1. SYNTHESIS, DECOMPOSITION And COMBUSTION
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SYNTHESIS, DECOMPOSITION
And COMBUSTION

2. Synthesis and Decomposition Reactions Used to Find Empirical Formulas
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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.

3. Simplest formula: synthesis reaction
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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 = mols
O: g ¸ 16.0 g/mol = mols
3: determine the ratio of moles
mol (reduced)
mol Mn O
0.0364
0.0356
0.0364/
= 1.022
0.0356/
0.0356= 1.000
4: the simplest formula is MnO

4. Simplest formula: decomposition reaction
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Simplest formula: decomposition reaction
A g sample of copper oxide is heated and the remaining copper is found to weigh g. What is the formula of the original oxide?
2: Cu: g ¸ 63.5 g/mol = mols
O: g ¸ 16.0 g/mol = mols
3: determine the ratio of moles
mol (reduced)
mol Cu O
0.0530
0.0265
0.0530/
= 2.00
0.0265/
0.0265= 1.000
4: the simplest formula is Cu2O

5. Combustion Reactions
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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.

6. Carbon-hydrogen analyzer

7. Combustion Reactions Continued
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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 non-metal oxide in a synthesis reaction to form Na2CO3 salt.

8. Combustion Reactions Continued
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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.

9. Empirical formula mass, efm)
1: A hydrocarbon is burned in a carbon-hydrogen analyzer 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 ¸ g/mol = mol CO2
H2O: 3.94 g ¸ g/mol = mol H20 3:
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H (2XH2O)
mol (reduced)
mol
C (1XCO2)
Empirical formula mass, efm)
0.218
0.438
0.218/ 0.218
= 1
0.438/ 0.218
= 2.01
{CH2= = 14.02}
4: the simplest formula is CH2; the molecular formula is (84/14=6) C6H12 (mm/efm=ratio to scale up)

10. 4: the simplest formula is the molecular formula is
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:
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mol (reduced)
mol
C (1XCO2)
H (2XH2O)
efm
4: the simplest formula is
the molecular formula is

11. 4: the simplest formula is CH3
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: mols CO2, mols H20 3:
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mol (reduced)
mol C H
efm
0.143
0.428
(1.01)= 15.04 1
2.99= 3
4: the simplest formula is CH3
the molecular formula is (mm/emf=2) …2XCH3= C2H6…ethene