Question

Question #0a578

Answer 1

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 `"A"` and of `"B"` present in those two compounds and the molar masses of the two elements.

You know that one mole of `"B"_2"A"_3` contains

• two moles of element `"B"`, `2 xx "B"`
• three moles of element `"A"`, `3 xx "A"`

Likewise, one mole of `"B"_2"A"` contains

• two moles of element `"B"`. `2 xx "B"`
• one mole of element `"A"`, `1 xx "A"`

This means that `0.05` moles of `"B"_2"A"_3` will contain

• `2 xx "0.05 moles" = "0.10 moles B"`
• `3 xx "0.05 moles" = "0.15 moles A"`

and `0.1` moles of `"B"_2"A"` will contain

• `2 xx "0.1 moles" = "0.20 moles B"`
• `1 xx "0.1 moles" = "0.10 moles A"`

Now, if you take `xcolor(white)(a)"g mol"^(-1)` to be the molar mass of element `"A"` and `ycolor(white)(a)"g mol"^(-1)` to be the molar mass of element `"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

`0.10y + 0.15x = 9" " " "color(orange)((1))`

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

`0.20y + 0.10x = 10" " " "color(orange)((2))`

Now use equations `color(orange)((1))` and `color(orange)((2))` to find `x` and `y`

`{ (0.10y + 0.15x = 9 | xx (-2)), (0.20y + 0.10x = 10) :}`
`color(white)(aaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaaaa)`

`-0.20x = -8 implies x = ((-8))/((-0.20)) = 40`

This will get you

`0.10y = 9 - 0.15 * 40`

`y = 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)|)))`