SYNTHESIS, DECOMPOSITION
And COMBUSTION
Synthesis and Decomposition
Reactions Used to Find Empirical
Formulas
An element is burned in oxygen to form an oxide
compound. The empirical formula of the
compound can be determined.
Step 1: determine the mass of the element and
the mass of oxygen added to form the oxide
Step 2: calculate the # of moles (mol = g  g/mol)
Step 3: express moles as the simplest ratio by
dividing through by the lowest number.
Step 4: write the simplest formula from mol ratios.
Simplest formula: synthesis reaction
A 2.00 g sample of manganese is burned in
oxygen and the product is found to weigh 2.57
g. What is the formula of the oxide formed?
2: Mn: 2.00 g  54.9 g/mol = 0.0364 mols
O: 0.57 g  16.0 g/mol = 0.0325 mols
3: determine the ratio of moles
Mn
O
mol
0.0364
0.0356
mol
0.0364/
0.0356/
(reduced) 0.0356
0.0356=
= 1.022
1.000
4: the simplest formula is MnO
Simplest formula: decomposition
reaction
A 3.787 g sample of copper oxide is heated and
the remaining copper is found to weigh 3.363
g. What is the formula of the original oxide?
2: Cu: 3.363 g  63.5 g/mol = 0.0530 mols
O: 0.424 g  16.0 g/mol = 0.0265 mols
3: determine the ratio of moles
Cu
O
mol
0.0530
0.0265
mol
0.0530/
0.0265/
(reduced) 0.0265
0.0265=
= 2.00
1.000
4: the simplest formula is Cu2O
Combustion Reactions
• There is one additional type of problem to solve
for a molecular formula. Combustion reactions
• When a combustion reaction occurs, the
products- CO2 and H2O can be captured by a
carbon hydrogen analyzer. This equipment
(p.163) is like a furnace with a two detachable
chimneys that are placed after the reaction.
• The first part of the chimney captures the water
and thus increases in mass and the second part
of the chimney captures the CO2 and increases
in mass. The increase in mass in both sections
can be used to calculate the moles of carbon
and moles of hydrogen in the original fuel.
Carbon-hydrogen analyzer
Combustion Reactions Continued
• The first part of the chimney captures the water
because the chimney contains a dehydrated
salt that becomes hydrated as the water from
the combustion reaction passes through.
• The second part of the chimney captures the
CO2 because it contains sodium hydroxide, a
metal oxide, that joins with the CO2, a nonmetal oxide in a synthesis reaction to form
Na2CO3 salt.
Combustion Reactions Continued
• In the first part, the number of moles of
hydrogen in the original compound is twice the
number of moles of water absorbed by the
chimney. (formula: H2O)
• In the second part, the number of moles of CO2
absorbed by the chimney is equal to the number
of moles of carbon in the fuel.
1: A hydrocarbon is burned in a carbon-hydrogen
analyzer. 3.94 g of water and 9.62 g of CO2 are
produced. The molar mass of the compound is
84.0 g/mol. Determine the molecular formula.
2: CO2: 9.62 g  44.00 g/mol = 0.218 mol CO2
H2O: 3.94 g  18.02 g/mol = 0.219 mol H20
Empirical
C (1XCO2) H (2XH2O)
formula
mol
0.438
0.218
mass, efm)
mol
0.218/
0.438/
3:
{CH2= 12
(reduced)
0.218
0.218
+2.02=
=1
= 2.01
14.02}
4: the simplest formula is CH2; the molecular
formula is (84/14=6) C6H12 (mm/efm=ratio to scale up)
1: A combustion reaction of a hydrocarbon is
known to produce 6.29 g of CO2 and 3.86 g
of water. If the molar mass is 30.0g/mol, what
is the molecular formula?
2:
3:
C (1XCO2) H (2XH2O)
mol
mol
(reduced)
4: the simplest formula is
the molecular formula is
efm
1: A combustion reaction of a hydrocarbon is
known to produce 6.29 g of CO2 and 3.86 g
of water. If the molar mass is 30.0g/mol, what
is the molecular formula?
2: 0.143 mols CO2, 0.214 mols H20
3:
H
C
efm
0.143
mol
0.428
12.01 +
mol
1
2.99= 3 3(1.01)=
(reduced)
15.04
4: the simplest formula is CH3
the molecular formula is (mm/emf=2)
…2XCH3= C2H6…ethene