1. A sample problem
3.43)
Allicin is the compound responsible for the characteristic smell of garlic. An analysis of the compound gives the following percent composition by mass: C: 44.4%, H: 6.21%, S: 39.5%, O: 9.86%. Calculate its empirical formula. What is its molecular formula given that its molar mass is about 162 g?
empirical formula = molecular formula = C6H10S2O
*Note that we arbitrarily chose a 100-g sample to make our math easier.
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2. Another sample problem
3.52)
The empirical formula of a compound is CH. If the molar mass of this compound is about 78g, what is its molecular formula?
Figure out molar mass of the empirical formula.
Divide the molar mass of the compound by the molar mass of the empirical formula.
Multiply each subscript on the empirical formula by this quotient to get the molecular formula.
78/13.01=6
C6H6
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3. Yet another sample problem
Practice exercise, p. 71)
A sample of a compound containing boron and hydrogen contains g of boron and g of hydrogen. The molar mass of the compound is about 30 g. What is its molecular formula?
Figure out molar mass of the empirical formula.
Divide the molar mass of the compound by the molar mass of the empirical formula.
Multiply each subscript in the empirical formula by this whole number to get the molecular formula.
B2H6
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4. A process in which one or more substances is changed into one or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what happens during a chemical reaction
3 ways of representing the reaction of H2 with O2 to form H2O:
reactants
products
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5. How to “Read” Chemical Equations
2 Mg + O MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg grams O2 makes 80.6 g MgO
2 grams Mg + 1 gram O2 makes 2 g MgO
Note: stoichiometric coefficients = atoms or moles… NOT grams!
*Note: Stoichiometric coefficients = atoms or moles… NOT grams!
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6. Balancing Chemical Equations
Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.
Ethane (C2H6) reacts with oxygen to form carbon dioxide & water
C2H6 + O2
CO2 + H2O
Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.
2C2H6
NOT
C4H12
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7. Balancing Chemical Equations
Start by balancing those elements that appear in only one reactant and one product.
C2H6 + O2
CO2 + H2O
start with C or H but not O
1 carbon
on right
2 carbon
on left
multiply CO2 by 2
C2H6 + O2
2CO2 + H2O
6 hydrogen
on left
2 hydrogen
on right
multiply H2O by 3
C2H6 + O2
2CO2 + 3H2O
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8. Balancing Chemical Equations
Balance those elements that appear in two or more reactants or products.
multiply O2 by 7 2
C2H6 + O2
2CO2 + 3H2O
2 oxygen
on left
4 oxygen
(2x2)
+ 3 oxygen
(3x1)
= 7 oxygen
on right
C2H O2
2CO2 + 3H2O 7 2
remove fraction
multiply both sides by 2
2C2H6 + 7O2
4CO2 + 6H2O
3.7
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9. Balancing Chemical Equations
5. Check your balanced eqn to make sure you have the same number of each type of atom on both sides of the eqn arrow.
2C2H6 + 7O2
4CO2 + 6H2O
final tally
reactants products
C 4 C 4
H 12 H 12
O 14 O 14
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10. Mass Changes in Chemical Reactions
Write balanced chemical equation
Convert quantities of known substances into moles
Use coefficients in balanced equation to calculate the number of moles of the sought quantity
Convert moles of sought quantity into desired units
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11. Balance a combustion reaction
Provide a balanced equation for the combustion of methanol (CH3OH).
CH4O + O2 CO H2O
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12. Balance a combustion reaction
Provide a balanced equation for the combustion of methanol (CH3OH).
2CH4O O2 2CO H2O
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13. Methanol burns in air according to the equation
2CH3OH + 3O CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
moles H2O
grams H2O
molar mass
CH3OH
coefficients
chemical equation
molar mass
H2O
1 mol CH3OH
32 g CH3OH x
4 mol H2O
2 mol CH3OH x
18 g H2O
1 mol H2O x =
209 g CH3OH
235 g H2O
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14. Limiting Reagents
When a chemist carries out a reaction, the reactants are
usually not present in stoichiometric amounts, that is, in
the proportions indicated by the balanced equation.
It is common to use the cheaper reactant in excess to ensure that all of the expensive reagent is consumed. The more expensive reagent is the limiting reagent and is often completely consumed in a reaction.
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15. Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3.
Which reactant is the limiting reagent?
2Al + Fe2O Al2O3 + 2Fe
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed OR
g Fe2O3
mol Fe2O3
mol Al needed
g Al needed
1 mol Al
27 g Al x
1 mol Fe2O3
2 mol Al x
160 g Fe2O3
1 mol Fe2O3 x =
124 g Al
367 g Fe2O3
Start with 124 g Al
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
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