Confusion regarding a stoichiometric formula

In a book, which contains only stoichiometry problems, I came across this formula.



> 
> For any given compound $\ce{M\_xN\_y}$, $x \text{ Moles of N} = y \text{ moles of M}$
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> 
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According to this rule if I take $\ce{H2O}$ then, $x = 2 \text{ and } y = 1$.


So by the above formula,


$2 \text{ moles of Oxygen} = 1 \text{ mole of Hydrogen}$


But surely,


$2 \* \mathrm{N\_A} \not= 1 \* \mathrm{N\_A}$, where $\mathrm{N\_A}$ is Avogadro's number.


However, in an example, the author has used this formula and successfully solved a problem? So, if it's correct, can anyone prove this and point out where my contradiction is wrong.


**Example as requested:**



> 
> What weight of oxygen will react with $1$g of calcium?($\ce{Ca}= 40\mathrm{g \over mol}$)
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> 
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Answer by author :
Since all the atoms of $\ce{Ca}$ have changed into $\ce{CaO}$, the amount of $\ce{Ca}$ in $\ce{CaO}$ is $1$g. Now from the formula of $\ce{CaO}$, we have,


$\text{Moles of Ca = Moles of O}$


$$\frac{\text{mass of Ca}}{\text{atomic mass of Ca}}= \frac{\text{mass of O}}{\text{atomic mass of O}}$$


therefore, $\text{mass of O} = {1\over40} \* 16 = 0.4$g




You misread the formula. For compound $\ce{M\_{x}N\_{y}}$


$$y\cdot \ce{mol M} = x\cdot \ce{mol N}$$


This is because the formula essentially tells you that the ratio between moles of $\ce{M}$ and $\ce{N}$ is:


$$\frac{\ce{mol M}}{\ce{mol N}} = \frac{x}{y}$$


This rearranges to the equation above.


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