Paige heated 3.00 g mercury (II) oxide (HgO, 216.59 g/mol) to form mercury (Hg, 200.59 g/mol) and oxygen (O2, 32.00 g/mol). She collected 0.195 g oxygen. What was the percent yield of oxygen?

Question

justjm · Answer

You need to convert grams of HgO to mols and then use the stoichiometric ratios to find mols of oxygen expected to be formed, and then convert mols of oxygen to grams again and compare. Then find the percent yield. I'll elaborate more in a few minutes.

justjm · Answer

First write the balanced equation:
$2HgO \rightarrow 2Hg+O_2$
Now use dimensional analysis to find the expect # of mols of $O_2$

$n_{O_{2}}=3.00~g~HgO\times\frac{1~mol~HgO}{200.59~g/mol}\times\frac{1~mol~O_2}{2~mol~HgO}\times\frac{32.00~g/mol}{1~mol~O_2}$

Calculate.

Once you get your value for $n_{O_2}$, you must find the percent yield.

$\%_{yield}=\frac{\text{actual yield}}{\text{theoretical yield}}$

The theoretical yield would be your value for $n_{O_2}$, and the actual yield is 0.185 g oxygen.

You can calculate and use 3 s.f. for your final answer.