How do I find the average mass % of oxygen with the data collected?

Question

Table 1 Potassium Chlorate Reaction
Titration	Mass of Sample before heating (g)	Mass of Sample after heating (g)	Moles of KCl (mols)	Moles of O (mols)
1	0.495	0.467	0.00664	0.0156
2	0.490	0.343	0.00657	0.00459
3	0.494	0.300	0.00657	0.00459

Answer 1

The reaction is...

2KClO₃ ==>2 KCl + 3O₂(g)

It's unusual that they would provide moles of O, and they are indeed variable. Since O₂ gas escapes during the reaction, not sure how these data were obtained. At any rate...

mass % O = mass O / mass original sample (x100%)

We have the mass of the original sample, so we need the mass of O₂ produced. This O₂ is a gas and will be driven off or lost during this reaction. So, the difference between the final mass of KCl and the original mass of KClO₃ should be the mass of O in the sample.

Looking at trial #1:

% O = 0.495 g - 0.467 g / 0.495 g (x100%) = 5.66%

Trial #2:

%O = 0.490 g - 0.343 g / 0.490 g (x100%) = 30.0%

Trial #3:

%O = 0.494 g - 0.300 g / 0.494 g (x100%) = 39.3%

Very variable data, but take the average: 5.66 + 30 + 39.3 /3 = 25.0%