# The elements A and Z combine to produce two different compounds: A_2Z_3 and AZ_2. If 0.15 mole of A_2Z_3 has a mass of 15.9 g and 0.15 mole of AZ_2 has a mass of 9.3 g, what are the atomic masses of A and Z?

Sep 5, 2016

${m}_{\text{a A" = "26 u}}$

${m}_{\text{a Z" = "18 u}}$

#### Explanation:

All you have to do here is set up a system of two equations with two unknowns, the molar mass of $\text{A}$ and the molar mass of $\text{Z}$.

Once you know the molar masses of the two elements, you can use a simple conversion factor to find their respective atomic masses.

So, you know that $0.15$ moles of ${\text{A"_2"Z}}_{3}$ have a mass of $\text{15.9 g}$. To make the calculations easier, calculate the mass of one mole of ${\text{A"_2"Z}}_{3}$

1 color(red)(cancel(color(black)("mole A"_2"Z"_3))) * "15.9 g"/(0.15color(red)(cancel(color(black)("moles A"_2"Z"_3)))) = "106 g"

Notice that one mole of ${\text{A"_2"Z}}_{3}$ contains

• two moles of $\text{A}$, $2 \times \text{A}$
• three moles of $\text{Z}$, $3 \times \text{Z}$

If you take $a$ to be the molar mass of $\text{A}$ and $z$ to be the molar mass of $\text{Z}$, you can say that

$2 \cdot a + 3 \cdot z = \text{106 g"" " " } \textcolor{\mathmr{and} a n \ge}{\left(1\right)}$

Now do the same for ${\text{AZ}}_{2}$. You have

1 color(red)(cancel(color(black)("mole AZ"_2))) * "9.3 g"/(0.15color(red)(cancel(color(black)("moles AZ"_2)))) = "62 g"

This time, in one mole of ${\text{AZ}}_{2}$ you have

• one mole of $\text{A}$, $1 \times \text{A}$
• two moles of $\text{Z}$, $2 \times \text{Z}$

You will thus have

$a + 2 \cdot z = \text{62 g"" " " } \textcolor{\mathmr{and} a n \ge}{\left(2\right)}$

Use equation $\textcolor{\mathmr{and} a n \ge}{\left(2\right)}$ to write

$a = 62 - 2 z$

Plug this into equation $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ to get

$2 \cdot \left(62 - 2 z\right) + 3 \cdot z = 106$

$124 - 4 z + 3 z = 106$

Rearrange to find

$z = 18$

This means that you have

$a = 62 - 2 \cdot 18 = 26$

So, you know that the molar masses of the two elements are

${\text{For A: " "26 g mol}}^{- 1}$

${\text{For B: " "18 g mol}}^{- 1}$

To convert these to atomic masses, use the conversion factor

$\textcolor{p u r p \le}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{1 u " = " 1 g mol}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You will have

$\text{For A: " m_"a A" = "26 u}$

$\text{For B: " m_"a Z" = "18 u}$