Michael W. answered • 07/09/16
Tutor
New to Wyzant
problem solving in general chemistry
here's the idea. If we think about the unbalanced equation
.. __ CxHyOz + __ O2 ---> __ CO2 + __ H2O
we can immediately see that
.. ALL the C in CO2 came from CxHyOz.... .right? O2 has no C
.. ALL the H in H2O came from CxHyOz... . again.. O2 has no H
so we can follow these steps
.. (1) convert mass CO2 to moles CO2 to moles C
.. (2) convert mass H2O to moles H2O to moles H
.. (3) mass O in the sample = mass O in products - 16.7g O2
.. (4) convert mass O to moles O
.. (5) simplify the mole ratios to get the formula
like this
*********
mole H = 6.25g H2O x (1mol H2O / 18.02g H2O)*(2mol H / 1mol H2O)
= 0.6937
.
mole C = 20.4g CO2 x (1mol CO2 / 44.01g CO2)*(1mol C / 1mol CO2)
= 0.4635
mass O in H2O = 6.25g H2O x (16.00g O / 18.02g H2O) = 5.549g O
mass O in CO2 = 20.4g CO2 x (32.0g O / 44.01g CO2) = 14.83g O
mass O in CxHyOz sample = (14.83g + 5.549g) - 16.7g = 3.67g
mole O in sample = 3.67g x (1mol / 16.00g) = 0.2294
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now we have these 3 mole values in our compound
.. mol C = 0.4635
.. mol H = 0.6937
.. mol O = 0.2294
we need to simplify those into the lowest whole numbers. We do this with a trick. We divide all 3 numbers by whichever is smallest. This forces 1 of the numbers to = 1. And the rest to be > 1 (easier to see that way).. We can do this because we don't care about the actual values... just the RATIOS. So 0.2294 is the smallest value and.
.. mol C = 0.4635 / 0.2294 = 2
.. mol H = 0.6937 / 0.2294 = 3
.. mol O = 0.2294 / 0.2294 = 1
and our empirical formula is.... C2H3O
******
note the extra sig fig during the intermediate steps. These empirical formula problems are very sensitive to calcs. Do make sure you carry 1 extra sig fig rather than rounding to the correct sig figs during those intermediate steps.
