Question

This a tough one folks here's the entire prompt:
Nicotine, the addictive ingredient in tobacco, contains the elements carbon, hydrogen, and nitrogen. When a 1.567g sample of nicotine is burned completely in oxygen, it produces 4.251g CO2 and 1.216g H20. What is the empirical formula for nicotine and the percent by mass of each element in the sample?

Answer 1

Check back tomorrow. I'll take care of this tomorrow morning. I promise.

Answer 2

First things first. Nicotine is composed of C, H, and N. You have data for C and H but nothing for N. But that can be determined.
Convert CO2 to g C, H2O to grams H. I will round my numbers but you should go through with a little more care. Note that mm stands for molar mass.
g C = 4.251 g C x (mm C/mmCO2) = 4.251 x (12/44) = about 1.159
g H = 1.216 g H2O x (2H/H2O) = 1.216 x (2/18) = about 0.1351
g N = g sample - g C - g H - 1.567- gC - gH = about 0.273

Now convert those g C, H, N to mols.
mols C = 1.159/12 = about 0.0965
mols H = 0.135/1 = 0.135
mols N = 0.273/14 = about 0.0195
Divide mols C, mols H, and mols N by the smallest number. I get
C = 4.9 which rounds to 5
H = 6.9 which rounds to 7
N = 1
So C5H7N is the empirical formula.

To find percent of each element. For example for C it is
%C = 100*(atomic mass C*12/mm C5H7N)
same process for H and N.
Post your work if you get stuck.