Two oxides of a metal X contain 50 % and 40 % | vedclass

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(C) In the first oxide $XO_2$,the percentage of oxygen is $50 \%$. This means the percentage of metal $X$ is also $50 \%$. Since the mass of $2$ oxyge

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$XO_3$

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(C) In the first oxide $XO_2$,the percentage of oxygen is $50 \%$. This means the percentage of metal $X$ is also $50 \%$. Since the mass of $2$ oxygen atoms is $2 	imes 16 = 32 \ g$,and this corresponds to $50 \%$ of the total mass,the mass of $X$ must also be $32 \ g$. In the second oxide,the percentage of oxygen is $40 \%$,which implies the percentage of metal $X$ is $60 \%$. Since the mass of $X$ remains $32 \ g$,we have $60 \% \equiv 32 \ g$. Therefore,$40 \% \equiv \frac{32}{60} 	imes 40 = \frac{64}{3} \approx 21.33 \ g$. Wait,let us re-evaluate: If $X$ is $60 \%$ and $O$ is $40 \%$,then $\frac{	ext{mass of } X}{	ext{mass of } O} = \frac{60}{40} = 1.5$. Given mass of $X = 32 \ g$,mass of oxygen $= \frac{32}{1.5} = 21.33 \ g$. Number of oxygen atoms $= \frac{21.33}{16} = 1.33$. This does not yield an integer. Let us re-read: 'Two oxides of $X$ contain $50 \%$ and $40 \%$ of non-metal'. If $X$ is the non-metal,then in $XO_2$,$X$ is $50 \%$. Mass of $X = 32 \ g$. In the second oxide,$X$ is $40 \%$,so oxygen is $60 \%$. Ratio $\frac{O}{X} = \frac{60}{40} = 1.5$. Mass of $O = 1.5 	imes 32 = 48 \ g$. Number of $O$ atoms $= \frac{48}{16} = 3$. Thus,the formula is $XO_3$.

Two oxides of a metal $X$ contain $50 \%$ and $40 \%$ of oxygen respectively. If the formula of the first oxide is $XO_2$,then the formula of the second oxide is:

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