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A chemical reaction shows the process in
which a substance (or substances) is
changed into one or more new substances
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
A chemical equation uses chemical symbols to
show what happens during a chemical reaction
reactants product
(g) (g) (l)
“Two molecules of hydrogen react with one molecule of
oxygen to yield two moles of water”
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
The Law of Conservation of Mass states that
matter is neither created nor destroyed
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
To conform with the Law of Conservation of Mass, there
must be the same number of each type of atom on both
sides of the arrow. Hence, we balance the equation by
adding coefficients before each chemical symbol
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
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Translate the statement
Balance the atoms
Specify states of matter
Adjust the coefficients
Check the atom balance
A quick note on balancing chemical equations
Calculating the amounts of
reactant and product
Stoichiometry of a double
cheeseburger
2 bun slices + 2 cheese slices
+ 2 burger patties =
In a balanced equation, the number of moles of
one substance is equivalent to the number of
moles of any of the other substances
2CO(g) + O2(g) 2CO2(g)
2 mol CO = 1 mol O2
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
2 mol CO
1 mol O2
= 11 mol O2
2 mol CO= 1
In a balanced equation, the number of moles of
one substance is equivalent to the number of
moles of any of the other substances
2CO(g) + O2(g) 2CO2(g)
2 mol CO = 2 mol CO2
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
2 mol CO
2 mol CO2
= 12 mol CO2
2 mol CO= 1
In a balanced equation, the number of moles of
one substance is equivalent to the number of
moles of any of the other substances
2CO(g) + O2(g) 2CO2(g)
1 mol O2 = 2 mol CO2
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
1 mol O2
2 mol CO2
= 12 mol CO2
1 mol O2
= 1
The amount of one substance in a reaction
is related to that of any other
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
All alkali metals react with water to produce
hydrogen gas and the corresponding
alkali metal hydroxide
2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)
How many moles of H2 will be formed by the
complete reaction of 6.23 moles of Li with water?
nH2 = x6.23 mol Li 1 mol H2
2 mol Li
= 3.12 mol H2
2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)
How many grams of H2 will be formed by the
complete reaction of 80.57 g of Li with water?
2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)
mH2 = x x80.57 g Li 1 mol Li 1 mol H2
6.941 g Li 2 mol Li
2.016 g H2x
1 mol H2
= 11.70 g H2
In a lifetime, the average American uses about
794 kg of copper in coins, plumbing, and wiring.
Copper is obtained from sulfide ores (such as
Cu2S) by a multistep process. After an initial
grinding, the first step is to “roast” the ore (heat it
strongly with O2) to form Cu2O and SO2
2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)
How many moles of oxygen are required
to roast 10.0 mol of Cu2S?
2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)
nO2 = x10.0 mol Cu2S 3 mol O2
2 mol Cu2S
= 15.0 mol O2
How many grams of SO2 are formed when
10.0 mol of Cu2S is roasted?
2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)
mSO2 = x x10.0 mol Cu2S 2 mol SO2 64.07 g SO2
2 mol Cu2S 1 mol SO2
= 641 g SO2
How many grams of O2 are required to form
2.86 kg of Cu2O?
2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)
mO2 = x x2.86 kg Cu2O 1000 g Cu2O 1 mol Cu2O
1 kg Cu2O 143.10 g Cu2O
3 mol O2 32.00 g O2x x
2 mol Cu2O 1 mol O2
= 960 g O2
Within the cylinders of a car’s engine, the
hydrocarbon octane (C8H18), one of many
components of gasoline, mixes with oxygen from
the air and burns to form carbon dioxide and water
vapor.
a. How much carbon dioxide ( in kg) is produced
when 2.00 kg of octane is burned ?
b. How much oxygen(in kg) is required to burn the
same amount of octane?
2C8H18(l) + 25O2 (g) 16CO2 (g) + 18H2O (g)
How many double cheeseburgers can you
make from 8 bun slices + 2 cheese slices + 6
burger patties ?
What is the limiting factor?
