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A combustion reaction occurs when a compound
that contains carbon and hydrogen is burned in
Oxygen gas (O2). It always produces carbon dioxide
and water vapor.
Example:
◦ CH4 + 2 O2  CO2 + 2 H2O
◦ Many times the compound also contains oxygen
◦ Example:
◦ 2C3H4O3 + 5 O2  6 CO2 + 4 H2O
◦ Notice the products remain the same.
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Vitamin C is a compound made up of carbon, hydrogen and
oxygen atoms. When we combust 1.00g of vitamin C we
produce 1.50g of carbon dioxide and 0.408g of water vapor.
Find the empirical formula of vitamin C.
#1 We know that it is a combustion reaction so:
◦ CxHyOz + O2  CO2 + H2O
◦ #2 We notice that all the carbon atoms from vitamin C
became part of carbon dioxide so;
◦ 1.50g CO2 x 1 mole CO2 / 44 g = 0.0341 moles of CO2
◦ Since there is only 1 atom of carbon in CO2 we have
0.0341 moles of C
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#3 We have 0.408 g of H2O so all the hydrogen atoms from
vitamin C became part of the water vapor.
0.408 g H2O x 1 mole H2O / 18 grams = 0.0227 moles of H2O
Notice thee are TWO atoms of hydrogen in H2O so;
We have 2x 0.0227= 0.0454 moles of Hydrogen
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#4 We now go back to Vitamin C, CxHyOz
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We know we have:
0.0341 moles of C x 12g/1 mole of C= .409g of C
0.0454 moles of H x 1g / 1mole of H = .0454g of H
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Oxygen = (1.00- .409- .0454)= .546 g of O
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#5 We now know the grams of each of the elements that make
up vitamin C
◦ C = .409g x 1mole C/ 12g = .0341 moles
◦ H = .0454 g x 1mole H/ 1g = .0454 moles
◦ O = .546 g x 1mole O/ 16 g = .0341 moles
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Divide all moles by smallest answer
C= .0341/.0341 =1
H= .0454/.0341 = 1.33
O = .0341/ .0341= 1
◦ Multiply all answers by 3 and we get
◦ C 3 H4 O3
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1.Balance the following reactions
◦ A) CaO + HNO3  Ca(NO3)2 + H2O
◦ B) NH3 + H2SO4  (NH4)2SO4
2.Given the following reaction, find the grams of H2O are
produced
◦ 2C6H6 + 15 O2  12 CO2 + 6 H2O
◦ You start with 125 grams of C6H6 and 875 g of O2
3. Find the empirical formula of a compound that is made up
of C,H and O. When you combust 2.00g of the compound in
excess oxygen gas you produced 3.83g of CO2 and 2.35g
of H2O