# Need help with stoichiometry problem

I'm pretty new to chemistry and I've been stuck on it for hours.

Question: $\pu{10g}$ of the hydroxide of a metal on ignition gave $\pu{8g}$ of oxide. The equivalent weight of the metal is:
a) $\pu{136g}$
b) $\pu{40g}$
c) $\pu{56g}$
d) $\pu{28g}$

I used the law of equivalence but the answer I got was $\pu{3g}$. what am I doing wrong?

10/E=8/16;
E=20;
E=M(equivalent mass of metal)+16+1;
M=3

I found your numbers confusing, but I think the first equation you wrote is not correct. This is how I'd solve it:

Let's say we have $\mathrm{n}$ equivalents of metal in the sample, and let's call the equivalent weight of the metal $\mathrm{e}$. For the hydroxide sample, we have:

$n(e+17)=10$

where $\mathrm{e+17}$ is the molar mass of the hydroxide $\ce{MOH}$.

For the oxide, we have:

$n/2(2e+16)=8$

where $\mathrm{2e+16}$ is the molar mass of the oxide $\ce{M_2O}$. Note that $\mathrm{n}$ becomes $\mathrm{n/2}$ because of stoichiometry:

$\ce{2MOH->M2O + H2O}$

So you have a system of two equations with two unkowns ($\mathrm{n}$ and $\mathrm{e}$), which you can solve normally. For instance, if you divide the equations:

$\cfrac{n(e+17)}{n/2(2e+16)}=\cfrac{10}{8}$

$\cfrac{e+17}{e+8}=\cfrac{10}{8}$

$8e+136=10e+80$

$2e=56$

$e=28$

So the equivalent weight of the metal is 28g.

Makes sense?

Nothing wrong with Vic's answer. I'd solve it differently, but hopefully you are familiar enough with algebra to understand that there are various ways to correctly come up with the correct answer...
Here's how I did it: There is a loss of 2 grams of what we assume is water. Now since we know we started from M?(OH)? we assume the reaction was 2OH → H2O + O
meaning ~34 grams of OH would produce 18 g of water and 16 g of oxide oxygen. Since we know that ~2 grams were lost (assumed to be only water, and all of the water) then we know 2g of water must have come from (2/18)*34 grams of OH (2:18 as X:34), or 34/9 grams of OH. Since we started with 10 grams total, we started with 90/9 grams total and of that 34/9 is OH. Subtract to determine how much metal: (90-34)/9 = 56/9 grams of metal. So far, so good. But what is its oxidation state? Since the question is asking for equivalent weight, we need some way to pin down the equivalency. Well, for every 18 grams of water lost, the residue retained (we assume) 16 grams of oxide. So, 2:18 as X:16 or x= 16/9. So now, we have the whole enchilada: 90/9ths grams = 56/9 M + 16/9 O + 18/9 OH (as H2O - meaning this treats the OH lost during ignition as H2O rather than OH ) leaving us, after ignition with 56/9 M and 16/9 O. For each equivalent of O= there should be one equivalent of M. This means that since O has an equivalent weight of of 8 that 16/9 ÷ 8 = 2/9 Eq are present (after ignition) so that there must also be 2/9 equivalents of M. 56/9 grams ÷ 2/9 equivalents = 56/2 g/eq = 28 g/eq which is the eq. wt of the Metal. (A quick search of the periodic table indicates that Fe with an AtWt of 55.845 and assuming we're talking about Fe(OH)2 would produce 2.00 grams of water, theoretically. So, it checks out.)