Question

Question #90e5d

Answer 1

I gave this a go and I got a number close to `50%`.

Let us name the two compunds `XO_A` and `XO_B`.

We know that the ratio of some element, let us say `X`, to oxygen in `XO_A` is `1:0.29`.

That is, for every `1g` of `X` we have `0.29g` of `O_A`. (1)

Now, taking into account the oxygen-to-oxygen ratio between `XO_A` and `XO_B` is `2:7`, we can say that

for every `2g` of `O_A` we have `7g` of `O_B` (2) or, for simplicity, for evey `1g` of `O_A` we get `3.5g` of `O_B`.

Therefore, using (1) and (2), we can say that

for every `1g` of `X` we have `0.29 * 3.5 g = 1.02g` `O_B` for the second coumpound.
An approximately `1:1` ratio of `X` and `O_B` in `XO_B` means that the percent composition of `X` to `XO_B` is

`percentcomposition = 1/(1+1) *100 = 50%`