I just did a lab that involved heating KClO3 (with MnO2 as a catalyst) to decompose it into KCl and oxygen gas. That part is easy. What's throwing me is the % yield calculations.

The theoretical yield of O should just be the mass percentage of O in KClO3 (48.00/122.55 = KClO3 is 39.17% O by mass).

The actual yield of O2 (measured by subtraction) was 0.840 g out of a initial mass of 2.062 g KClO3, or 40.7%.

My question is this: Given that the theoretical yield was 39.17% based on moles of O (not O2) , why is the actual yield 40.7% (well with margins of error outlined by the professor) when it's based on O2? Shouldn't it be half as much? Or did I mess my calculations up?

Please feel free to edit tags, I'm out of my depth on this particular SE site in that area.


1 Answer 1


Your calculations are correct. The key is that you are measuring percentage by mass: the mass of oxygen in 2.062 g of $\mathrm{K}\mathrm{Cl}\mathrm{O}_3$ is the same whether it is considered the mass of individual oxygen atoms or of $\mathrm{O}_2$ (or of any other allotrope of oxygen, for that matter). You have half as many moles of $\mathrm{O}_2$, but the mass is the same (since the mass of 1 mol $\mathrm{O}_2$ is twice the mass of 1 mol O).


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