Question

Question #383ec

Answer 1

`"43.3 g mol"^(-1)`

Explanation:

For starters, you can use the mass of the compound and the mass of oxygen to determine how many grams of element `"A"` you have in your sample

`"mass of compound = mass of A + mass of O"`

In your case, you will have

`"mass of A" = "0.359 g" - "0.128 g" = "0.231 g"`

Now, the problem wants you to determine the molar mass of substance `"A"`, which implies that you must figure out the mass of exactly `1` mole of `"A"`.

Use the molar mass of oxygen to calculate the number of moles of oxygen present in the sample

`0.128 color(red)(cancel(color(black)("g"))) * "1 mole O"/(16.0 color(red)(cancel(color(black)("g")))) = "0.00800 moles O"`

According to its chemical formula, the oxide contains `2` moles of `"A"` for every `3` moles of oxygen. This means that the sample must contain

`0.00800 color(red)(cancel(color(black)("moles O"))) * "2 moles A"/(3color(red)(cancel(color(black)("moles O")))) = "0.00533 moles A"`

Since you know that `0.00533` moles of `"A"` have a mass of `"0.231 g"`, you can say that `1` mole of `"A"` will have a mass of

`1 color(red)(cancel(color(black)("mole A"))) * "0.231 g"/(0.005333color(red)(cancel(color(black)("moles A")))) = "43.3 g"`

Therefore, you can say that element `"A"` has a molar mass of

`color(darkgreen)(ul(color(black)("molar mass A = 43.3 g mol"^(-1))))`

The answer is rounded to three sig figs.