Question

Magicians often use ‘flash powder’ in their acts. Flash powder contains 69.0% potassium chlorate (31.9% K, 28.9% Cl, 39.2% O) and the remainder is aluminium. Calculate the percentage composition by mass of each element in flash powder

Answer 1

`\rm K`: approximately
`22.0\%`.

`\rm Cl`: approximately
`19.9\%`.

`\rm O`: approximately
`27.0\%`.

`\rm Al`:
`31.0\%`.

Step-by-step explanation:

Consider a
`100\; \rm g` sample. There would be
`69.0\% * 100\; \rm g = 69.0\; \rm g` of
`\rm KClO_(3)` and
`100.0\; \rm g - 69.0\; \rm g = 31.0\; \rm g` of
`\rm Al`.

Out of that
`69.0\; \rm g` of
`\rm KClO_(3)`:

`31.9\%` (by mass) would be
`\rm K`:
`31.9\% * 69.0\; \rm g \approx 22.0\; \rm g`.

`28.9\%` (by mass) would be
`\rm Cl`:
`28.9\% * 69.0\; \rm g \approx 19.9\; \rm g`.

`39.2\%` (by mass) would be
`\rm O`:
`39.2\% * 69.0\; \rm g \approx 27.0\; \rm g`.

Overall, the composition (by mass) of each element in this mixture would be:

`\rm K`:

`\begin{aligned} & (31.9\% * 69.0\; \rm g)/(100\; \rm g) \\ =\; & 31.9\% * 69.0\% \\ \approx \; & 22.0\%\end{aligned}`.

`\rm Cl`:

`\begin{aligned} & (28.9\% * 69.0\; \rm g)/(100\; \rm g) \\ =\; & 28.9\% * 69.0\% \\ \approx \; & 19.9\%\end{aligned}`.

`\rm O`:

`\begin{aligned} & (39.2\% * 69.0\; \rm g)/(100\; \rm g) \\ =\; & 39.2\% * 69.0\% \\ \approx \; & 27.0\%\end{aligned}`.

`\rm Al`:

`\begin{aligned}& (100\; \rm g - 69.0\; \rm g)/(100\; \rm g) \\ =\; & 100\% - 69.0\% \\ =\; & 31.0\%\end{aligned}`.