# Question 6a3c8

Feb 17, 2017

$\text{MO}$

#### Explanation:

The idea here is that the metal oxide contains only atoms of the unknown metal, $\text{M}$, and atoms of oxygen, $\text{O}$.

This means that the mass of the metal oxide can be expressed as the sum of the mass of the metal atoms and the mass of the oxygen atoms

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

Now, you know for a fact that the metal oxide contained $\text{0.64 g}$ of metal. This means that it also contained

${m}_{\text{O" = "0.72 g" - "0.64 g" = "0.08 g}}$

of oxygen. Since elemental oxygen has a molar mass of ${\text{8 g mol}}^{- 1}$, you can say that the initial sample of metal oxide contained

0.08 color(red)(cancel(color(black)("g"))) * "1 mole O"/(8color(red)(cancel(color(black)("g")))) = "0.01 moles O"

The unknown metal has a molar mass of ${\text{64 g mol}}^{- 1}$, which means that the initial sample of metal oxide also contained

0.64 color(red)(cancel(color(black)("g"))) * "1 mole M"/(64color(red)(cancel(color(black)("g")))) = "0.01 moles M"#

You now know for a fact that the initial sample contained

$\left\{\begin{matrix}\text{0.01 moles M" \\ "0.01 moles O}\end{matrix}\right.$

This shows that the metal and the oxygen are present in a $1 : 1$ mole ratio in the metal oxide. Since this is the smallest whole number ratio that can exist between the two elements, you can say that the empirical formula of the metal oxide is

$\text{MO } \to$ the empirical formula of the metal oxide