A solution of 2.50 g of a compound having the empirical formula C6H5P in 25.0 g of benzene(C6H6) is observed to freeze at 4.3 degrees C. Calculate the molar mass of the solute and its molecular formula.
I wonder how you came up with 119 g/mol?
delta T = Kf*m
5.5-4.3 = 1.2 for delta T.
I found Kf = 4.90
1.2 = 4.90*m
m = 1.2/4.90 = 0.2449
m = moles/kg solvent
moles = m*kg solvent = 0.2449*0.025 = 0.00612 moles.
moles = grams/molar mass or
molar mass = grams/moles = 2.50g/0.00612 = about 400 or so. Check my work.
Explanation:
The idea here is that you need to use the freezing-point depression equation to determine what the molality of the solution.
Once you have the solution's molality, use it to find the number of moles of solute it contains.
As you know, the equation for freezing-point depression looks like this
ΔTf=i⋅Kf⋅b
, where
ΔTf - the freezing-point depression;
i - the van't Hoff factor
Kf - the cryoscopic constant of the solvent;
b- the molality of the solution.
The cryoscopic constant of benzene is equal to 5.12∘C kg mol−1
You're dealing with a non-electrolyte, which means that the van't Hoff factor will be equal to 1.
The freezing-point depression is defined asΔTf=T∘f−Tf, where
T∘f - the freezing point of the pure solvent
Tf- the freezing point of the solution
Pure benzene freezes at
5.5∘C, which means that the freezing-point depression will be
ΔTf=5.5∘C−4.3∘C=1.2∘C
Plug in your values and solve for b, the molality of the solution
ΔTf=i⋅Kf⋅b⇒b=ΔTfi⋅Kf
b=1.2∘C1⋅5.12∘Ckg mol−1=0.2344 mol kg−1
As you know, molality is defined as moles of solute per kilograms of solvent.
b=n solute/m solvent
In your case, you have a mass of
25.0 g
of benzene, which means that the solution will contain
b=n solute/m solvent⇒n solute=b×m solvent
n solute=0.2344 molkg−1⋅25.0⋅10−3kg
n solute=0.00586 moles
Now, molar mass is defined as the mass of one mole of a substance. In your case, the
2.50-g
of solute contain a total of
0.00586
moles, which means that the molar mass will be
MM=2.50 g/0.00586 moles=426.6 g/mol
As you know, a compound's empirical formula tells you what the smallest whole number ratio that exists between the elements that make up said compound is.
The molecular formula, which tells you exactly how many atoms of each element are needed to form the compound, will always be a multiple of the empirical formula.
In this case, calculate the molar mass of the empirical formula by adding the molar masses of all of its constituent elements
6×12.011 g/mol+5×1.00794 g/mol+1×30.974 g/mol=108.08 g/mol
You can thus say that
108.08g/mol⋅n=426.6g/mol
This will get you
n=426.6/108.08=3.95≈4
The compound's molecular formula will be
(C6H5P)4⇒C24H20P4→ 1, 2, 3, 4 - tetraphenyltetraphosphetane
To calculate the molar mass and molecular formula of the solute, we need to follow these steps:
Step 1: Calculate the number of moles of benzene (C6H6)
We can do this by using the formula: moles = mass / molar mass
Given:
Mass of benzene (C6H6) = 25.0 g
Molar mass of benzene (C6H6) = 78.11 g/mol
Using the formula, we have:
Moles of benzene (C6H6) = 25.0 g / 78.11 g/mol ≈ 0.32 mol
Step 2: Calculate the number of moles of the solute by using the freezing point depression equation:
ΔT = K_f * m
where:
ΔT = change in freezing point
K_f = cryoscopic constant (for benzene = 5.12 °C·kg/mol)
m = molality of the solution = moles of solute / mass of solvent (in kg)
Given:
Change in freezing point (ΔT) = 4.3 °C
Cryoscopic constant (K_f) = 5.12 °C·kg/mol
Mass of solvent (benzene) = 25.0 g = 0.025 kg
Using the equation, we can solve for moles of solute:
4.3 = 5.12 * (moles of solute / 0.025)
moles of solute = 4.3 * 0.025 / 5.12 ≈ 0.0209 mol
Step 3: Calculate the molar mass of the solute by using the formula:
molar mass = mass / moles
Given:
Mass of solute = 2.50 g
Moles of solute = 0.0209 mol
Using the formula, we can solve for the molar mass:
molar mass = 2.50 g / 0.0209 mol ≈ 119.62 g/mol
Step 4: Calculate the empirical formula mass (EFM) of the solute:
The empirical formula mass can be calculated by summing up the atomic masses of the empirical formula.
The empirical formula C6H5P has:
(6 * atomic mass of carbon) + (5 * atomic mass of hydrogen) + (1 * atomic mass of phosphorus) = (6 * 12.01) + (5 * 1.01) + (1 * 30.97) ≈ 102.17 g/mol
Step 5: Calculate the molecular formula:
The molecular formula mass (MFM) is the mass of the actual molecule.
The molecular formula mass can be calculated by finding the ratio of the molar mass to the empirical formula mass:
MFM = molar mass / EFM = 119.62 g/mol / 102.17 g/mol ≈ 1.17
Since the molecular formula must be a whole number ratio of the empirical formula, we need to determine the nearest whole number multiplier that will give us approximately 1.17. In this case, it is close to 1.
Hence, the molecular formula is approximately equal to the empirical formula, C6H5P.
To summarize:
- The molar mass of the solute is approximately 119.62 g/mol.
- The molecular formula of the solute is C6H5P.
I think I got the answer.
Molar Mass = 119 g/mol
Molecular formula is the same = C6H5P