Question

Two oxides of metal M have $27.6$% and $30$% oxygen by weight . If the formula of the first oxide is $M_3{O}_4$, what is the formula of second oxide?

Answer 1

Given that, formula of first oxide $=M_3{O}_4$

Let mass of the metal $=x$

% of metal in $M_3{O}_4=(3x/3x+64)\times 100$

but as given % $age=(100-27.6)=72.4$ %

so,$(3x/3x+64)\times 100=72.4$

or $x=56$

in $2nd$ oxide,

oxygen $=30$%

So metal $=70$%

so, the ratio is

$M:O$

$\frac{70}{56}:\frac{30}{16}$

$1.25:1.875$

$2:3$

so, $2nd$ oxide is $M_2{O}_3$.