A sample of a white solid contains 0.212 g of magnesium and 0.140 g oxygen. What is the empirical formula of this compound?

1 Answer
Jun 27, 2016




The idea here is that you can find the empirical formula of the compound by calculating how many moles of each constituent element you have in that sample.

Once you know that, the smallest whole number ratio that exists between magnesium and oxygen will give you the compound's empirical formula.

So, you can convert the two masses to moles by using the molar masses of the two elements

#"For Mg: " 0.212 color(red)(cancel(color(black)("g"))) * "1 mole Mg"/(24.3050 color(red)(cancel(color(black)("g")))) = "0.008722 moles Mg"#

#"For O: "0.140 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "0.008750 moles O"#

To find the ratio of the elements, simply divide both values vy the smallest one

#"For Mg: " (0.008722 color(red)(cancel(color(black)("moles"))))/(0.008722color(red)(cancel(color(black)("moles")))) = 1#

#"For O: " (0.008750color(red)(cancel(color(black)("moles"))))/(0.008722color(red)(cancel(color(black)("moles")))) = 1.003 ~~ 1#

Since #1:1# is already the smallest whole number ratio that can exist between the two elements, the empirical formula of the compound will be

#"Mg"_ 1 "O"_ 1 implies color(green)(|bar(ul(color(white)(a/a)color(black)("MgO")color(white)(a/a)|)))#

Since you're dealing with an ionic compound, you can say that the empirical formula is the same as the ionic formula, which implies that your unknown ionic compound is magnesium oxide, #"MgO"#.