[td04]

This is a question that I got last week for homework that I did not get the answer to because I was away.  Any help would be appreciated.

A 12.00g sample of a mixture of KClO₃ and KCl is heated until all the oxygen has been removed from the sample.  The product, entirely KCl, has a mass of 9.00 g.  What is the percentage of KClO₃ in the original mixture?

[mike]

You have lost 3 grams of O, or 0.1875 moles of O, or 0.0625 moles of O₃. Therefore there must have been 0.0625 moles of KClO₃, or 7.66 grams of KClO₃, or 7.66/12*100 = 64% (I think, maybe someone can double check this for me?)

[enahs]

(I think, maybe someone can double check this for me?)

You are correct.

You can also solve it another way, by calculating the number of moles of each component (since we have to relate things by moles and not mass). You know the molecular weight of each component, so the masses you have are just X and 12-X, which gives you the unit of moles = moles. So then you just have to solve for X to get the original mass.

So

Xg KClO₃       +        12-Xg KCl      =             9.00g KCl
122.55 g/mol            74.55 g/mol               74.55 g/mol

Where X = 7.66
7.66 g KClO₃
12-7.66 = 4.34 KCl (in original mixture).
Same numbers as Mike got.