# Question 90e5d

Dec 5, 2014

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

Let us name the two compunds $X {O}_{A}$ and $X {O}_{B}$.

We know that the ratio of some element, let us say $X$, to oxygen in $X {O}_{A}$ is $1 : 0.29$.

That is, for every $1 g$ of $X$ we have $0.29 g$ of ${O}_{A}$. (1)

Now, taking into account the oxygen-to-oxygen ratio between $X {O}_{A}$ and $X {O}_{B}$ is $2 : 7$, we can say that

for every $2 g$ of ${O}_{A}$ we have $7 g$ of ${O}_{B}$ (2) or, for simplicity, for evey $1 g$ of ${O}_{A}$ we get $3.5 g$ of ${O}_{B}$.

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

for every $1 g$ of $X$ we have $0.29 \cdot 3.5 g = 1.02 g$ ${O}_{B}$ for the second coumpound.
An approximately $1 : 1$ ratio of $X$ and ${O}_{B}$ in $X {O}_{B}$ means that the percent composition of $X$ to $X {O}_{B}$ is

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