The molar mass of a compound is 88 g/mol. It contains 54.53% carbon, 9.15% hydrogen and 36.32% oxygen. What is the molecular formula of the compound?

1 Answer
Mar 2, 2016

Answer:

#"C"_4"H"_8"O"#

Explanation:

One strategy to use when given the molar mass of the compound is to pick a sample that would correspond to one mole and use it to find the molecular formula without finding the empirical formula first.

In this case, a molar mass of #"88 g mol"^(-1)# tells you that one mole of this compound has a mass of #"88 g"#.

Use the percent composition of the compound to determine how many grams of each element you'd get in this #"88-g"# sample. Remember, you can convert between percentages and grams by using a sample of #"100 g"#.

You will thus have

#"For C: " 88color(red)(cancel(color(black)("g sample"))) * "54.53 g C"/(100color(red)(cancel(color(black)("g sample")))) = "47.986 g C"#

#"For H: " 88color(red)(cancel(color(black)("g sample"))) * "9.15 g H"/(100color(red)(cancel(color(black)("g sample")))) = "8.052 g H"#

#"For O: " 88 color(red)(cancel(color(black)("g sample"))) * "36.32 g O"/(100color(red)(cancel(color(black)("g sample")))) = "31.962 g O"#

Now use the molar mass of each element to determine how many moles of each are present in one mole of this compound

#"For C: " 47.986color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = 3.995 ~~ "4 moles C"#

#"For H: " 8.052color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = 7.990 ~~ "8 moles H"#

#"For O: " 31.962 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = 1.998 ~~ "2 moles O"#

Since this is how many moles of each element are present in one mole of this compound, you can say that one molecule of the compound will contain

  • #4# atoms of carbon
  • #8# atoms of hydrogen
  • #2# atoms of oxygen

Therefore, the compound's molecular formula, which tells you exactly how many atoms of each element make up one molecule of the compound, will be

#color(green)("C"_4"H"_8"O"_2)#