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Topic: Practice problem 5  (Read 4037 times)

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Rutherford

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Practice problem 5
« on: April 30, 2014, 01:28:40 PM »
A mixture of carbon dioxide and hydrogen enriched with a heavier isotope (deuterium) has been used to synthesize a sample of methanol. Measurements carried out by means of a mass spectrometer have shown that in the sample the amount of methanol molecules containing three atoms of deuterium and one atom of light hydrogen is 1.55 times larger than the amount of methanol molecules containing two atoms of deuterium and two atoms of light hydrogen. Calculate the percentage of deuterium (in atomic percent) in the hydrogen sample used for the synthesis.

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Re: Practice problem 5
« Reply #1 on: April 30, 2014, 01:57:35 PM »
Hard to say for sure whether I understand the problem correctly, but I got 69.9% (3sf). This was on the basis that the fractions of each isotope of hydrogen in the methanol would be the same as the fractions of each isotope of hydrogen in the "hydrogen sample used for synthesis".

Rutherford

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Re: Practice problem 5
« Reply #2 on: April 30, 2014, 02:01:17 PM »
Statistically it should. The answer is correct. If you want, you can explain a bit the calculation for others. And it's your turn to post a problem.

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Re: Practice problem 5
« Reply #3 on: April 30, 2014, 02:27:05 PM »
The fraction of moles of C(1H)(2H)3O is equal to 1.55 times the fraction of moles of C(1H)2(2H)2O (out of total number of moles of methanol).

These fractions are given by statistical analysis based on the probability of a particular H atom being 1H or 2H, which is the same as the fraction of H atoms which are 1H or 2H respectively.

Rutherford

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Re: Practice problem 5
« Reply #4 on: April 30, 2014, 02:54:49 PM »
And the equation would be: 1.55 : 1 = 4p(1H)·p(2H)3 : 6p(1H)2·p(2H)2. Probability of 1H can be expressed through probability of 2H and then it's easy to calculate the percentage.