Calculating the molecular formula of caffeine? Caffeine, contains 49.48% Carbon, 5.15% Hydrogen, 28.87% nitrogen, 16.49% oxygen. It has a molar mass of 194.2 gmol -1.

The way I tackle these problems is using a process that goes like this:

Percent to mass; mass to mole; divide by small; multiply 'til whole

So first we want to take the percents and make them masses. We assume we have a 100g sample of caffeine, and so 49.48% Carbon is equal to 49.48 g of Carbon in the sample, 5.15% Hydrogen is equal to 5.15 g of Hydrogen in the sample, and so on.

Now we take those numbers and convert them to moles of the element we are dealing with, so:

#(49.48 g)/(12.01 (g/("mol")))=4.12 "mol"# of Carbon
#(5.15 g)/(1.01 (g/("mol"))) = 5.1 "mol"# of Hydrogen
#(28.82 g)/(14.01 (g/("mol"))) = 2.06 "mol"# of Nitrogen
#(16.49 g)/(16 (g/("mol"))) = 1.03 "mol"# of Oxygen

Now we divide all the moles by the smallest number of moles, in this case the 1.03 from Oxygen. So:

#4.12/1.03=4# for Carbon
#5.1/1.03=5# for Hydrogen
#2.06/1.03=2# for Nitrogen
#1.03/1.03=1# for Oxygen

Since all the numbers are already whole we can skip the last part and go to an empirical formula:

#C_4H_5N_2O#

Now we have to check the molar mass of the empirical formula to make sure it matches up with the molar mass given:

#(12*4)+(1*5)+(14*2)+(16*1)=97# which is not what we were given, so we take what we were given and divide it by this number:

#194.2/97~~2#, so we have to multiply the numbers in the empirical formula by 2 to get our molecular formula:

#C_8H_10N_4O_2#

Hope this helped!

For a start we establish the #"empirical formula"# of caffeine...

And to this end we assume #100*g# of compound and interrogate its atomic composition by dividing the mass of the individual elements by the atomic mass of each element:

#"Moles of carbon"=(49.48*g)/(12.011*g*mol^-1)=4.12*mol*C#

#"Moles of hydrogen"=(5.15*g)/(1.00794*g*mol^-1)=5.11*mol*H#

#"Moles of nitrogen"=(28.87*g)/(14.01*g*mol^-1)=2.06*mol*H#

#"Moles of oxygen"=(16.49*g)/(15.999*g*mol^-1)=1.03*mol*H#

We normalize the result by dividing thru by the LOWEST molar quantity, that of #"oxygen..."#

#C_((4.12*mol)/(1.03*mol))H_((5.11*mol)/(1.03*mol))N_((2.06*mol)/(1.03*mol))O_((1.03*mol)/(1.03*mol))-=C_4H_5N_2O-="empirical formula"#

And we know that the #"molecular formula"# is a whole number multiple of the #"empirical formula"#. And we were quoted the molecular mass of the beast..

#194.2*g*mol^-1={4xx12.011+5xx1.00794+2xx14.01+15.999}_n*g*mol^-1#...and thus we solve for #n#...

#n-=(194.2*g*mol^-1)/(97.1*g*mol^-1)-=2#

And thus we got....#C_8H_10N_4O_2-="molecular formula"#.

This approach is standard for the calculation of (i) the #"empirical formula"#, the simplest whole number ratio defining constituent atoms in a species, and (ii), the #"molecular formula"#. The former can be assessed by #"combustion analysis"#, and then we need an estimate of #"molecular mass....."#