1. Nicotine,a poisonous compound found in tobacco leaves,is 74% carbon,8,65% hydrogen and 17,35% nitrogen (a)calculate the empirical formula of nicotine (b)what is the molecular formula of nicotine if it has a molar mass of 165 g.mol^-1

Answer 1

Take a 100 g sample which will give you
74 g C
8.65 g H
17.35 g N
Now, mols = g/atomic mass.
Convert these to mols and you redo these because I will estimate.
74/12 = about 6.2
8.65/1 = 8.65
17.35/14 = 1.24

Now find the ratio of these numbers to one another with the smallest being 1.0. The easy way to do that is to divide all of them by the smallest number in the lot. That is 1.24; therefore,
6.2/1.24 = about 5
8.65/1.24 = about 7
1.24/1,24 = 1.00 (because we made it that way).
So the empirical formula is C5H7N.
What is the empirical formula mass? That's (5*12) + (7*1) + (1*14) = about 81
The molar mass is 165; we want to know how may of the "empirical formula pieces" fill fit within that 165. That's simply 165/81 = 2.04. We round to a whole number of 2.00 so the molecular formula is (C5H7N)2 since two of the empirical formulas make up 1 molecular formula. The problem didn't ask but you can find the "real" molar mass (the 165 is an estimate) by adding the atomic masses together. You can do it but I found 81*2 = 162. (The molar mass is determined by boiling point elevation or freezing point depression. Those methods give good estimates of the molar mass but there is some error involved in the "chemistry" of the process due to interactions that the formulas do not take into account. So they are useful for problems like this and that's why we can round the answer to a whole number as above.

Answer 2

Thank u