Measuring Molecular Mass with Freezing Point Depression

[Vavi Ask]

Homework Statement

6g of non-volatile solute dissolved in 500g of water freezes at 272.64K. What is the molar mass of the solute?

Data given-
T_solution = 272.64K
K_f = 1.8
w_solute = 6g
w_solvent = 500g

2. Homework Equations
∆T_f = T₀ -T_solution
∆T_f = K_f × molality (m)

The Attempt at a Solution

∆T_f = T₀-T_solution
= 273-272.64
=0.36

∆T_f = K_f x m
⇒0.36 = 1.8 × m
∴m = 0.36 / 1.8
= 0.05 m

molality = ( w_solute / m_solute ) × (1000 / w_solvent )
⇒0.05 = ( 6 / m_solute ) × ( 1000 / 500 )
∴m_solute = ( 6 / 0.05 ) × ( 1000 / 500 )
= 12 / 0.05
= 240 g
Is this correct?

 

[Vavi Ask]

Someone please help me. Please check whether I'm right. It's important. I usually get immediate replies. Why not today?

 

[epenguin]

Assuming it is roughly right, which I have not checked, you would need to correct two numbers to get it quite right:
Since the depression is a small temperature change you need to use a more accurate value, 273.16 for freezing point of water, and I also see here http://www.chemteam.info/Solutions/FP-depression.html a value for K_f different enough to make a significant difference in the result. You'd practically want that accuracy for useful conclusions in many cases. The measurements look only to within about 3% and you don't want to make it worse. I hope 6 g means 6.00 g not roughly 6 g!

 

[mjc123]

0.36/1.8 ≠ 0.05

 

[Vavi Ask]

Assuming it is roughly right, which I have not checked, you would need to correct two numbers to get it quite right:
Since the depression is a small temperature change you need to use a more accurate value, 273.16 for freezing point of water, and I also see here http://www.chemteam.info/Solutions/FP-depression.html a value for K_f different enough to make a significant difference in the result. You'd practically want that accuracy for useful conclusions in many cases. The measurements look only to within about 3% and you don't want to me it worse. I hope 6 g means 6.00 g not roughly 6 g!

Thank you so much sir!

 

[epenguin]

You're welcome. I have two further points to make.

Firstly the, at first sight, small error, or oversight causes quite a significant error in the conclusion. Now this is not just a pedantry – I deduce that the question was designed to test your understanding of this point. Because I can't see why anyone would want to quote a freezing point in the context of freezing point depression relative to that of water, in absolute temperature (°K) rather than centigrade if not for this purpose!

I have further scientific points, but I have to go out right now so come back later.

 

[epenguin]

So we were saying… the question was trying to test your appreciation of the points mentioned in the last post and test or teach appreciation of precision in science. Which does not mean this stereotype of I'm a scientist I have to be terribly precise. I have seen students do things like very carefully weigh out things to the nearest milligram for use in experiments in which there was a 20% error down the line. So the precision has to be that appropriate for what you're trying to achieve. But here a small fraction of adegree made a significant difference.

The problem set me thinking. At school decades ago we were drilled in endless calculations of freezing point depression (and BP elevation) - it was the kind of thing you could pick up good marks in an exam. But I do not remember that we did a single relevant experiment, which maybe means we didn't really think about what was involved.

Freezing point depression by one molal concentration of the FP of water is only about half a degree. And one molal is a fairly hefty concentration – that would take you well out of ideality I would think for many substances, anyway there would be solubility limitations, you would want to use less I imagine. Only now I wondered how easy it is to measure considerably smaller temperature changes. How would I do it (how would you do it)?

OK it's fairly easy to measure to 0.1° C - a common clinical thermometer does that and we consider a measured variation of 0.1° is telling us something significant that we rely on. Most people will still be familiar with with mercury-in-glass thermometers. But I can't remember ever seeing one that measured better than. 0.1°. I looked up catalogues on the net and saw numbers of thermometers that measured from -1 to +5° but none were given better then 0.1° precision. However I would imagine that some have been manufactured just for this scientific purpose. The problem is if you want to measure the molecular mass of a substance of which this is only 100 to within one dalton at 1 molal you need precision of 1% of 0.6° = 0.006°. Seems difficult.

I guess part of the answer is: they didn't do it precisely. And That they needed only an approximate measurement, and then got the exact molecular mass from knowing the composition of the compound and atomic masses. That they only needed to distinguish whether a compound was C_xH_yO_z or C_2xH_2yO_2z or something else with the same atomic ratios.

I guess I might use an electrical resistance thermometer - electrical resistance varies with temperature and can be measured quite precisely. But I don't know and do doubt that this is what Raoult did. For the mercury thermometer the sensitivity and the min-max range depend on bulb volume and diameter of the mercury column. And also its length–but I have never seen one longer than about 40 cm at most.

You'd think they would have wanted to get past the limitations of instruments that I have sketched. I read (never heard of it before) that soon after the start of the 20th century an instrument now called, after its inventor, the Beckman thermometer came into use. This makes it "possible to estimate temperature changes to 0.001 °C". https://en.wikipedia.org/wiki/Beckmann_thermometer (From the description which says it has a 5° range I struggle to understand how you can get it as sensitive as claimed, but perhaps one can be made so). It is pointed out that it does not measure temperatures, but only temperature changes. It was not very obvious to me what was the point of it at all unless it is intended for the same instrument to be used with different solvents. (If one is concerned with FP depression with water solvent then temperature centigrade is a temperature difference.) If I were doing the experiments in the 1880s or so I would not care what the temperature of anything was, I would just compare quantitatively FP depression by my unknown substance with that by substance of known molecular mass (so all the calculations we dutifully did at school were pointless).

Checking background I realized that my knowledge of the history of chemistry of this period was sketchy. I was surprised that Raoult’s FP depression work was published as late as 1878. I had imagined it earlier, for by that time molecular theory was well advanced and very many organic chemical structures known. After all qualitatively the FP depression phenomenon had been studied more than a century earlier by Fahrenheit and Black.

I also wondered what explanations or theories the scientists of those times had of these observations. Real molecular masses are something different from just composition proportions - it must have been baffling to know how you could get them. Did they regard to these laws is just empirical and being so useful not question their luck? (Which is how we regarded it in the school lessons I have mentioned, the great thing being able to get the right answer in the calculations. I never heard of such a thing as 'colligative properties’ till later.) Gibbs work was 'published'around the same time as Raoult's but I thought to many years past before it was understood or even widely known of. However according to a link in the biog sketch https://en.wikipedia.org/wiki/François-Marie_Raoult Goldberg had predicted the phenomenon just a few years earlier, with no doubt less perfected and more understandable account than Gibbs’. Somewhere I read that that following Raoult, FP depression was the main method of measuring molecular masses for many decades.