If a .500g sample of citric acid containing only carbon, hydrogen, and oxygen is burned to produce .687g #CO_2# and .187g #H_2O# what is the percent composition and how do you calculate it?

The idea here is that you can use the masses of carbon dioxide, #"CO"_2#, and water, #"H"_2"O"#, produced by the reaction to determine how much carbon and how much hydrogen you had in the given sample of citric acid.

Your assumption here is that all the carbon that was initially present in the acid is now present in the carbon dioxide, and all the hydrogen present in the acid is now present in the water.

Let's start with carbon dioxide. Use its molar mass to determine how many moles of carbon dioxide were produced by the reaction

#0.687color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "0.01561 moles CO"_2#

Now, one mole of carbon dioxide contains one mole of carbon, which means that the sample of citric acid contained #0.01561# moles of carbon.

Use carbon's molar mass to convert this to grams

#0.01561color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = "0.1875 g C"#

Do the same for water. You will have

#0.187color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.01038 moles H"_2"O"#

Keep in mind that one mole of water contains two moles of hydrogen, which means that the initial sample of citric acid contained

#0.01038color(red)(cancel(color(black)("moles H"_2"O"))) * "2 moles H"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "0.02076 moles H"#

Use hydrogen's molar mass to convert this to grams

#0.02076color(red)(cancel(color(black)("moles H"))) * "1.00794 g"/(1color(red)(cancel(color(black)("mole H")))) = "0.02092 g H"#

Use the initial mas of the sample to determine how much oxygen it contained

#m_(O) = m_"citric acid" - (m_(C) + m_(H))#

#m_(O) = "0.500 g" - ("0.1875 g" + "0.02092 g")#

#m_(O) = "0.2916 g"#

To get the percent composition of citric acid, simply divide the mass of each constituent element by the total mass of the sample and multiply the result by #100#.

#"For C: " (0.1875color(red)(cancel(color(black)("g"))))/(0.500color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"37.5% C"color(white)(a/a)|)))#

#"For H: " (0.02092color(red)(cancel(color(black)("g"))))/(0.500color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"4.20% H"color(white)(a/a)|)))#

#"For O: " (0.2916color(red)(cancel(color(black)("g"))))/(0.500color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"58.3% O"color(white)(a/a)|)))#

The answers are rounded to three sig figs.