[prncess23]

i have NO idea how to do this. please help.

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.

i am thinking i should use these formulas....there are a lot of variations....
Freezing Point_total = Freezing Point_solvent - 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)

[spock]

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.

[prncess23]

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.

thank you so much...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...?

[prncess23]

anyone???

[valdorod]

DTf is 1.83

DTf is the change in freezing point

the problem states that it is 1.83 below the original freezing point.  That is how much it changed.

[prncess23]

thank you so much. i always over-analyze things.

[prncess23]

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.

ook...so i did all the steps and i got a weird answer.

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?

[spock]

Well, you have a problem in Step 2.  If you are using the molality formula
                 (molality=moles solute/ kg solvent)
and solve for moles you get this equation:

                  moles of solute = kg solvent x molality
Plugging in you should get
                  moles of solute = .200 kg  x .35742     (200 grams of benzene = .200 kg)
That is not equal to 1.7871 moles!