Daniel L. Reger
Scott R. Goode
David W. Ball http://academic.cengage.com/chemistry/reger

• Stoichiometry is the study of the quantitative relationships involving the substances in chemical reactions.
• A chemical equation describes the identities and relative amounts of reactants and products in a chemical reaction.
• Just like a chemical formula, a chemical equation expresses quantitative relations.

• A chemical equation is a shorthand notation to describe a chemical reaction.
2Mg + O
2

2MgO magnesium react to form magnesium and oxygen oxide

• Reactants are the substances consumed.
• Products are the substances formed.
• Coefficients are numbers before the formula of a substance in an equation.
• A balanced equation has the same number of atoms of each element on both sides of the equation.

•
•
Write the correct formula for each substance.
H
2
+ Cl
2

HCl (unbalanced)
Add coefficients so the number of atoms of each element are the same on both sides of the equation.
H
2
+ Cl
2

2 HCl (balanced)

•
•
•
•
•
Write the correct formula for each substance.
C
5
H
12
+ O
2

CO
2
+ H
2
O
Assume one molecule of the most complicated substance balance C.
C
5
H
12
. Adjust the coefficient of CO
+ O
2

5 CO
2
+ H
2
O
2 to
Adjust the coefficient of H
2
O to balance H.
C
5
H
12
+ O
2

5CO
2
+ 6 H
2
O
Adjust the coefficient of O
C
5
H
12
+ 8 O
2

5CO
2
2 to balance O.
+ 5H
2
O
Check the balance by counting the number of atoms of each element.

Test Your Skill
• Balance the equation
C
4
H
9
OH + O
2

CO
2
+ H
2
O

Test Your Skill
• Balance the equation
C
4
H
9
OH + O
2

CO
2
+ H
2
O
• Answer:
C
4
H
9
OH + 6O
2

4CO
2
+ 5H
2
O

•
•
Sometimes fractional coefficients are obtained.
C
5
H
10
C
5
H
10
C
5
H
10
C
5
H
10
+ O
+ O
2
2

CO
2

5 CO
2
+ H
2
O
+ H
2
O
+ O
+
2
15/2

5CO
2
O
2
+ 5

5CO
2
H
2
O
+ 5H
2
O
Multiply all coefficients by the denominator.
2 C
5
H
10
+ 15 O
2

10 CO
2
+ 10 H
2
O

• Neutralization is the reaction of an acid with a base to form a salt and water.
• An acid is a compound that dissolves in water to produce hydrogen ions.
• A base dissolves in water to produce hydroxide ions. It is usually a soluble metal hydroxide.
• A salt is an ionic compound consisting of the cation of a base and the anion of an acid.

HCl + NaOH

NaCl + H
2
O
H
2
SO
4
2HClO acid
4
+ 2KOH

K
2
SO
4
+ 2H
2
O
+ Ca(OH)
2

Ca(ClO
4
)
2
+ 2H
2
O base
 salt water

• The number of hydrogen ions provided by the acid must equal the number of hydroxide ions provided by the base.
H
3
PO
4
+ 3NaOH

3H + + 3OH -

Na
3
PO
4
+ 3H
3H
2
2
O
O

• A combustion reaction is the process of burning.
• Most combustion reactions are the combination of a substance with oxygen.
• When an organic compound burns in oxygen, the carbon is converted to CO
2
, and the hydrogen forms water, H
2
O .

Test Your Skill
• Identify the type of the reaction, then balance it.
(a) (C
2
H
5
(b) H
2
SO
)
2
4
O + O
2

CO
2
+ Ca(OH)
2
+ H

CaSO
4
2
O
+ H
2
O

• Oxidation is the loss of electrons by a chemical process.
• When sodium forms a compound, Na + is formed. Sodium is oxidized.
• Reduction is the gain of electrons by a chemical process.
• When Cl ions are formed from elemental chlorine, chlorine is reduced.

• An oxidation-reduction reaction , or redox reaction , is one in which electrons are transferred from one species to another.
• In every redox reaction, at least one species is oxidized and at least one species is reduced.
• 2Na(s) + Cl
2
(g) → 2NaCl(s) is a redox reaction because Na is oxidized and Cl is reduced.

