# Question 77d3f

Feb 1, 2016

${\text{40 g mol}}^{- 1}$

#### Explanation:

The interesting thing about this problem is that you don't really need to know the compound's molecular formula, but you do need to know its empirical formula.

As you know, a compound's empirical formula tells you the smallest whole number ratio that exists between the elements that make up said compound.

In this case, if the molecular formula is said to ${\text{M"_4"O}}_{8}$, the empirical formula will be ${\text{M"_1"O}}_{2}$, since $1 : 2$ is the smallest whole number ratio that exists between $4$ and $8$.

So, your compound contains two moles of oxygen for every one mole of $\text{M}$.

Now, a $\text{18.00-g}$ sample of this compound is said to contain a total of $\text{10 g}$ of $\text{M}$. This of course implies that it contains

${m}_{\text{compound" = m_"M" + m_"O}}$

${m}_{\text{O" = "18.00 g" - "10 g" = "8 g oxygen}}$

At this point, all you really need to know if how many moles of oxygen you have in the sample. Use oxygen's molar mass to help you with that

8 color(red)(cancel(color(black)("g"))) * "1 mole O"/(16.0 color(red)(cancel(color(black)("g")))) = "0.5 moles O"

This means that sample must contain

0.5 color(red)(cancel(color(black)("moles O"))) * "1 mole M"/(2color(red)(cancel(color(black)("moles O")))) = "0.25 moles M"

Finally, to get the molar mass of element "M", divide its mass by the number of moles it contains

M_M = "10 g"/"0.25 moles" = color(green)("40 g mol"^(-1))#