# Question f0888

Jun 9, 2017

$6.0 \cdot {10}^{24}$

#### Explanation:

For starters, you should know that the oxide anion, ${\text{O}}^{2 -}$, is formed when a neutral oxygen atom, $\text{O}$, takes in $2$ electrons.

As you know, a neutral atom has equal numbers of protons inside its nucleus and electrons surrounding its nucleus.

This means that the number of electrons present in an oxygen atom is equal to the element's atomic number, $Z$, which, as you know, tells you the number of protons located inside the nucleus.

Oxygen's atomic number is equal to

${Z}_{{\text{O}}_{2}} = 8$

which implies that a neutral oxygen atom has $8$ electrons surrounding its nucleus. Consequently, an oxide anion will have

${\text{8 e"^(-) + "2 e"^(-) = "10 e}}^{-}$

surrounding its nucleus.

Now, use the molar mass of elemental oxygen, not of oxygen gas, to calculate the number of moles of oxide anions present in your sample--keep in mind that you can do this because the mass of an oxide anion is essentially equal to the mass of a neutral oxygen atom

16 color(red)(cancel(color(black)("g"))) * "1 mole O"^(2-)/(16.0color(red)(cancel(color(black)("g")))) = "1.0 moles O"^(2-)

Next, use Avogadro's constant to calculate the number of oxide anions present in the sample

1.0 color(red)(cancel(color(black)("moles O"^(2-)))) * (6.022 * 10^(23)color(white)(.)"O"^(2-)"anions")/(1color(red)(cancel(color(black)("mole O"^(2-))))) = 6.022 * 10^(23)color(white)(.)"O"^(2-)"anions"#

Since you know that every oxide anion has $10$ electrons surrounding its nucleus, you can say that the total number of electrons present in your sample will be

$6.022 \cdot {10}^{23} \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{O"^(2-)"anions"))) * "10 e"^(-)/(1color(red)(cancel(color(black)("O"^(2-)"anion")))) = color(darkgreen)(ul(color(black)(6.0 * 10^(24)color(white)(.)"e}}^{-}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass of the sample.