Question #90e5d

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Question #90e5d

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Answer 1 · 2014-12-05T12:14:27

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

Let us name the two compunds $X O_{A}$ $XO_A$ and $X O_{B}$ $XO_B$ .

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

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

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

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

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

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

$p e r c e n t c o m p o s i t i o n = \frac{1}{1 + 1} \cdot 100 = 50 %$ $percentcomposition = 1/(1+1) *100 = 50%$

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Question #90e5d

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