Question #5ce05

1 Answer
May 29, 2017

Answer:

#"C"_8"H"_{16}"O"_4#.

Explanation:

First figure out how much carbon there is per mole of the substance X:

#({54.5% "C"}/{100%})×{176" g"}/{"mol X"}={96" g C"}/{"mol X"}#

Then one mole of carbon is #12# grams (atomic mass), so:

#{96" g C"}/{"mol X"}×{1" mol C"}/{12"g C"}={8" mol C"}/{"mol X"}#

So we have #"C"_8# in the formula.

Do the same with hydrogen:

#{9.09% "H"}/{100%}×{176" g"}/{"mol X"}={16" g H"}/{"mol X"}#

#{16" g H"}/{"mol X"}×{1" mol H"}/{1"g H"}={16" mol H"}/{"mol X"}#

Thus #"C"_8 "H"_{16}#.

Now for the tricky part. When chemists report an elemental analysis and it adds up to less than 100%, the difference is generally attributed to oxygen. Our analytical methods, constrained by working in an oxygenated and water-laden world, do not in general detect or measure oxygen directly. Instead we infer it by difference after accounting for the other elements.

Here we found that #176" g X"# contains #96" g C"# and #16" g H"#. Subtracting then gives:

#176" g X"- 96" g C" - 16" g H" = #64" g O"#

So then:

#{64" g O"}/{"mol X"}×{1" mol O"}/{16"g O"}={4" mol O"}/{"mol X"}#

So one mole of X contains four moles of oxygen to go with eight moles of carbon and 16 miles of hydrogen. Thus #C"_8"H"_{16}"O"_4#.