• An oxidizing agent is the reactant that accepts electrons, causing an oxidation to occur.
• The oxidizing agent is reduced.
• A reducing agent is the reactant that supplies electrons, causing a reduction to occur.
• The reducing agent is oxidized.
• In the reaction of sodium with chlorine, Na is the reducing agent and Cl agent.
2 is the oxidizing

• In a half-reaction , either the oxidation or reduction part of a redox reaction is given, showing the electrons explicitly.
• Half-reactions emphasize the transfer of electrons in a redox reaction.
• For 2Na(s) + Cl
2
(g) → 2NaCl(s) :
• Na → Na + + 1e oxidation half-reaction
• Cl
2
+ 2e → 2Cl reduction half-reaction

• The oxidation state is the charge on the monatomic ion, or the charge on an atom when the shared electrons are assigned to the more electronegative atom.
• Electron pairs shared by atoms of the same element are divided equally.
• In CaCl
2
, an ionic compound:
• calcium has an oxidation state of +2.
• chlorine has an oxidation state of -1.

Rules for Assigning Oxidation Numbers
1.
Oxidation numbers for atoms in their elemental form are 0.
2.
The oxidation number of a monatomic ion is equal to the charge on the ion.
3.
In compounds, F is always -1. Other halogens are also -1 unless they are combined with a more electronegative element (O or a halogen above it in the periodic table).

Rules for Assigning Oxidation Numbers
(cont’d)
4. In compounds, O is -2 except for peroxides (where it is -1) or when combined with F.
5. In compounds, H is +1 except in metal hydrides, where it is -1.
6. The sum of all the oxidation numbers of the atoms in a substance must sum to the charge on the substance.

• Assign oxidation numbers for each atom in K
2
CrO
4
.
• K = +1 by rule 2.
• O = -2 by rule 5.
• Cr = +6 by rule 6.
• 2(+1) + 6 + 4(-2) = 0, the overall charge on the substance.

Test Your Skill
• Assign oxidation numbers to each atom in the following substances.
(a) PF
3
(b) CO (c) NH
4
Cl

• Determine oxidation numbers that are changing and write the skeleton halfreactions.
• Balance each half-reaction separately.
• Balance element being oxidized or reduced.
• Balance all other elements except H and O.
• Balance O by adding H
2
O as needed.
• Balance H by adding H + as needed.
• Balance charges by adding e as needed.

Balancing Redox Equations (cont’d)
• Multiply one or both reactions by an integer so that the number of electrons in both half-reactions is the same.
• Add the two half-reactions, canceling out the electrons and any other species that appears on both sides of the equation.

Balancing Redox Equations (cont’d)
• Balance the following equation.
• Cu + NO
3
→ Cu 2+ + NO
• Cu is oxidized from 0 to +2
• Write a skeleton equation for Cu:
• Cu → Cu 2+
• No other step necessary except to balance charges with electrons:
• Cu → Cu 2+ + 2e -

Balancing Redox Equations (cont’d)
• N is reduced from +5 to +2.
• Write a skeleton equation for N:
• NO
3
→ NO
• Balance O by adding water:
• NO
3
→ NO + 2H
2
O
• Balance H by adding H + :
• 4H + + NO
3
→ NO + 2H
2
O
• Balance charge by adding e :
• 3e + 4H + + NO
3
→ NO + 2H
2
O

Balancing Redox Equations (cont’d)
• Make number of electrons in both reactions the same by multiplying to least common multiple (6, in this case):
• 3 ×(Cu → Cu 2+ + 2e )
• 2 ×(3e + 4H + + NO3 → NO + 2H2O)
• Combine the two reactions and cancel as appropriate:
• 3Cu + 8H + + 2NO3 → 3Cu 2+ + 2NO + 4H2O
• Reaction is balanced.

• In basic solutions, add OH to each side of the reaction to eliminate any H + by combining them to make H
2
O.

Test Your Skill
• Balance the following equation in acid solution:
• Cr
2
O
7
2+ C
2
H
5
OH → Cr 3+ + CO
2

Test Your Skill
• Balance the following equation in basic solution:
• Zn + ClO → Zn(OH)
4
2+ Cl -

•
•
One mole is the amount of substance that contains as many entities as the number of atoms in exactly 12 grams of the 12 C isotope of carbon.
Avogadro’s number is the experimentally determined number of 12 C atoms in 12 g, and is equal to
6.022 x 10 23 .

