If the equivalent weight of a metal (M) is x | vedclass

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(C) For the oxide $M_mO_n$,let the atomic weight of $M$ be $A$.The mass of $M$ in the compound is $(m 	imes A) 	ext{ g}$ and the mass of oxygen is $

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$\frac{2nx}{m}$

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(C) For the oxide $M_mO_n$,let the atomic weight of $M$ be $A$.The mass of $M$ in the compound is $(m 	imes A) 	ext{ g}$ and the mass of oxygen is $(n 	imes 16) 	ext{ g}$.According to the law of equivalence:$\frac{	ext{Mass of } M}{	ext{Equivalent weight of } M} = \frac{	ext{Mass of } O}{	ext{Equivalent weight of } O}$Given that the equivalent weight of $M$ is $x$ and the equivalent weight of oxygen is $8$:$\frac{m 	imes A}{x} = \frac{n 	imes 16}{8}$$\frac{m 	imes A}{x} = 2n$$A = \frac{2nx}{m}$

Isotope	$\%$ Natural abundance	Molar mass
$^1H$	$99.985$	$1$
$^2H$	$0.015$	$2$

If the equivalent weight of a metal $(M)$ is $x$ and the formula of its oxide is $M_mO_n$,then the atomic weight of $M$ is ......

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