Calculate the molecular formula of nicotine (CxHyNz, xyz being unknown)if 4.38 mg of this compound burns to form 11.9 mg of carbon dioxide and 3.41mg of water, and 3.62 g of nicotine changes the freezing point of 73.4g of water by .563C. Show all work.

Question

I am confused on how to even attempt this problem, I was hoping someone would be able to help me.

Answer 1

Start by computing the percent CO2 and H2O, converting to percent C and H, then subtract from 100 to obtain percent O.
Take a 100 g sample of that and each percent translates directly to grams C, H, and O.
Convert g C, H, and O to moles C, H, and O. You want to find the ratio of the elements to each other; the easy way to do this is to divide the smallest number of moles by itself, then divide the other two numbers by that same small number. Round to whole numbers EXCEPT don't round too much; i.e., a ratio of 1 to 1.5 would become 2 to 3. This will give you the empirical formula.
The other data is to determine the approximate molar mass.
Delta T = kf*molality and
molality = moles/kg solvent and
moles = g/molar mass.
Equation 1 give you the molality, equation 2 using m and kg solvent to obtain moles and equation 3 with moles and grams give you the molar mass. That will tell you the molecular formula.
Post your work if you get stuck. I'll not be on for a couple of hours but I can help you through after that if you need more help.

Answer 2

I feel really stupid but I can even figure out the first part, with the percents, but after that I feel like I could do it.

Answer 3

It may be just as well that you didn't put in a lot of work because I made an error. I read the problem as CxHyOz but it actually is CxHyNz. So in the prose above, substitute N everywhere I have O.
I'll get you started.
First, we take the g CO2 and g H2O and convert them to g C and g H. I wrote above to do it in two steps but I'll do it here in one step. Here are the numbers.
11.9 mg CO2, 3.41 mg H2O.

11.9 mg CO2 x (1 mole C/1 mole CO2) x (12.01 g C/1 mole C) x (1 mol CO2/44.01 g CO2) =3.247 mg carbon.

3.41 mg H2O x (2 mole H/1 mole H2O) x (2 g H/1 mole H2O) x (1 mole H2O/18.015 g H2O) = 0.3786 m33g hydrogen.

Nitrogen is determined by difference.
4.38 mg - 3.247 mg C - 0.3786 mg H = 0.7544 mg nitrogen. I've carried more places than I'm allowed but I'll round later.

%C = (mg C/mg sample)*100 = (3.247/4.38)*100 = 74.1% (rounded)

%H = (mg H/mg sample)*100 = (0.3786/4.38)*100 =8.64% (rounded)

%N = (mg N/mg sample)*100 = (0.7544/4.38)*100 = 17.2% (rounded).

Now you take a 100 g sample and these percents convert directly to g C, H, and N.
74.1% C gives 74.1 g C.
8.64% H gives 8.64 g H.
17.2% N gives 17.2 g N.

Now convert those to moles.
74.1/12.01 = ??
8.64/1 = ??
17.2/14 = ??

Now find the empirical formula as outlined in my first post and continue to the molar mass determination by the depression of the freezing point.
You may want to repost at the beginning since this post is so far down the list.