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?

1 Answer
May 31, 2018

Answer:

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....