# 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 $1.00$ $\mu \text{g}$ of magnesium nitrate?

Jun 15, 2017

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) $\text{g}$

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))))

$= 4.061 \cdot {10}^{15}$ "f. units Mg"("NO"_3)_2

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

• one magnesium cation, $1 \times {\text{Mg}}^{2 +}$
• two nitrate anions, $2 \times {\text{NO}}_{3}^{-}$

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))))

$= \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{8.12 \cdot {10}^{15} \textcolor{w h i t e}{.} \text{NO"_3^(-)color(white)(.)"anions}}}}$

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, $1 \times \text{N}$
• three oxygen atoms, $3 \times \text{O}$

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"))))

$= \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{2.44 \cdot {10}^{16} \textcolor{w h i t e}{.} \text{atoms of O}}}}$

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

$2 \times 3 = \text{6 moles of O}$

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))))

$= 4.046 \cdot {10}^{- 8}$ $\text{moles O}$

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"))))

$= \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{6.47 \cdot {10}^{- 7} \textcolor{w h i t e}{.} \text{g}}}}$

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