a compound containing only boron, nitrogen and hydrogen was found to be 40.3% B, 52.2%N, and 7.5% H by mass. If 3.301g of this compound is dissolved in 50.00g of benzene, the solution produced freezes at 1.30 degrees celsius. If Kf(freezing point depression constant) for benzene is 5.12 degrees celsius per metre and the freezing point of benzene is 5.48 degrees celsius, what is the molecular weight of this compound?



Responses

chem - DrBob222, Friday, November 7, 2008 at 4:24pm
You can use the percents to determine the empirical formula.
Take 100 g sample which will give you
40.3 g C, 52.2 g N and 7.5 g H. Now divide each by the atomic mass of C, N, and H respectively, find the ratio in small whole numbers and you have the empirical formula.
Now use delta T = kf*molality to determine molality. Finally, knowing molality = # mols/kg solvent, you can determine the molar mass and from that the molecular formula and from that the molar mass.
Post your work if you get stuck.

ok i was able to find the ratio to be 1:1:2
and i also found the molality from the equation delta T=Kf*molality to be 1.07/m. I get lost about what to do after

The ratio of B1N1H2 is correct.

The molality is not.
delta T = Kf*m (by the way, Kf is 5.12 degrees C/molal [not metre].)
(5.48-1.30)=5.12*m.
molality = 0.816 but you should confirm this.
molality = # mols/kg solvent.
Solve for mols.
0.816 = # mols/0.050 = 0.04
# mols = g/molar mass
0.04 = 3.301/molar mass
molar mass approximately 81. These calculations need to be redone for I have estimated and rounded here and there. Also, it is worth noting here that molar masses determined this way are not exact, usually within 5 or so. Therefore, there must be a better method BUT this gives an approximate number to work with.
Now we go through the B,N,H think which you have done and obtained B1N1H2.
Lets do the empircal formula mass. It is
10.8 + 14 + 2 = 26.8.
So how many formula masses must be in the molecular formula. That would be 81/26.8 = 3.02, we round that to 3.000 and say the molecular formula is EXACTLY (BNH2)3 = B3N3H6. Now look up the atomic masses of B, N, and H, multiply by the subscripts and add them together to obtain the molar mass of the compound.
Summary of what we've done.
Freezing point work gives an approximate molar mass.
B,N,H analysis gives an exact EMPIRICAL formula.
We divide the approximate molar mass by the exact empirical formula mass to get a number, round that to a whole number, and multiply that whole number by each of the subscripts in the empircal formula to obtain the molecular formula. Then look up atomic masses on periodic table and determine the exact molar mass. Check my work. Check for typos. Redo the calculations to make sure the numbers are correct.

To determine the molecular weight of the compound, you can proceed with the following steps:

1. Find the molality (m) of the solution:
- Given delta T = 1.30 degrees Celsius
- Given Kf = 5.12 degrees Celsius per mole
- From the equation delta T = Kf * molality, we can rearrange it to solve for molality:
molality = delta T / Kf = 1.30 / 5.12 = 0.2531 mol/kg

2. Calculate the moles of the compound:
- Given that 3.301 g of the compound is dissolved in 50.00 g of benzene, you can convert the mass of the compound to moles using its molar mass.
- Since the compound contains only B, N, and H, and the empirical formula is 1:1:2, we can determine the molar mass as follows:
- Molar mass = (B molar mass) + (N molar mass) + 2 * (H molar mass)
= (10.81 g/mol) + (14.01 g/mol) + 2 * (1.01 g/mol)
= 10.81 g/mol + 14.01 g/mol + 2.02 g/mol
= 26.84 g/mol
- Moles of the compound = Mass of the compound / Molar mass
= 3.301 g / 26.84 g/mol
= 0.123 mol

3. Determine the molecular formula:
- Now, we need to find the molecular formula of the compound based on the empirical formula and the moles calculated above.
- The empirical formula tells us that the ratio of atoms is 1:1:2, so the empirical formula is BN2.
- To find the molecular formula, we need to know the actual ratio of atoms in the compound. Divide the moles of the compound by the moles of the empirical formula to find the ratio:
- Moles ratio = Moles of the compound / Moles of the empirical formula
= 0.123 mol / 0.123 mol
= 1

4. Calculate the molecular weight:
- The molecular weight can be calculated by multiplying the molar mass of the empirical formula by the moles ratio:
- Molecular weight = Molar mass of the empirical formula * Moles ratio
= 26.84 g/mol * 1
= 26.84 g/mol

Therefore, the molecular weight of the compound is 26.84 g/mol.

To find the molecular weight of the compound, you will need to follow these steps:

1. Calculate the empirical formula: From the given percentages, the empirical formula is B: N: H = 1:1:2.

2. Determine the molality (m) of the solution: Use the given delta T = 1.30 degrees Celsius and the Kf value for benzene (5.12 degrees Celsius per molality) to calculate the molality.
delta T = Kf * molality
1.30 = 5.12 * molality
molality = 1.30 / 5.12 = 0.2539 m

3. Convert molality to the number of moles of solute: As the molality is given in m, which represents moles of solute per kg of solvent, you need to convert the grams of solvent to kilograms.
Given mass of benzene = 50.00 g
Mass of benzene in kg = 50.00 g / 1000 = 0.05000 kg
Moles of solute = molality * mass of solvent in kg
Moles of solute = 0.2539 * 0.05000 = 0.0127 mol

4. Calculate the molar mass of the compound: Divide the mass of the compound (3.301 g) by the number of moles of the compound (0.0127 mol).
Molar mass = 3.301 g / 0.0127 mol = 259.37 g/mol

5. Determine the molecular formula: To find the molecular formula, you need to compare the molar mass (259.37 g/mol) with the empirical formula weight (B3N4H8 = 55.81 g/mol). Divide the molar mass by the empirical formula weight to find the ratio of the molecular formula to the empirical formula weight.
Molecular formula ratio = Molar mass / Empirical formula weight
Molecular formula ratio = 259.37 g/mol / 55.81 g/mol
Molecular formula ratio ≈ 4.65

6. Adjust the molecular formula: The ratio obtained should be close to a whole number. In this case, it is close to 5. Multiply the empirical formula (B3N4H8) by this ratio to obtain the molecular formula.
Molecular formula = B3N4H8 * 5 = B15N20H40

Therefore, the molecular weight of the compound is approximately 15(10) + 14(20) + 1(40) = 750 g/mol (rounded to the nearest whole number).