Question

Nicotine, a component of tobacco, is composed of C, H, and N. A 3.150-mg sample of nicotine was combusted, producing 8.545 mg of co and 2.450 mg of H20. what is the empirical formula for nicotine?

Answer 1

I make it `C_5H_7N`...

Explanation:

And so we got `C_xH_yN_n`

We combust a `3.159*mg` mass of the stuff to get....

`(8.545*mg)/(44.01*g*mol^-1)=1.94xx10^-4*mol` `CO_2`...i.e. a mass with respect to carbon of `2.33*mg`....(`CO` would not result from the combustion.)

`(2.450*mg)/(18.01*g*mol^-1)=1.36xx10^-4*mol` `H_2O`...i.e. a mass with respect to hydrogen of `0.274*mg`, `0.272*mg`..

The balance of the mass was due to NITROGEN....`{3.159-2.33-0.272}*mg=0.557*mg-=3.98xx10^-5*mol`

We divide the molar quantities thru by the LEAST molar quantity to get an empirical formula of....

`C_((1.94xx10^-4*mol)/(3.98xx10^-5*mol))H_((2.74xx10^-4*mol)/(3.98xx10^-5*mol))N_((3.98xx10^-5*mol)/(3.98xx10^-5*mol))-=C_(4.87)H_(6.88)N~=C_5H_7N`

`"Whew, arithmetic.....!!"` I take it this is first year....?

Ordinarily combustion analysis gives you `%N`...here we had to interpolate the nitrogen mass by subtracting the calculated masses of carbon, and hydrogen, from the starting mass....