# Calculating molar mass of an oxidant by iodometry

I am stuck on a question from the Australian Chemistry Olympiad involving molar mass, shown here:

BTW, this whole paper is available online for free, and is distributed by its owners for free. Just search up: Australian Chemistry Olympiad Papers and look at the past papers.

Anyways, in this question, I am able to solve part a and part b, but part c is a little tricky. I understand how to solve for molar mass: it is simply the mass of the substance divided by the number of moles in the same amount of substance. However, in the solution they use the number of moles of the metal cation M, as well as the subscript variable x. Can someone please describe in detail what this means. If there are any more questions, feel free to comment below. Thanks in advance.

P.S. The answers are not my own, but those from the answer sheet.

• What you did is correct. Why do you have doubts? – Alchimista Jul 22 '17 at 2:50
• The answers are not my own, but from the answer sheet – Jainil Shah Jul 22 '17 at 2:59
• I don’t understand how the person got part c – Jainil Shah Jul 22 '17 at 2:59
• @Abhigyan the answers are from the answer sheet – Jainil Shah Jul 22 '17 at 3:01

The molar mass of a substance, as you said earlier, is:

simply the mass of the substance divided by the number of moles in the same amount of substance

In this question, you can see that as per part b), we found that there were $7.553\times10^{-4} \text{ moles}$ of $\ce{M^{n+}}$ present.

Hence, the salt, i.e., $\ce{M_xA_y.ZH2O}$ contains $\ce{M_x}$ = $7.553\times 10^{-4} \text{ moles}$ $\ce{M^{n+}}$.

=> $\ce{M}$ = $\frac{7.553\times 10^{-4}}{x} \text{ moles}$ of $\ce{M^{n+}}$

$\therefore$ taken moles of $\ce{M_xA_y.ZH2O}$ = $\frac{7.553\times 10^{-4}}{x} \text{ moles}$

But we know that the taken moles have a mass of $0.2642 g$.

Now, divide $0.2642 g$ by $\frac{7.553\times 10^{-4}}{x} \text{ moles}$ to get the molar mass, in terms of x, which is $349.8x$