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Author Topic: Freezing point depression and molecular formula and weight  (Read 12312 times)

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prncess23

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Freezing point depression and molecular formula and weight
« on: August 28, 2008, 09:55:09 PM »

Hello.

I dont even know how to approach this problem. Help please.

Molecular weights can be determined by freezing point depression measurements. A 6.0 g sample of a compound whose empirical formula is (CH2)x is dissolved in 200 g of benzene. the freezing point of the solution is 1.83 degrees C below that of pure benzene. the molal freezing point constant for benzene is 5.12 degrees C/m. Determine a) molecular weight of the compound and b) the correct molecular formula for compound
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Dan

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Re: Freezing point depression and molecular formula and weight
« Reply #1 on: August 29, 2008, 02:00:00 AM »

Start by reading about colligative properties. eg.

http://en.wikipedia.org/wiki/Colligative_properties#Freezing_point_depression

That link gives a formula you can use. You have been given the "freezing point constant" (aka cryoscopic constant).

Can you estimate the Van 't Hoff factor?
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prncess23

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Re: Freezing point depression and molecular formula and weight
« Reply #2 on: August 30, 2008, 03:57:44 PM »

Start by reading about colligative properties. eg.

http://en.wikipedia.org/wiki/Colligative_properties#Freezing_point_depression

That link gives a formula you can use. You have been given the "freezing point constant" (aka cryoscopic constant).

Can you estimate the Van 't Hoff factor?

how do i figure out the van't hoff factor when i dont know the change in temperature?
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Borek

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Re: Freezing point depression and molecular formula and weight
« Reply #3 on: August 30, 2008, 10:17:22 PM »

how do i figure out the van't hoff factor when i dont know the change in temperature?

Van't Hoff factor is a property of the dissolved substance, temperature change doesn't matter.
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Dan

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Re: Freezing point depression and molecular formula and weight
« Reply #4 on: August 31, 2008, 04:52:57 AM »



how do i figure out the van't hoff factor when i dont know the change in temperature?
[/quote]

As Borek said, you don't need to know the change in temperature to estimate the Van 't Hoff factor. You should read the quesion again though because you have been given the change in temperature.

Read about the Van 't Hoff factor (eg. here) and try to grasp the concept and you will not need to calculate the value.
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prncess23

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Re: Freezing point depression and molecular formula and weight
« Reply #5 on: August 31, 2008, 10:56:06 PM »

ok so the van 't Hoff factor, i is the number of moles of solute actually in solution per mole of solid solute added.

umm....still hecka lost.

need step by step answers.

i=6.00g unknown/200g benzene....but how do i get the moles for the unknown since i dont know what it is????

i have these 2 formulas:
Freezing Pointtotal = Freezing Pointsolvent - ΔTf

where :ΔTf = molality * Kf * i, (Kf = cryoscopic constant, which is 5.12 degrees Celsius C/m for the freezing point of benzene; i = Van 't Hoff factor)
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Borek

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Re: Freezing point depression and molecular formula and weight
« Reply #6 on: August 31, 2008, 11:18:12 PM »

ok so the van 't Hoff factor, i is the number of moles of solute actually in solution per mole of solid solute added.

Van't Hoff factor differs from 1 only for substances that dissociate or polymerize. Your substance seems to be a hydrocarbon. Will it dissociate? Polymerize?

Quote
i=6.00g unknown/200g benzene....but how do i get the moles for the unknown since i dont know what it is????

i have these 2 formulas:
Freezing Pointtotal = Freezing Pointsolvent - ΔTf

where :ΔTf = molality * Kf * i, (Kf = cryoscopic constant, which is 5.12 degrees Celsius C/m for the freezing point of benzene; i = Van 't Hoff factor)

You can calculate molality from these data, then take a look at definition of molality - there is only one unknown and it is a molar mass of the substance. Solve for it.
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prncess23

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Re: Freezing point depression and molecular formula and weight
« Reply #7 on: September 03, 2008, 07:38:25 PM »

i was told by another source that i could do the following, which seems less complicated than what you are telling me to do, but i appreciate the help.

Step 1:  Use    DTf =   Kf  x molality to solve for the molality (both Kf and the change in freezing point (DTf) are known.

Step2: Use the molality formula (molality = moles of solute/kg of solvent) to solve for moles of solute.  The molality is known from Step 1 and kg of solvent = .200 kg from the problem.

Step 3: Use the fact that  molecular weight = mass of solute/moles of solute to find the mol weight of the solute.  The mass of solute is given as 6.0g and the moles were calculated in step 2.

Step 4:  Determine the number of empirical formula units in a molecule by dividing the mol weight (determined in step 3) by the emp. formula mass (CH2 = 14g/mol).  The results of this division will give you x in (CH2)x.

i understand everything except step one. i dont have DTf...i only have Kf= 5.12 degrees C/m for benzene and that the freezing point of the solution is 1.83 degrees C below that of pure benzene. The freezing point of pure benzene is 5.5°C, which i looked up. So to get DTf, do i subtract 5.5 - 1.83= 3.67°C...right?? or am i missing something...?
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Borek

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Re: Freezing point depression and molecular formula and weight
« Reply #8 on: September 03, 2008, 08:57:16 PM »

What is DTf?

What you were told to do is just a way of calculating reciprocal of Van't Hoff factor. There are many ways to skin that cat.
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prncess23

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Re: Freezing point depression and molecular formula and weight
« Reply #9 on: September 06, 2008, 10:48:52 PM »

ok. so i used the steps below to get my answer....but it doesnt seem right. please help.

step 1: i got 0.35742 for molality
step 2: 1.7871 for moles solute
step 3: 3.3574 for molecular weight
step 4: 2398 for x in(CH2)x

what am i doing wrong?

Step 1:  Use    DTf =   Kf  x molality to solve for the molality (both Kf and the change in freezing point (DTf) are known.

Step2: Use the molality formula (molality = moles of solute/kg of solvent) to solve for moles of solute.  The molality is known from Step 1 and kg of solvent = .200 kg from the problem.

Step 3: Use the fact that  molecular weight = mass of solute/moles of solute to find the mol weight of the solute.  The mass of solute is given as 6.0g and the moles were calculated in step 2.

Step 4:  Determine the number of empirical formula units in a molecule by dividing the mol weight (determined in step 3) by the emp. formula mass (CH2 = 14g/mol).  The results of this division will give you x in (CH2)x.
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Borek

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Re: Freezing point depression and molecular formula and weight
« Reply #10 on: September 07, 2008, 01:19:40 AM »

step 1: i got 0.35742 for molality

OK

Quote
step 2: 1.7871 for moles solute

Step2: Use the molality formula (molality = moles of solute/kg of solvent) to solve for moles of solute.  The molality is known from Step 1 and kg of solvent = .200 kg from the problem.

How did you get 1.79?
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Dan

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Re: Freezing point depression and molecular formula and weight
« Reply #11 on: September 07, 2008, 01:50:27 AM »

Quote
step 2: 1.7871 for moles solute

Step2: Use the molality formula (molality = moles of solute/kg of solvent) to solve for moles of solute.  The molality is known from Step 1 and kg of solvent = .200 kg from the problem.

How did you get 1.79?

By dividing molality by mass of solvent instead of multiplying...

prncess23, consider the units of molality:

molality (mol/kg) = moles (mol) / solvent mass (kg)

rearrange the equation above and you get:

molality (mol/kg) x solvent mass (kg) = moles (mol)
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