3
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.