2. Looks OK. Since the mixture of x and y had a lower m.p., x and y are different compounds. Since the m.p. of the mixture of y and z was unchanged, y = z as you said.
>> What would you expect as an approx. melting range for 2:1 mix. of z and y? <<
If y and z are the same compound, their mixture melting point should be unaffected whatever the ratio of y and z is. Perhaps they meant a 2:1 mixture of x and z. In that case, the melting point should be the same as the 2:1 mixture of x and y, 105-130°.
3. If the rate of heating is the same, a large, solid mass will take longer to melt than a powdered sample, and thus will have a broader range and won't finish melting until the temperature is higher, since the temperature is increasing constantly during melting. But in this example, the remelted solid has a higher but sharp m.p. (only 1°), indicating a fairly pure substance. Thus a chemical change, such as a rearrangement, is a possibility, but the change would have to be a quantitative one (100%), or you would have a mixture that would likely have a lower and broader m.p. range. This is common when decomposition accompanies heating. Solvates such as hydrates can lose some or all of the solvent molecules on heating, leaving a less solvated compound which subsequently has a higher melting point. But I'm inclined to think that some kind of structural change is occurring here. For example, the compound flavaspidic acid has two forms, a and b. The a-form melts at 92° and solidifies again at 110°. The melting point of this solid is now 156°, the same as the b-form of the compound. (From the Merck Index.)
4. That's what I'd expect too.
5. The effect of too rapid of heating depends on the response time of the thermometer compared to the sample capillary tube and sample. In our lab we use Mel-Temp apparatuses, which have an electrically-heated metal block in contact with the sample capillary and an ordinary glass thermometer. With these, I've noticed that too rapid of heating gives a low m.p. with a very narrow range. What happens is the sample heats almost instantaeously, but the large thermometer lags behind because it takes longer to heat. We don't have Gallenkamp apparatuses, but I get the impression that the response of the temperature sensor is very rapid. So, if the temperature response of the sensor is immediate and the sample capillary and sample lags a little behind, then I would expect a similar result as yours; a slightly higher melting point with a broader range.
Hope this helps a little! Good questions, not as straightforward as they first appear!
Steve
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