How can I calculate the empirical formula of magnesium oxide?

For example, you might heat a known mass of magnesium in a crucible and determine the mass of oxide formed.

EXAMPLE

Assume that you heated 0.297 g of magnesium and obtained 0.493 g of the oxide. What is the empirical formula of magnesium oxide?

Solution

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of #"Mg"# to #"O"#.

#"Mass of Mg = 0.297 g"#

#"Mass of magnesium oxide = mass of Mg + mass of O"#

#"0.493 g = 0.297 g + mass of O"#

#"Mass of O = (0.493 – 0.297) g = 0.196 g"#

#"Moles of Mg" = 0.297 color(red)(cancel(color(black)("g Mg"))) × "1 mol Mg"/(24.3color(red)(cancel(color(black)( "g Mg")))) = "0.012 22 mol Mg"#

#"Moles of O "= 0.196 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "0.012 25 mol O"#

To get this into an integer ratio, we divide both numerator and denominator by the smaller value.

From this point on, I like to summarize the calculations in a table.

#"Element"color(white)(Mg) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(mll)"Integers"#
#stackrel(—————————————————-——)(color(white)(m)"Mg" color(white)(XXXm)0.297 color(white)(X)"0.012 22" color(white)(X)1color(white)(Xmmmm)1#
#color(white)(m)"O" color(white)(XXXXll)0.196 color(white)(m)"0.012 25" color(white)(X)1.002 color(white)(XXX)1#

There is 1 mol of #"Mg"# for 1 mol of #"O"#.

The empirical formula of magnesium oxide is #"MgO"#.

Here is a video that illustrates how to determine an empirical formula.