Question

Question #dc7c8

Answer 1

`A_("r M") = 64`

Explanation:

The idea here is that you need to use the mass of oxygen to figure out how many moles of oxygen were present in the original sample.

This will then allow you to figure out the number of moles of the metal and its relative atomic mass.

So, you know that the sample has a total mass of `"16.0 g"`. The mass of oxygen that was removed from the sample is equal to

`"16.0 g " - " 12.8 g" = "3.2 g"`

Oxygen has a relative atomic mass equal to

`A_("r O") = 16`

This means that the mass of oxygen present in the original sample is equivalent to

`"3.2 g" / 16 = "0.2 moles O"`

Since the chemical formula of the oxide is `"MO"`, it follows that the sample must have contained equal numbers of moles of the two constituent elements.

This means that you have

`"moles of M " = " 0.2 moles"`

Consequently, the relative atomic mass of the metal is

`"12.8 g"/"0.2 moles" = 64`

SIDE NOTE `color(white)(.)`Don't worry about units. The relative atomic mass is calculated by taking the atomic mass, which is measured in grams per mole, `"g mol"^(-1)`, and dividing it by `1/12"th"` of the mass of a single carbon-12 atom, which is equal to `"1 g mol"^(-1)`.

In other words, you have

`"12.8 g"/"0.2 moles" = "64 g mol"^(-1) ->` the atomic mass of the metal

and

`(64 color(red)(cancel(color(black)("g mol"^(-1)))))/(1color(red)(cancel(color(black)("g mol"^(-1))))) = 64 ->` the relative atomic mass of the metal