What is the empirical formula for a compound which contains 80.2 g zinc and 19.8 g oxygen ?

1 Answer
zhirou
Feb 21, 2018

#ZnO#.

Explanation:

First, let's list the information we have! :)
These are the elements in our compound:

  • Zinc. There's #"80.2g"# of it.
  • Oxygen. There's #"19.8g"# of it.

To find out their empirical formula, or the simplest ratio of the moles of zinc to oxygen, we need to calculate the number of moles in #"80.2g"# of zinc and #"19.8g"# of oxygen.

We'll need to use this formula:

#"number of moles" = "mass of sample"/"mass of 1 mole"#

For zinc, the number of moles in #"80.2g"# would be #"80.2g"# divided by the mass of #1# mole of zinc, which is #"65.38g/mol"#.
#"80.2g"/"65.38g/mol"="1.23 mol"#

For oxygen, the number of moles in #"19.8g"# would be #"19.8g"# divided by the mass of #1# mole of oxygen, which is #"16.00g/mol"#.
#"19.8g"/"16.00g/mol"="1.24 mol"#

In our compound, the moles of zinc and oxygen are roughly the same—#1.23:1.24=1:1#

This tells us that the empirical formula for the compound would be #ZnO#.

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