Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: needhelp22 on January 27, 2008, 12:39:13 PM
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okay so im dong this lab using freezing point depression to find Freezing point of Naphthalene, Kf value of naph (change in T = molality * Kf) , and determining freezing point of sulfur when mixed with naph
the end of the lab asks me to find molar mass of sulfur. I do and i get 93 g / mol....my freinds tell me its suppose to be arond 250 g /mol... our teacher doesnt grade corectness since messing up on labs isn' too hard. I just need to understand the concept. Why does putting naph in a test tube, adding sulfur, melting the two powders, and finding their freezing point depression, change the molar mass of sulfur?
also we are required to find out the # of atoms of Sulfur per molecule....now i can do this if i had a compound given but i dont know the ratio of naph to sulfur at all. I found the mole of sulfur and therefore the atoms of sulfur in teh solution..but im stuck on where to go from there
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What kind of Sulfur are we talking about here (I.E. was it some kind of sulfate)?
As the molar mass of sulfur is 32g/mol....
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regular sulfur...but i got 93 g/mol as a molar mass at the end of it
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Your question is all over. You do not melt two powders.
Start here (http://en.wikipedia.org/wiki/Freezing-point_depression).
Then maybe list your data and calculations.
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What if sulfur doesn't dissolve as atomic?
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This is very common lab for second semester general chemistry in the US. The purpose of the lab is for the student to come to the conclusion that Borek did. If done correctly the student ends up with a molar mass that is then divided by the one given in the periodic table (32 g/mol) and S8 is obtained. However, for your results 93/32 you end up with S3. As your instructor/professor said, sometimes, it does not work as expected.
For this experimnet, graphing of data and correct extrapolation is crucial. Also, this experiment, typically has two parts, where the first part has the student calculate the freezing points constant (Kf) of naphthalene first. This value is used in part two of the calculations to find the molar mass of the solute (sulfur in this case) If Kf is calculated incorrectly then the molar mass of sulfur would also be incorrect.
Students are usually taught abou the diatomic elements, H2, O2, N2, etc, in this experiment they find that other elements such as sulfur and phosphorus are not monoatomic in the free state.
Thus the student should arrives at the question/answer that Borek is proposing. You can see James F. Hall Experimental Chemistry 6th edition for a description of this lab.
Valdo
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http://www.bibliobase.com/Chemistry/Hall/pdf_doc/23.pdf
thats the lab we did...we only did CHIOCE 1
my calcs and data are right here.
Data
Data
Temperature (Degrees Celsius) Drops
Time Naphthalene Naphthalene/Dichlorobenzene Naphthalene/Sulfur
30 Sec 85.1 79.8 81.0
60 Sec 80.1 76.5 76.2
90 Sec 77.2 75.1 73.9
120 Sec 75.3 74.8 73.1
150 Sec 74.9 74.2 72.9
180 Sec 74.3 73.9 72.8
210 Sec 74.1 73.0 72.8
240 Sec 73.9 72.5 72.8
270 Sec 73.5 72.0 72.8
300 Sec 73.2 71.0 72.1
330 Sec 73.0 70.2 71.8
360 Sec 73.0 68.9 71.0
390 Sec 72.9 67.5 70.9
420 Sec 72.9 66.0 70.4
450 Sec 72.4 64.5 69.7
480 Sec 72.1 62.9 68.0
510 Sec 71.8 61.0 66.5
540 Sec 70.8 69.2 64.2
570 Sec 70.8 60.9
600 Sec 70.8 56.8
630 Sec 70.8
660 Sec 70.7
690 Sec 68.4
720 Sec 64.0
750 Sec 60.0
Part A Data Collected
Mass Of Test Tube 30.98 g
Mass of Test Tube + Naphthalene 40.10 g
Mass of Naphthalene Used 9.120 g
Freezing Point of Naphthalene (observation) 75.3 Degrees Celsius
Part B Data Collected
Mass of Test Tube and Naphthalene 40.20 g
Mass of Naphthalene 9.220 g
Mass of Test Tube, Naphthalene, Dichlorobenzene 41.04 g
Mass of Dichlorobenzene .8400 g
Moles of Dichlorobenzene 5.753 X 10-3 mol C6H4Cl2
Freezing point of Naphthalene/ Dichlorobenzene (observation) 74.2 Degrees Celsius
DT from Part A and Part B 1.10 Degrees Celsius
Molality of Dichlorobenzene Solution 5.753 X 10-3 mol C6H4Cl2
Kf for Naphthalene 1.76 C Kg/ Mol
Part C Data Collected
Mass of Test Tube 30.97 g
Mass of Naphthalene and Test tube 40.07 g
Mass Naphthalene 9.100 g
Mass of Test Tube, Naphthalene, Sulfur 40.61 g
Mass of Sulfur .5400 g
Freezing point of Naphthalene/ Sulfur (observation) 73.9 Degrees Celsius
DT from Part A and Part C 1.40 Degrees Celsius
Moles of Sulfur 7.23 X 10-2 mol
Molality of Sulfur Solution .795 mol/kg
Molar Mass of Sulfur 74.6 g/mol
Number of Sulfur Atoms Per Molecule 4.35 X 1022 atoms of Sulfur
Calculations
Mass of Naphthalene Used in Part A:
Mass of Test Tube and Naphthalene – Mass of Test Tube = Mass of Naphthalene
40.10 grams – 30.98 grams = 9.120 grams of Naphthalene
Mass of Naphthalene In Part B:
40.20 g – 30.98 g = 9.220 grams of Naphthalene
Mass of Dichlorobenzene in Part B:
41.04 g – 40.20 grams = .8400 grams of Dichlorobenzene
DT from Part A and Part B