8 bun slices
2 cheese slices
6 burger patties
# Cheeseburgers=
2 cheese slices [1 cheeseburger/
2 cheese slices]
Limiting Reactants
The reactant used up first in a chemical
reaction is called the limiting reactant. Excess
reactants are present in quantities greater than
necessary to react with the quantity of the
limiting reactant.
A + B C + D
Given the amounts of A and B, which is the
limiting reactant? How much C and D are
produced?
Urea is prepared by reacting ammonia with
carbon dioxide:
2NH3(g) + CO2(g) (NH2)2CO(aq) + H2O(l)
In one process, 637.2 g of NH3 are allowed to
react with 1142 g of CO2.
(a) Which is the limiting reactant?
(b) How much urea (in grams) is produced?
(c) How much of the excess reactant (in grams)
is left at the end of the reaction?
Strategy
• Check if the equation is balanced.
• Convert mass of each reactant to moles.
• Calculate the amount of product formed
from the each of the reactants.
• The reactant the produces the less
amount is the limiting reactant.
1. The reaction between aluminum and iron (III) oxide can generate temperatures around 3000⁰C and is
used in welding metals:
Al + Fe2O3 -- Al2O3 + 2Fe
In one process, 124 g of Al are reacted with 601 g of
ferric oxide.
(a) Which is the limiting reactant?
(b) How much Al2O3 (in grams) is produced?
(c) How much of the excess reactant (in grams) is left at
the end of the reaction?
2. Titanium is a strong & light metal used in rockets
& aircrafts. It is prepared by the reaction between
titanium (IV) chloride with molten magnesium at around 1000⁰C:
TiCl4 + Mg Ti + 2MgCl2
In a certain industrial operation, 3.54 x 107g of TiCl4are reacted with 1.13 x 107 g of magnesium.
(a)Which is the limiting reactant?
(b)How much Ti (in grams) is produced?
(c)How much of the excess reactant (in grams) is left
at the end of the reaction?
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Mass % of element X =
atoms of X in formula x atomic mass of X (amu)
molecular (or formula) mass of compound (amu)
x 100
Mass % of element X =
moles of X in formula x molar mass of X (g/mol)
mass (g) of 1 mol of compound
x 100
Mass Percent from the Chemical Formula
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Sample Problem 3.3 Calculating Mass Percents and Masses of
Elements in a Sample of a Compound
PLAN:
SOLUTION:
PROBLEM: In mammals, lactose (milk sugar) is metabolized to glucose
(C6H12O6), the key nutrient for generating chemical potential
energy.
(a) What is the mass percent of each element in glucose?
(b) How many grams of carbon are in 16.55 g of glucose?
We have to find the total mass of
glucose and the masses of the
constituent elements in order to
relate them.
Per mole glucose there are 6 moles of
C, 12 moles of H, 6 moles of O(a)
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Sample Problem 3.3 Calculating the Mass Percents and Masses of
Elements in a Sample of a Compoundcontinued
6 mol C x 12.01 g C
mol C = 72.06 g C 12 mol H x
1.008 g H
mol H = 12.096 g H
6 mol O x 16.00 g O
mol O= 96.00 g O
M = 180.16 g/mol
(b)mass percent of C =
72.06 g C
180.16 g glucose= 0.4000 x 100 = 40.00 mass % C
mass percent of H =12.096 g H
180.16 g glucose= 0.06714 x 100 = 6.714 mass % H
mass percent of O =96.00 g O
180.16 g glucose= 0.5329 x 100 = 53.29 mass % O
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Empirical and Molecular Formulas
Empirical Formula -
Molecular Formula -
The simplest formula for a compound that agrees with
the elemental analysis and gives rise to the smallest set
of whole numbers of atoms.
The formula of the compound as it exists; it may be a
multiple of the empirical formula.
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Sample Problem 3.4 Determining an Empirical Formula from Masses
of Elements
PROBLEM:
PLAN:
Elemental analysis of a sample of an ionic compound showed
2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the empirical
formula and name of the compound?
Once we find the relative number of moles of each element,
we can divide by the lowest mol amount to find the relative
mol ratios (empirical formula).
