# Question e8b20

Oct 4, 2015

"C: "40.0%
"H: "6.70%
"O: "53.3%

#### Explanation:

The idea here is that you need to use the mass of carbon dioxide and the mass of water to determine how much carbon and how much hydrogen the original acetic acid sample contained.

Start with the amount of carbon. Assuming that all the carbon that was initially part of the acetic acid is now a part of carbon dioxide produced by the raction, you can use carbon's dioxide percent composition to find how much carbon you ahve in 6.21 mg of ${\text{CO}}_{2}$

Use carbon and carbon dioxide's molar masses to get

(12.011 color(red)(cancel(color(black)("g/mol"))))/(44.01color(red)(cancel(color(black)("g/mol")))) xx 100 = "27.29% C"

This means that 6.21 mg of ${\text{CO}}_{2}$ will contain

6.21color(red)(cancel(color(black)("mg CO"""_2))) * "27.29 mg C"/(100color(red)(cancel(color(black)("mg CO"""_2)))) = "1.695 mg C"

Do the same for water and hydrogen to get hydrogen's percent composition in water

(2 xx 1.008color(red)(cancel(color(black)("g/mol"))))/(18.015color(red)(cancel(color(black)("g/mol")))) xx 100 = "11.19% H"

This means that 2.54 mg of water contain

2.54color(red)(cancel(color(black)("mg H"_2"O"))) * "11.19 mg H"/(100color(red)(cancel(color(black)("mg H"_2"O")))) = "0.2842 mg H"

The original sample of acetic acid also contained oxygen, which means that you can write

${m}_{\text{total" = m_"C" + m_"H" + m_"O}}$

${m}_{\text{O" = 4.24 - 1.695 - 0.2842 = "2.261 mg O}}$

The percent composition of acetic acid will thus be

(1.695color(red)(cancel(color(black)("mg"))))/(4.24color(red)(cancel(color(black)("mg")))) xx 100 = color(green)(40.0% "carbon")

(0.2842color(red)(cancel(color(black)("mg"))))/(4.24color(red)(cancel(color(black)("mg")))) xx 100 = color(green)(6.70%"hydrogen")

(2.261color(red)(cancel(color(black)("mg"))))/(4.24color(red)(cancel(color(black)("mg")))) xx 100 = color(green)(53.3%"oxygen")#