we divided by the smallest mole which is 1.Then we still have our mole value intact.

So to get our empirical formula we apply the values to the elements

Our empirical formula is C4H5N2O

B)now to find out molecular formula we have;

(C4H5N2O)x=194

Then adding the atomic masses together and multiplying by x we have;

97x=194

x=2 so substituting our x to be 2 we have our molecular formula to be:

C8H10N4O2

C)Equation for combustion of caffeine is;

C8H10N4O2 + 19/2O2--------8CO2+5H2O+2N2

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