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Sample Problem 3.4 Determining an Empirical Formula from Masses
of Elements
SOLUTION:2.82 g Na x
mol Na
22.99 g Na= 0.123 mol Na
4.35 g Cl xmol Cl
35.45 g Cl= 0.123 mol Cl
7.83 g O xmol O
16.00 g O= 0.489 mol O
Na1.00 Cl1.00 O3.98 NaClO4
NaClO4 is sodium perchlorate.
continued
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Sample Problem 3.5 Determining a Molecular Formula from Elemental
Analysis and Molar Mass
PROBLEM:
PLAN:
During physical activity, lactic acid (M = 90.08 g/mol) forms in
muscle tissue and is responsible for muscle soreness.
Elemental analysis shows that this compound contains 40.0
mass % C, 6.71 mass % H, and 53.3 mass % O.
(a) Determine the empirical formula of lactic acid.
(b) Determine the molecular formula.
a) Assume 100 g of lactic acid and find the mass of each
element. Convert mass of each to moles, get a ratio and
convert to integer subscripts.
b) Divide molar mass by empirical mass to get the multiplier
then write the molecular formula accordingly.
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Sample Problem 3.5 Determining a Molecular Formula from Elemental
Analysis and Molar Mass
continued
SOLUTION: In 100.0 g of lactic acid, there are:
40.0 g C 6.71 g H 53.3 g O
40.0 g C x
6.71 g H x
53.3 g O x
mol C
12.01 g C
mol H
1.008 g H
mol O
16.00 g O
= 3.33 mol C
= 6.66 mol H
= 3.33 mol O
We convert the grams to moles and get a ratio for the empirical formula.
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Sample Problem 3.5 Determining a Molecular Formula from Elemental
Analysis and Molar Mass
continued
SOLUTION:
3.33 mol C 6.66 mol H 3.33 mol O
C3.33 H6.66 O3.33
3.33 3.33 3.33
CH2O empirical formula
mass of CH2O
molar mass of lactate 90.08 g
30.03 g3 C3H6O3 is the
molecular formula
We now divide by the smallest number and get a ratio for the empirical
formula.
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Combustion apparatus for determining formulas
of organic compounds.Figure 3.5
CnHm + (n + ) O2 = n CO(g) + H2O(g)m
2m
2
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Sample Problem 3.6 Determining a Molecular Formula from
Combustion Analysis
PLAN:
PROBLEM: Vitamin C (M = 176.12 g/mol) is a compound of C,H, and O
found in many natural sources, especially citrus fruits. When a
1.000-g sample of vitamin C is placed in a combustion chamber
and burned, the following data are obtained:
mass of CO2 absorber after combustion = 85.35 g
mass of CO2 absorber before combustion = 83.85 g
mass of H2O absorber after combustion = 37.96 g
mass of H2O absorber before combustion = 37.55 g
What is the molecular formula of vitamin C?
The difference in absorber mass before and after combustion
is the mass of oxidation product of the element. Find the mass
of each element from its combustion product, convert each to
moles and determine the formula. Using the molar mass and
empirical mass, determine the molecular formula.
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SOLUTION:
CO2 85.35 g - 83.85 g = 1.50 g H2O 37.96 g - 37.55 g = 0.41 g
There are 12.01 g C per mol CO2. 1.50 g CO2 x12.01 g C
44.01 g CO2
= 0.409 g C
0.41 g H2O x 2.016 g H
18.02 g H2O= 0.046 g H
There are 2.016 g H per mol H2O.
O must be the difference: 1.000 g - (0.409 + 0.046) = 0.545 g O
0.409 g C
12.01 g C
0.046 g H
1.008 g H
0.545 g O
16.00 g O
= 0.0341 mol C = 0.0456 mol H = 0.0341 mol O
C1.00H1.3O1.00 C3H4O3 176.12 g/mol
88.06 g= 2.000 = 2 C6H8O6
Sample Problem 3.6 Determining a Molecular Formula from Combustion
Analysiscontinued
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