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?

All you have to do here is set up a system of two equations with two unknowns, the molar mass of #"A"# and the molar mass of #"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 #"A"_2"Z"_3# have a mass of #"15.9 g"#. To make the calculations easier, calculate the mass of one mole of #"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 #"A"_2"Z"_3# contains

• two moles of #"A"#, #2 xx "A"#
• three moles of #"Z"#, #3 xx "Z"#

If you take #a# to be the molar mass of #"A"# and #z# to be the molar mass of #"Z"#, you can say that

#2 * a + 3 * z = "106 g"" " " "color(orange)((1))#

Now do the same for #"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 #"AZ"_2# you have

• one mole of #"A"#, #1 xx "A"#
• two moles of #"Z"#, #2 xx "Z"#

You will thus have

#a + 2 * z = "62 g"" " " "color(orange)((2))#

Use equation #color(orange)((2))# to write

#a = 62 - 2z#

Plug this into equation #color(orange)((1))# to get

#2 * (62 - 2z) + 3 * z= 106#

#124 - 4z + 3z = 106#

Rearrange to find

#z = 18#

This means that you have

#a = 62 - 2 * 18 = 26#

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

#"For A: " "26 g mol"^(-1)#

#"For B: " "18 g mol"^(-1)#

To convert these to atomic masses, use the conversion factor

#color(purple)(bar(ul(|color(white)(a/a)color(black)("1 u " = " 1 g mol"^(-1))color(white)(a/a)|)))#

You will have

#"For A: " m_"a A" = "26 u"#

#"For B: " m_"a Z" = "18 u"#