How many oxygen atoms are contained in 223.4 grams of lithium carbonate and how do I get this answer?

1 Answer
Truong-Son N.
Jun 6, 2018

There are #5.462 xx 10^24# oxygen atoms in #"223.4 g"# of #"Li"_2"CO"_3#.

You'd have to recognize the chemical formula of lithium carbonate first.

  • Carbonate has a #-2# charge and is #"CO"_3^(2-)#, something you must memorize for general chemistry.

  • Lithium is an alkali metal, which has a typical oxidation state of #+1#.

Hence, it forms #"Li"_2"CO"_3#, which has a molar mass of

#overbrace(2 xx "6.941 g Li"/"mol")^(2 xx "Li") + overbrace("12.011 g C"/"mol")^("C") + overbrace(3 xx "15.999 g O"/"mol")^(3 xx "O")#

#=# #"73.890 g/mol"#

So, the number of mols of #"Li"_2"CO"_3# contained in #"223.4 g"# is:

#223.4 cancel("g Li"_2"CO"_3) xx "1 mol"/(73.890 cancel("g Li"_2"CO"_3)) = "3.023 mols"#

For every #"1 formula unit of Li"_2"CO"_3#, there are #"3 atoms of O"#. Hence, we have #"9.069 mols O atom"#.

As a result,

#9.069 cancel"mols atoms" xx (6.022 xx 10^23 "things")/cancel("1 mol")#

#=# #color(blue)(5.462 xx 10^24color(white)(.)"O atoms")#

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