[Rutherford]

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.

[Big-Daddy]

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]

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.

[Big-Daddy]

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]

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.