Question

How many nitrate ions, `"NO"_3^(-)`, and how many oxygen atoms are present in `1.00` `mu"g"` of magnesium nitrate, `"Mg" ("NO"_3)_2`?

Also, what mass (in grams) of oxygen is present in `1.00` `mu"g"` of magnesium nitrate?

Answer 1

Here's what I got.

Explanation:

Start by converting the mass of magnesium nitrate

`1.00color(red)(cancel(color(black)(mu"g"))) * "1 g"/(10^6color(red)(cancel(color(black)(mu"g")))) = 1.00 * 10^(-6)` `"g"`

to moles by using the molar mass of the compound.

`1.00 * 10^(-6) color(red)(cancel(color(black)("g"))) * ("1 mole Mg"("NO"_3)_2)/(148.3color(red)(cancel(color(black)("g")))) = 6.743 * 10^(-9)` `"moles Mg"("NO"_3)_2`

Next, convert the number of moles of formula units of magnesium nitrate by using Avogadro's constant

`6.743 * 10^(-9) color(red)(cancel(color(black)("moles Mg"("NO"_3)_2))) * (6.022 * 10^(23)color(white)(.)"f. units Mg"("NO"_3)_2)/(1color(red)(cancel(color(black)("mole Mg"("NO"_3)_2))))`

`= 4.061 * 10^(15)` `"f. units Mg"("NO"_3)_2`

Now, you know that one formula unit of magnesium nitrate contains

• one magnesium cation, `1 xx "Mg"^(2+)`
• two nitrate anions, `2 xx "NO"_3^(-)`

This means that your sample will contain

`4.061 * 10^(15) color(red)(cancel(color(black)("f. units Mg"("NO"_3)_2))) * ("2 NO"_3^(-)color(white)(.)"anions")/(1color(red)(cancel(color(black)("f. unit Mg"("NO"_3)_2))))`

`= color(darkgreen)(ul(color(black)(8.12 * 10^(15)color(white)(.)"NO"_3^(-)color(white)(.)"anions")))`

The answer is rounded to three sig figs, the number of sig figs you have for the mass of magnesium nitrate.

Now, you know that each nitrate anion contains

• one nitrogen atom, `1 xx "N"`
• three oxygen atoms, `3 xx "O"`

This means that your sample contains

`8.12 * 10^(15)color(red)(cancel(color(black)("NO"_3^(-)"anions"))) * "3 atoms of O"/(1color(red)(cancel(color(black)("NO"_3^(-)"anion"))))`

`= color(darkgreen)(ul(color(black)(2.44 * 10^(16)color(white)(.)"atoms of O")))`

The answer is rounded to three sig figs.

Finally, to find the mass of oxygen present in the sample, use the fact that each mole of magnesium nitrate contains

`2 xx 3 = "6 moles of O"`

This means that you sample contains

`6.743 * 10^(-9) color(red)(cancel(color(black)("moles Mg"("NO"_3)_2))) * "6 moles O"/(1color(red)(cancel(color(black)("mole Mg"("NO"_3)_2))))`

`= 4.046 * 10^(-8)` `"moles O"`

Use the molar mass of oxygen to convert this to grams

`4.046 * 10^(-8) color(red)(cancel(color(black)("moles O"))) * "15.9994 g"/(1color(red)(cancel(color(black)("mole O"))))`

`= color(darkgreen)(ul(color(black)(6.47 * 10^(-7)color(white)(.)"g")))`

Once again, the answer is rounded to three sig figs.