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


  For any given compound $\ce{M_xN_y}$, $x \text{ Moles of N} = y \text{ moles of M}$


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?($\ce{Ca}= 40\mathrm{g \over mol}$)


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.