T of Naphthalene – T of Naphthalene and Dichlorobenzene
75.3 Degrees Celsius – 74.2 Degrees Celsius = 1.10 Degrees Celsius
Molality of Dichlorobenzene Solution in Part B
Molality = moles of solute / kg of solvent
- Moles of Dichlorobenzene
.8400 g C6H4Cl2 1 Mol C6H4Cl2 5.753 X 10-3 mol C6H4Cl2
146 g C6H4Cl2
- Kg of Naphthalene
9.220 g Naphthalene 1 kg Naphthalene .009220 kg of Naphthalene
1000 g Naphthalene
Molality = 5.753 X 10-3 mol C6H4Cl2 / .009220 kg of Naphthalene
Molality = .6240 mol/kg
Kf Value in Part B
DT = mKf
1.10 Degrees Celsius = .6240 mol/kg (Kf)
Kf = 1.76 C Kg/ Mol
Mass of Naphthalene in Part C
40.07 g – 30.97 g = 9.100 grams of Naphthalene
Mass of Sulfur in Part C
40.61 g – 40.07 g = .5400 g of Sulfur
DT from Part A and Part C
75.3 Degrees Celsius – 73.9 Degrees Celsius = 1.4 Degrees Celsius
Molality of Sulfur Solution in Part C
DT = Kfm (Kf is used from Part B)
1.40 Degrees Celsius = 1.76 C Kg/ Mol (m)
Molality = .795 mol/kg
Molar Mass of Sulfur
Molar Mass = grams of sulfur / moles of sulfur
- Moles of Sulfur
o Molality = moles of solute/ kg of solvent
o .795 mol/kg = moles of sulfur / .0091 kg of Naphthalene
o moles of Sulfur = 7.23 X 10-2 mol
Molar Mass = .5400g Sulfur / 7.23 X 10-2 mol Sulfur
Molar Mass = 74.6 g / mol
Number of Sulfur Atoms Per Molecule
7.23 X 10-2 mol Sulfur 6.022 X 1023 atoms of Sulfur 4.35 X 1022 atoms of Sulfur
1 mol Sulfur
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thanks valdorod...really helps...whts this s8 and s3 your talkin about...can u give me a general place where i can read about it. thanks.
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S3 & S8. Like H2 or O3.
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ohhhh wow..im stupid.....arite thnks for your help guys. so when sulfur is dissolved, theres 8 moles of it thats suppose to dissolve? but i only got 3 based on my data. I'm having a little hard time grasping hte why. when sulfur dissolves, we are saying that each molecule has 3 moles of sulfur? so solving for the # of atoms of sulfur per molecule would be 3 times 6.022 X10 to the 23rd.
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we are saying that each molecule has 3 moles of sulfur?
>:(
What is a mole, what is a molecule, what is an atom?
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okay...this is what im thinking (sry for the unclear stuff)
a molecule in this lab contains naph and sulfur. Theres is some compound that tells you. and in my case, that compound would have S3 in it. Which tells you that in that compound, theres 3 moles of sulfur compared to ___ moles of naph.
now im using compound and molecule as the same thing...is that wrong?
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a molecule in this lab contains naph and sulfur
No. There are separate molecules of naphthalen and separate molecules of sulfur. They are mixed. Do you know what is a chemical compound and what is a mixture?
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From the data that you have posted, I can tell that the experiment did not go as it should have gone. The drop in freezing point for your mixtures was too small. This has led to your calculation of Kf to be too small. Nonetheless, you should always work with the data that you obtain. From just skimming through your data and calculations everything seems ok, except at the end.
After you calculate the molar mass of sulfur from your data, you obtained 74.6 g/mol, all you need to do is divide by the mass of one mole of elemental sulfur.
Basically you have calculated the molar mass of Sx, and you are looking for "x". To find x just divide 74.6(your mass) by 32 ( the mass of sulfur from PT) and you get 2. Thus your result is S2. So forget that S3 mentioned before, that was from the data on your first post.
If the experiment had worked as planned, the falue for Kf should have been about 7, and the molar mass of sulfur about 250. However, always calculate using your data, do not work backwards in labs.
Valdo
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a molecule in this lab contains naph and sulfur
No. There are separate molecules of naphthalen and separate molecules of sulfur. They are mixed. Do you know what is a chemical compound and what is a mixture?
mixture doesnt react with each other, compound does. I think i understand how many atoms there are now....i still dont understand why teh mass of sulfur increaseed or how it did, idk i just can't think of it in my mind correctly.
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i still dont understand why teh mass of sulfur increaseed
So we are back at the beginning. Because the solution (in which naphthalene is a solvent) contains not atomic sulfur (S1, molar mass 32) but molecules of sulfur (S8, molar mass 8*32). You are determining molar mass of the DISSOLVED substance, and this dissolved substance is MOLECULAR sulfur.
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okay i understand that, only question i have left is why it should be 250/ s8, any specific reason or is that just the way experimenst ahve gone in the past
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S8 is confirmed experimentally on many occasions. Just like O2 is O2 and H2O is H2O.