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(C) The mass of the element is $A_1 \ g$.The mass of the oxide is $A_2 \ g$.Therefore,the mass of oxygen in the oxide is $(A_2 - A_1) \ g$.According t

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$\frac{A_1}{A_2 - A_1} 	imes 8$

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(C) The mass of the element is $A_1 \ g$.The mass of the oxide is $A_2 \ g$.Therefore,the mass of oxygen in the oxide is $(A_2 - A_1) \ g$.According to the law of equivalent proportions,the equivalent mass of the element is given by the formula:$	ext{Equivalent mass of element} = \frac{	ext{Mass of element}}{	ext{Mass of oxygen}} 	imes 	ext{Equivalent mass of oxygen}$.Since the equivalent mass of oxygen is $8$,we have:$	ext{Equivalent mass} = \frac{A_1}{A_2 - A_1} 	imes 8$.

If $A_1 \ g$ of an element gives $A_2 \ g$ of its oxide,then the equivalent mass of the element is .....

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