What is the approximate Molecular Weight of A?

Question

The condensate from a steam distillation contains 10g of compound A and 18g of Water. At the temperature of the distillation, the vapor pressure of Water is 660 Torr.

Answer 1

Hi Halie! This is a nice lab question. I assume that your course instructor provided some instruction about how to do this calculation, perhaps as a part of the lab procedure or in an accompanying textbook. If you have access to a lab-focused textbook (I grabbed "Microscale and Macroscale Techniques in the Organic Laboratory" by Pavia, Lampman, Kriz, and Engel), head to the section/chapter on steam distillation for a more complete description.

First of all, when a mixture of immiscible liquids is boiled (which is what a steam distillation is), the overall vapor pressure is the sum of the vapor pressures of the two liquids (Ptot = PA + PB). As such, if the vapor pressure of the water was 660 torr at the temperature of distillation, that means the vapor pressure of the organic liquid was 100 torr (since atmospheric pressure is 760 torr, and the mixture will boil when its vapor pressure reaches atmospheric pressure).

Next, the mole ratio of the distillate is equal to the ratio of the partial pressures of the two liquids: [Moles A / Moles B] == [Pressure A / Pressure B]. This is because, from the ideal gas law, we know that all other things being equal (T, V), pressure is proportional to moles. Solving for Moles A (where B is water) gives:

Moles A == [PA/PH2O] * Moles H2O

The distillation formed 1.0 mole of water (18 g being one mole), so:

Moles A == [100 / 660] * 1 == 0.15 moles

Combining that with the mass of A formed (10 g) to give molecular weight is simple division:

10 g A / 0.15 moles A == 67 g/mol

Note on sig figs. I did this calculation using 2 sig figs, despite the fact that in order to justify them, the problem should have stated "10. g of compound A" -- with the decimal place (the other numbers are all given to two sig figs). Without the decimal after the 0, the mass of compound A really only has 1 sig fig, and you should technically round to 70 g/mol.

Also, I'll note that in the above example, I rounded the number of moles of A down to 0.15 from 0.151515.... Good computational practice (I think) is to keep extra digits during intermediate calculations, then round to the appropriate number of sig figs after all calculations are done. Doing that changes the result by 1 g/mol, as 10/.151515151515... == 66 g/mol, whereas 10/.15 = 66.6 g/mol, which rounds to 67 for two sig figs. (You can also obtain that exact result by combining the two calculations, and doing 10 / [(100/660) * 1] which equals exactly 66.) That said, the whole idea with sig figs is that there is always some computational uncertainty in the last digit (which reflects the uncertainty in the measurements of the experimentally determined data), so it's not surprising that doing the calculation slightly differently could affect the last digit by 1. Yes, even organic chemists should know and understand how to appropriately use and report only digits which are significant!

One final note.... This molecular weight seems moderately reasonable. If you got something back to front somewhere during the calculation (like, had your pressures upside down with 660/100, rather than 100/660), you are likely to have gotten a molecular weight of a very different order of magnitude... It's always prudent to make sure your calculated answer passes the "sniff test". If, say, you calculated that the molecular weight of the compound was 6.7 g/mol, instead of 67 g/mol, that would be a clue that you screwed something up, and should go back and check your work. A single atom of carbon weighs 12 g/mol, so 6.7 g/mol doesn't make sense as the MW for any organic compound. On the other hand, if you calculated its MW to be 670 g/mol, that's harder to rule out based on a "sniff test".... Many organic compounds are that large, though very few of them would likely have a vapor pressure of 100 torr at a temperature below 100 C. At the least, though, a calculated MW that large would give me pause and suggest I check my work. On the other hand, a MW of 66 g/mol is perfectly reasonable for an organic compound which has a vapor pressure of 100 torr just below 100 C. I think the data presented here probably do not correspond to any actual organic compound, and were just "made up" for computational simplicity, but they are at least reasonable. And it's always nice when your calculated result seems reasonable (ie, passes the "sniff test"). (Making the mistake I described above -- having the pressures upside down -- actually gives the result of 1.5 g/mol for the molecular weight of A. That does NOT pass the sniff test!)