A compound has a molar mass of 142.00g/mol and the percent composition is 50.7% Carbon, 4.2% Hydrogen and 45.1% Oxygen. What is the molecular formula?

1 Answer
Dec 12, 2015



Every time a problem provides you with a compound's molar mass, you can use a little trick to help you determine the compound's molecular mass faster.

More specifically, instead of determining the empirical formula first, then using the molar mass to get the molecular formula, you can skip the empirical formula altogether.

As you know, a compound's molar mass tells you what the mass of one mole of that substance is. In this case, one mole of your compound has a mass of #"142.00 g"#.

This means that if you pick a sample of #"142.00 g"# of this compound, you can use its percent composition to get the exact number of moles of each element you get per mole of compound.

So, you know that this compound's percent composition is as follows

  • carbon #-> 50.7%#
  • hydrogen #-> 4.2%#
  • oxygen #-> 45.1%#

This means that it contains

#142.00 color(red)(cancel(color(black)("g compound"))) * "50.7 g C"/(100color(red)(cancel(color(black)("g compound")))) = "71.994 g C"#

#142.00color(red)(cancel(color(black)("g compound"))) * "4.2 g H"/(100color(red)(cancel(color(black)("g compound")))) = "5.964 g H"#

#142.00color(red)(cancel(color(black)("g compound"))) * "45.1 g O"/(100color(red)(cancel(color(black)("g compound")))) = "64.042 g O"#

Now all you have to do is use the molar masses of these three elements to determine how many moles of each you get in one mole of your compound

#"For C: " 71.994 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = 5.994 ~~ 6#

#"For H: " 5.964 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = 5.92 ~~6#

#"For O: " 64.042 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = 4.003 ~~4#

Therefore, the molecular formula for this compound will be


SIDE NOTE I recommend finding the empirical formula first, then using the molar mass to find the molecular formula. The result will be exactly the same.