A 13.03 g sample contains only ethylene glycol (C2H6O2) and propylene glycol (C3H8O2). When the sample is added to 100.0g of pure water, the resulting solution has a freezing point of -3.5C. What is the percent composition of ethylene glycol and propylene glycol in the original sample?
delta T = Kf*m
Solve for m
m = moles/kg solvent
Solve for moles.
moles = grams/molar mass
Solve for molar mass.
Then let X = fraction of C2H6O2
and 1-X = fraction of C3H8O2
--------------------------
X(molar mass C2H6O2) + (1-x)(molar mass C3H8O2) = molar mass of the mixture.
Then convert X to percent and 1-X to percent. I think the answer is about 48% and 52% respectively but these are approximations.Post your work if you get stuck.
Well, if the solution is freezing, it sounds like those glycols have got some cold feet! But let's figure it out.
To find the percent composition of each glycol, we need to calculate how much of each glycol is present. Let's assume x g of ethylene glycol (C2H6O2) and y g of propylene glycol (C3H8O2) are present in the original sample.
The freezing point depression can be calculated using the formula ΔTF = KF × m, where ΔTF is the change in freezing point, KF is the freezing point depression constant (1.86 °C/m for water), and m is the molality of the solution.
Since we know the freezing point depression (ΔTF = -3.5 °C) and the mass of water (100.0 g), we can calculate the molality of the solution (m) using the formula m = moles of solute / mass of solvent in kg.
Now, let's calculate the moles of each glycol using their respective molar masses:
Molar mass of ethylene glycol (C2H6O2) = (2 × 12.01 g/mol) + (6 × 1.01 g/mol) + (2 × 16.00 g/mol) = 62.07 g/mol
Molar mass of propylene glycol (C3H8O2) = (3 × 12.01 g/mol) + (8 × 1.01 g/mol) + (2 × 16.00 g/mol) = 76.10 g/mol
Now, we can calculate the moles of each glycol:
moles of ethylene glycol = x / 62.07
moles of propylene glycol = y / 76.10
Since the moles of solute and solvent are directly proportional to the freezing point depression, we can set up the following ratio:
(ΔTF/sample mass) = (moles of solute / (moles of solute + moles of solvent))
Substituting the values we know:
(-3.5 / 100.0) = ((x / 62.07) / ((x / 62.07) + (y / 76.10)))
Now, you'll just have to solve this equation for x, the mass of ethylene glycol, and y, the mass of propylene glycol. Once you have the values, you can calculate the percent composition by dividing the mass of each glycol by the total sample mass (13.03 g) and multiplying by 100.
Good luck, math whiz! I believe in you!
To find the percent composition of ethylene glycol and propylene glycol in the original sample, we can use the freezing point depression equation.
First, let's determine the molality of the ethylene glycol and propylene glycol in the solution using the freezing point depression equation:
ΔTf = Kf * m
Where ΔTf is the freezing point depression, Kf is the molal freezing point depression constant of water (-1.86°C/m), and m is the molality of the solution.
Given that the freezing point depression (ΔTf) is -3.5°C and the mass of the water is 100.0g, we can rearrange the equation to solve for the molality (m):
m = ΔTf / Kf
m = -3.5°C / -1.86°C/m
m ≈ 1.88 molal
Now, let's calculate the moles of ethylene glycol (C2H6O2) and propylene glycol (C3H8O2) in the solution:
moles = molality * kg solvent
Since the mass of the water is 100.0g, the mass of the solvent is also 100.0g.
moles ethylene glycol = 1.88 molal * 0.100 kg
moles ethylene glycol ≈ 0.188 mol
moles propylene glycol = 1.88 molal * 0.100 kg
moles propylene glycol ≈ 0.188 mol
Now, let's calculate the mass percent of ethylene glycol and propylene glycol in the original sample:
mass percent = (mass of component in sample / total mass of sample) * 100
mass of ethylene glycol = (0.188 mol * 62.07 g/mol) ≈ 11.65 g
mass of propylene glycol = (0.188 mol * 76.10 g/mol) ≈ 14.31 g
total mass of sample = 11.65 g + 14.31 g ≈ 25.96 g
mass percent of ethylene glycol = (11.65 g / 25.96 g) * 100 ≈ 44.90%
mass percent of propylene glycol = (14.31 g / 25.96 g) * 100 ≈ 55.10%
Therefore, the percent composition of ethylene glycol and propylene glycol in the original sample is approximately 44.90% and 55.10%, respectively.