# Calculating yield percentage of magnetite synthesis

I’ve made magnetite $$\ce{Fe3O4}$$ in the lab and now I need to calculate yield percentage. How can I do it? I have these data.

\begin{align} \ce{FeSO4.7H2O(aq) &-> Fe^2+(aq) + SO4^2- (aq) + 7H2O(l)}\\ \ce{KOH(aq) &-> K+(aq) + OH- (aq)}\\ \ce{KNO3(aq) &-> K+(aq) + NO3- (aq)}\\ \ce{Fe^2+(aq) + 2OH- (aq) &-> Fe(OH)2(s)}\\ \ce{3Fe(OH)2(s) + NO3- (aq) &-> Fe3O4(s) + NO2- (aq) + 3H2O(l)}\\ \end{align}

\begin{align} \ce{FeSO4.7H2O &->} 4.20~\mathrm{g}\\ \ce{KOH &->} 1.90~\mathrm{g}\\ \ce{KNO3 &->} 0.12~\mathrm{g}\\ \text{Magnetite}~\ce{&->} 0.36~\mathrm{g}\\ \end{align}

I know how to calculate the yield for simple reactions but I don’t know how to do it when there are many reactions involved.

• Ok thanks, as I said I know how to calculate yield for simple reactions but not for multistep synthesis. (ac) state symbol is the same as (aq) except that in spanish it is common to write (ac) ("aqueous" = "acuosa"), english is not my native language sorry. – Ragnar Jun 18 '15 at 0:30
• then share the result with us -post an answer – Jaroslav Kotowski Jun 18 '15 at 9:20

## 1 Answer

Add up the reactions (double the second one so there is sufficient hydroxide) to get a net reaction with your starting materials as reactants and your desired product as one of the products:

$$\ce{3FeSO4.7H2O(aq) + 6KOH(aq) + KNO3(aq) -> Fe3O4(s) + NO2-(aq) + ...}$$

Then, find the limiting reactant, calculate the theoretical yield from it, and compare it to the amount of product to figure out the percent yield.

• I think it should be of 3 mol of FeSO4.7H2O and 6 mol KOH since based on the equations, 3 mol of Fe(OH)3 react with 1 KNO3 – Simon-Nail-It May 5 at 4:17
• @Simon-Nail-It Yup, you nailed it. – Karsten Theis May 9 at 12:44