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July 25, 2016
Posted by **DrBob222** on Saturday, February 3, 2007 at 5:20pm.

Take a 100 g sample.

That means we have

19.9% C = 19.9 g C

46.7% N = 46.7 g N

6.67% H = 6.67 g H

26.7% O = 26.7 g O

Find mols of these amounts.

mols C = 19.9/12 = 1.66

mols N = 46.7/14 = 3.33

mols H = 6.67/1 = 6.67

mols O = 26.7/16 = 1.67

To determine the empirical formula (it will be in small whole numbes), divide the smallest number of moles by itself (thereby forcing that to be 1.00), then to keep everything equal, divide all the other numbers by the same thing. The small whole numbers you obtain will be the empirical formula.

To determine the molar mass, use the boiling point constant.

delta T = K

delta T is the difference between the normal boiling point of benzene and the boiling point of the solution of unknown and benzene. K

Knowing that molality = # mols/kg solvent you can calculate # mols.

Knowing # mols=grams/molar mass you can calculate molar mass.

Post your work if you get stuck. If you don't know how to get started on 3 and 4, tell us what you don't understand about.

Determine the forumula and molecular mass for an organic compound from this information:

-5.00g of the organic solid was dissolved in 100.0g of benzene.

-Boiling point of solution was 82.3 degrees Celsius.

-The organic compound is 46.7% nitrogen, 6.67% hydrogen, 26.7% oxygen, and the remainder is carbon.

-The boiling temperature of pure benzene is 80.2 degrees C; Kb = 2.53 degrees CKg/mol

1. determine the molecular mass of the organic solid

2. determine the molecular formula of organic solid

3. determine the mole fraction of the organic solid in solution

4. if density of solution is 0.8989 g/ml, calculate molarity of solution.

This is what I got:

molecular formula: CN2H4O

molecular mass: 1.21 x 10^3

I got the molecular mass by subtracting 80.2 from 82.3 to get delta T which is 2.1 degrees celsius.

molality = delta T/kb=2.1/2.53=0.83 molal

mol = molal x kg= 0.83 x .005 kg=.00415 mol

mol = g/M therefore M=g/mol = 5/.00415 = 1205 M = 1.21 x 10^3 M

is that correct?

The empirical formula is correct. Good work. The molar mass is not quite right but almost. 0.83 molal is ok.

BUT, when I said molality = mols/kg I guess I SHOULD have said kg solvent but I assumed you would know you didn't use sample weight for BOTH g and kg solvent. You used kg of the sample as 0.005 and not 0.1 (for 100 g benzene solvent).

When you redo that one small part you will have CLOSE to (but not exact) molar mass. To find the exact molar mass is a 2-3 step procedure.

1. Add atomic masses of CN2H4O to obtain the exact mass of the empirical formula.

2. Divide the approximate molecular mass by what you have in #1. You should get very close to a whole number.

3. That whole number will be the number of units to multiply the empirical formula by. Usually these things are 1, 2, or 3.

4. Then recalculate the molecular mass from the molecular formula.

ok so now I got mol=molal x kg = 0.83 x 0.1 = 0.083 mol

M = 5.00g/0.083 mol = 60.24 = 60.2...when tryying to find the exact molar mass, the whole number was 1.002 which is very close to one so I just left it as 60.02...i think that should be right.

I need the molecular formula not the empirical formula, but when i divide the approximate molecular mass buy the exact, it's 1.002 so wouldn't the molecular formula and empirical formula be the same in this case?

i dont know give me the answer

- Chemistry -
**Anonymous**, Wednesday, January 5, 2011 at 9:39pmyes it would