How do I calculate the mass percent in the original mixture?
Looks close. Your math looks sound, but the reasoning is messy. $\ce{O2}$ is the only oxygen here.

$$\ce{2KClO3 -&gt; 2KCl + 3O2}$$

You might have skipped writing out that reaction — it helps a lot.

If $0.700~\mathrm{g}$ of residue remained, that means $0.950\ \mathrm{g} - 0.700\ \mathrm{g} = 0.250\ \mathrm{g}$ of $\ce{O2}$ left the container (which means that $0.250\ \mathrm{g}$ of $\ce{O2}$ was made!)

We can calculate the number of moles, similar to what you did.

$0.250\ \mathrm{g} \times \frac{1\ \mathrm{mol}}{32\ \mathrm{g}} = 0.00781\ \mathrm{mol}\ \ce{O2}$

However, we have no reason to divide by 2 — $\ce{O2}$ is the only way (well, at least for this purpose) oxygen can exist in nature. Treat it as "oxygen".

Let's apply a mole ratio, now. There were $\ce{2 KClO3}$ molecules used for every $\ce{3 O2}$ molecules we got.

$0.00781\ \mathrm{mol}\ \ce{O2} \times \frac{2~\mathrm{mol}\ \ce{KClO3}}{3~\mathrm{mol}~\ce{O2}} = 0.00521\ \mathrm{mol}\ \ce{KClO3}$ which we had at the beginning. Note that we can do this because it says all of it reacted completely. If only $95~\%$ of it reacted, we would multiply by $0.95$, for example.

Find the grams of $\ce{KClO3}$ expended in reaction.

$0.00521\ \mathrm{mol} \cdot \frac{122.55\ \mathrm{g}}{1\ \mathrm{mol}} = x~\mathrm{g}\ \ce{KClO3}$

Your percent is pretty easy from there.

$\frac{x\ \mathrm{g}}{0.950\ \mathrm{g}} \times 100~\% = y~\%$

Where I'm sure you can figure out $x$ and $y$. (Remember to find $x$ to 3 significant digits).

Leave a comment if I wasn't clear on something!