Question

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.

Answer 1

`C_8H_10N_4O_2` is the molecular formula for caffeine.

Explanation:

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!

Answer 2

`C_8H_10N_4O_2-="caffeine...."`

Explanation:

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....."`