# Question 0a578

##### 1 Answer
May 9, 2016

Here's what I got.

#### Explanation:

The idea here is that you need to find two equations that establish a relationship between the number of moles of $\text{A}$ and of $\text{B}$ present in those two compounds and the molar masses of the two elements.

You know that one mole of ${\text{B"_2"A}}_{3}$ contains

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

Likewise, one mole of $\text{B"_2"A}$ contains

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

This means that $0.05$ moles of ${\text{B"_2"A}}_{3}$ will contain

• $2 \times \text{0.05 moles" = "0.10 moles B}$
• $3 \times \text{0.05 moles" = "0.15 moles A}$

and $0.1$ moles of $\text{B"_2"A}$ will contain

• $2 \times \text{0.1 moles" = "0.20 moles B}$
• $1 \times \text{0.1 moles" = "0.10 moles A}$

Now, if you take $x \textcolor{w h i t e}{a} {\text{g mol}}^{- 1}$ to be the molar mass of element $\text{A}$ and $y \textcolor{w h i t e}{a} {\text{g mol}}^{- 1}$ to be the molar mass of element $\text{B}$, you can say that you have

0.10 color(red)(cancel(color(black)("moles B"))) * ycolor(white)(a) "g"/(color(red)(cancel(color(black)("mole B")))) + 0.15 color(red)(cancel(color(black)("moles A"))) * xcolor(white)(a)"g"/(color(red)(cancel(color(black)("mole A")))) = "9 g"

This is equivalent to

$0.10 y + 0.15 x = 9 \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(1\right)}$

Do the same for the second compound

0.20 color(red)(cancel(color(black)("moles B"))) * ycolor(white)(a) "g"/(color(red)(cancel(color(black)("mole B")))) + 0.10 color(red)(cancel(color(black)("moles A"))) * xcolor(white)(a)"g"/(color(red)(cancel(color(black)("mole A")))) = "10 g"

This is equivalent to

$0.20 y + 0.10 x = 10 \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(2\right)}$

Now use equations $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ and $\textcolor{\mathmr{and} a n \ge}{\left(2\right)}$ to find $x$ and $y$

$\left\{\begin{matrix}0.10 y + 0.15 x = 9 | \times \left(- 2\right) \\ 0.20 y + 0.10 x = 10\end{matrix}\right.$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a}}$

$- 0.20 x = - 8 \implies x = \frac{\left(- 8\right)}{\left(- 0.20\right)} = 40$

This will get you

$0.10 y = 9 - 0.15 \cdot 40$

$y = \frac{3}{0.10} = 30$

Therefore, the molar masses of the two elements are

"For A: " color(green)(|bar(ul(color(white)(a/a)"40 g mol"^(-1)color(white)(a/a)|)))

"For B: " color(green)(|bar(ul(color(white)(a/a)"30 g mol"^(-1)color(white)(a/a)|)))#