Question #f0888
1 Answer
Explanation:
For starters, you should know that the oxide anion,
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,
Oxygen's atomic number is equal to
#Z_ ("O"_ 2) = 8#
which implies that a neutral oxygen atom has
#"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
#6.022 * 10^(23) color(red)(cancel(color(black)("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.