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.
I am confused on how to even attempt this problem, I was hoping someone would be able to help me.
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.
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.
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.
Sure, I can help you solve this problem step-by-step.
Step 1: Calculate the moles of carbon in the carbon dioxide produced.
Given:
Mass of carbon dioxide = 11.9 mg = 0.0119 g
Molar mass of carbon dioxide (CO2) = 44.01 g/mol
The formula weight of carbon dioxide is (12.01 g/mol * 1) + (16.00 g/mol * 2) = 44.01 g/mol.
Moles of carbon dioxide = (0.0119 g) / (44.01 g/mol)
Step 2: Calculate the moles of carbon in the nicotine compound.
Using the stoichiometry of the balanced combustion reaction, the coefficients of carbon dioxide and nicotine can be compared to determine the ratio.
The balanced combustion reaction equation for nicotine is:
CxHyNz + (x + (1/2)y - z)/2O2 --> xCO2 + y/2H2O + z/2O2
From the equation, we can see that "x" moles of carbon dioxide are formed from 1 mole of nicotine.
So, Moles of carbon in nicotine = Moles of carbon dioxide from Step 1
Step 3: Calculate the empirical formula of nicotine.
We have the following molecular formula for nicotine: CxHyNz
Since we have the moles of carbon as per Step 2, now we need the ratio of moles of hydrogen to carbon, and the ratio of moles of nitrogen to carbon.
To calculate the ratios, we need the mass of hydrogen and nitrogen in the nicotine compound, which is obtained from the mass of water and nicotine, respectively.
Given:
Mass of water = 3.41 mg = 0.00341 g
Molar mass of water (H2O) = 18.02 g/mol
The formula weight of water is (1.01 g/mol * 2) + (16.00 g/mol * 1) = 18.02 g/mol
Moles of hydrogen = (0.00341 g) / (18.02 g/mol)
Now, we move on to the second piece of information:
Given:
Mass of nicotine = 3.62 g
Moles of nicotine = (3.62 g) / (molar mass of nicotine)
Step 4: Determine the empirical formula of nicotine.
Using the ratio of moles of hydrogen and nitrogen to moles of carbon, we can determine the empirical formula of nicotine.
The empirical formula shows the simplest whole-number ratio of the elements in the compound.
Divide the moles of hydrogen and nitrogen by the moles of carbon to get their ratio.
Empirical formula = CxHyNz (with whole-number ratios)
Step 5: Calculate the molecular formula of nicotine.
To determine the molecular formula, we need to know the molar mass of the empirical formula and the molar mass of nicotine.
Given the molar mass of the empirical formula and the molar mass of nicotine, we can calculate the ratio and find the values of x, y, and z.
Given:
Molar mass of the empirical formula = (empirical formula weight)
Molar mass of nicotine = 3.62 g/mol (calculated in Step 3)
Molar mass of the empirical formula / Molar mass of nicotine = (x/y)/(3.62 g/mol)
To find the exact molecular formula, you need to know the molar mass of the empirical formula. If you provide more information about the molar mass, we can proceed with calculating the molecular formula.
I hope this helps you understand how to approach the problem. Let me know if there is anything else I can assist you with.