# Question #0a578

##### 1 Answer

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 **molar masses** of the two elements.

You know that **one mole** of

two molesof element#"B"# ,#2 xx "B"# three molesof element#"A"# ,#3 xx "A"#

Likewise, **one mole** of

two molesof element#"B"# .#2 xx "B"# one moleof element#"A"# ,#1 xx "A"#

This means that **moles** of

#2 xx "0.05 moles" = "0.10 moles B"# #3 xx "0.05 moles" = "0.15 moles A"#

and **moles** of

#2 xx "0.1 moles" = "0.20 moles B"# #1 xx "0.1 moles" = "0.10 moles A"#

Now, if you take **molar mass** of element **molar mass** of element

#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

#{ (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)|)))#