# Question b2080

Nov 3, 2015

Calcium

#### Explanation:

The idea here is that you need to use balanced chemical equation to find the mole ratio that exists between the metal nitrate and the remaining solid compound.

Now, the nitrate anion, ${\text{NO}}_{3}^{-}$, has a $\left(1 -\right)$ charge. Group 2 metals can form $\left(2 +\right)$ cations, which means that you'll need two nitrate anions to balance the positive charge of the cation.

The chemical formula of the compound can thus be written as

"M"("NO"_3)_2

Now, you know that the sample has a total mass of $\text{5 g}$. When you heat metal nitrates, the decomposition reaction that takes place produces a metal oxide, nitrogen dioxide, ${\text{NO}}_{2}$, and oxygen gas, ${\text{O}}_{2}$

${\text{M"("NO"_3)_2 -> "MO" + "NO"_2 + "O}}_{2}$

The key here is to balance this chemical equation by using the oxygen and nitrogen atoms.

$2 {\text{M"("NO"_3)_text(2(s]) -> 2"MO"_text((s]) + 4"NO"_text(2(g]) uarr + "O}}_{\textrm{2 \left(g\right]}} \uparrow$

Notice that the nitrogen dioxide and the oxygen gas are evolved during the reaction. This means that the remaining sample will consist of the metal oxide.

Now it's simply a matter of using the $1 : 1$ mole ratio that exists between the metal nitrate and the metal oxide to find the molar mass of the metal.

The number of moles of metal nitrate that ract is

5color(red)(cancel(color(black)("g"))) * ("1 mole M"("NO"_3)_2)/((x + 124.01) color(red)(cancel(color(black)("g")))) = 5/(x+ 124.01)"moles M"("NO"_3)_2

Here $\text{124.01 g/mol}$ represents the molar mass of two nitrate ions, how many you get per formula unit of "M"("NO"_3)_2.

You know that $\text{3.29 g}$ are lost ,which means that the mass of the oxide will be

${m}_{\text{oxide" = "5 g" - "3.29 g" = "1.71 g}}$

The number of moles of metal oxide will be

1.71color(red)(cancel(color(black)("g"))) * "1 mole MO"/((x + 15.994)color(red)(cancel(color(black)("g")))) = 1.71/(x + 15.9994)"moles MO"#

Since these two numbers must be equal, it follows that you have

$\frac{5}{x + 124.01} = \frac{1.71}{x + 15.9994}$

This is equivalent to

$5 \cdot \left(x + 15.9994\right) = 1.71 \cdot \left(x + 124.01\right)$

$5 x - 1.71 x = 212.057 - 79.997$

$3.29 x = 132.06 \implies x = \frac{132.06}{3.29} = \text{40.14 g/mol}$

This is approximately equal to the molar mass of calcium, $\text{Ca}$, which is listed as

${M}_{\text{M calcium" = "40.078 g/mol}}$