Question

Calculate the percentage by mass of each element in the following compound?

MgO, Al2O3, MgSO4

Answer 1

Well... I could do some of these, but the process is the same for all of them. I will do one example, `"MgSO"_4`, and you can follow the process for the rest.

You could also make it easier on yourself by looking up these compounds on wikipedia, and use their molar masses alongside the atomic masses from the periodic table...

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`"MgSO"_4`:

First, collect the atomic masses of each atom.

`M_("Mg") = "24.305 g/mol"`

`M_("S") = "32.065 g/mol"`

`M_("O") = "15.999 g/mol"`

Then count how many of THAT atom are in the compound. There are one `"Mg"`, one `"S"`, and four `"O"` atoms in magnesium sulfate.

The percent by mass is simply the ratio of the total atomic mass to the total molar mass.

Thus:

`color(blue)(%"Mg") = color(green)"24.305 g Mg/mol"/("24.305 g Mg/mol" + "32.065 g S/mol" + 4 xx "15.999 g O/mol") xx 100%`

`= color(blue)(20.2% "Mg")`

And to save time, evaluate the denominator first, and re-use it for the percent mass of `"S"`.

Now, simply switch out the atomic mass for that of sulfur.

`color(blue)(%"S") = color(green)"32.065 g S/mol"/("24.305 g Mg/mol" + "32.065 g S/mol" + 4 xx "15.999 g O/mol") xx 100%`

`= color(blue)(26.6% "S")`

Let's make oxygen easy... It's the third of three atoms, so...

`color(blue)(%"O") = 100.0% - 20.2% - 26.6%`

`= color(blue)(53.2% "O")`

Or if you want to do oxygen in full:

`color(blue)(%"O") = (color(green) (4 xx "15.999 g O/mol"))/("24.305 g Mg/mol" + "32.065 g S/mol" + 4 xx "15.999 g O/mol") xx 100%`

`= color(blue)(53.2% "O")`