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Difficult Question

Can Somebody please solve this chemistry question It came as a part of a previous test and i wasn't able to find a way to use ny/nx since all my answers would be negative. It wasn't a homework problem.

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    $\begingroup$ While this was not a homework, it is still a homework-like question. From our standpoint, there is no difference between homework problems, self-study problems, end-of-chapter problems, exam problems, example problems, etc. Whether it was assigned for a grade or not is not the key feature of such questions - they exist so that you can learn how to do these problems. $\endgroup$
    – Ben Norris
    Apr 27, 2014 at 10:38

1 Answer 1

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Non-cyclic (or acyclic) alkanes have the general formula $\ce{C_{n}H_{2n + 2}}$, and corresponding alkenes have the general formula $\ce{C_{n}H_{2n}}$. So in this problem let's say our alkane (x) is $\ce{C_{n}H_{2n + 2}}$, then our alkene (y) with one more carbon atom will be $\ce{C_{n + 1}H_{2(n + 1)}}$. The molecular weight of the alkane will be $$\ce{MW(x) = (12*n) + (1*(2n +2 )) = 14n + 2}$$ and $$\ce{MW(y) = ((12*(n + 1)) + (1*(2n + 1 ))) = 14n + 14}$$ We also know that grams(x) = grams(y), and that moles(x)/moles (y) = 7/5. The number of moles of a substance is equal to its weight in grams divided by its molecular weight. Therefore $$\frac{N_x}{N_y}=7/5=\frac{G(x)/14n+2}{G(y)/14n+14}$$ But in this problem grams(x) = grams(y) so our equation simplifies to $$7/5=\frac{14n+14}{14n+2}$$ Solving for "n" gives you the number of carbon atoms in the alkane, and "n+1" will be the number of carbon atoms in the alkene. I'll leave the mathematical solution and the naming of the two moleculesto you.

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  • $\begingroup$ It's not an assignment btw its is just a practice of a previous test the teacher gave us but didn't give its explanation. I wasn't able to solve it since i didn't understand how to use the mol ratio, and all the answers got to be negative which is impossibly so. But thanks for the answer. $\endgroup$
    – Jack
    Apr 26, 2014 at 12:17

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