Argon in green balloons
H g Cu
Na
Fe
Al

• One mole of anything contains
6.022 x 10 23 entities.
• 1 mol H = 6.022 x 10 23 atoms of H
•
•
•
1 mol H
2
1 mol CH
CH
4
= 6.022 x 10 23 molecules of H
2
4
= 6.022 x 10 23 molecules of
1 mol CaCl
2
CaCl
2
= 6.022 x 10 23 formula units of

Avogadro’s number allows the interconversion of moles and numbers of atoms or molecules.
Moles of substance
Avogadro’s number
Number of atoms or molecules

• How many atoms are present in 0.35 mol of Na?
• How many moles are present in 3.00 x
10 21 molecules of C
2
H
6
?

• The molar mass ( M ) of any atom, molecule or compound is the mass (in grams) of one mole of that substance.
• The molar mass in grams is numerically equal to the atomic mass or molecular mass expressed in u .
• Molar mass converts from mass (in grams) to amount (in moles) or the reverse.

Substance
Ar
C
2
H
6
NaF
Atomic Scale
Name Mass atomic mass 39.95 u molecular mass
30.07 u formula mass 41.99 u
Lab Scale
Molar Mass
39.95 g/mol
30.07 g/mol
41.99 g/mol

Moles of substance
Molar mass of substance
Mass of substance

Molar Mass Conversion
• What is the mass of 0.25 moles of CH
4
?
• Molar mass of CH
4
= 16.0 g/mol.

• How many moles of ethylene (C
2
H
4
,
M = 28.0 g/mol) are present in 16 g of that compound?
• What is the mass, in grams, of 0.178 moles of Fe?

Test Your Skill
• What mass of compound must be weighed out to have a 0.0223 mol sample of H
2
C
2
O
4
( M = 90.04 g/mol)?

• Use the formula of the compound to calculate the mass of each element in the compound and use those numbers to calculate percentage composition.

• What is the mass percentage composition of H
2
C
2
O
4
?

The sample burns in excess O
2
: the H
2
O is trapped by the CaCl
2 and CO
2 is trapped by the NaOH.

• The masses of C and H in the sample are calculated from the masses of H
2
O and
CO
2 formed in the combustion reaction.

• A compound contains only C, H, and O.
A 0.1000 g-sample burns completely in oxygen to form 0.0930 g of water and
0.227 g of CO
2
. Calculate the mass of each element in this sample.

• What is the empirical formula of a compound that contains 0.799 g C and
0.201 g H in a 1.000 g sample?

Example: Empirical Formula
• What is the empirical formula of a chromium oxide that is 68.4% Cr and
31.6% O?

Test Your Skill
• What is the empirical formula of a compound that is 59.9% Ti and 40.1%
O?

• The molecular formula must be a whole number multiple of the empirical formula.
• If the empirical formula is CH
2
, the molecular formula is (CH
2
) n where n
 molar mass of compound molar mass of empirical formula
• The molecular mass must be measured experimentally.

Example: Molecular Formula
• An empirical formula calculation based on a combustion analysis experiment shows a new compound has the empirical formula C
4
H
8
O. A mass spectrometry experiment determines that the molar mass of the compound is 216 g/mol. What is the molecular formula?

• Write the balanced equation.
• Calculate the number of moles of the species for which the mass is given.
• Use the coefficients in the equation to convert the moles of the given substance into moles of the substance desired.
• Calculate the mass of the desired species.

Example: Stoichiometry
• What mass of SO
3 forms from the reaction of 4.1 g of SO
2 with an excess of
O
2
?

Test Your Skill
• Given the equation
4FeS
2
+ 11O
2

2Fe
2
O
3
+ 8SO
2 what mass of SO
2 is produced from reaction of 3.8 g of FeS
2 oxygen?
with excess

• The previous calculation showed that given the equation
4FeS
2
4.1 g SO
2
+ 11O
2

2Fe
2
O
3
+ 8SO
2 is produced from reaction of
3.8 g of FeS
2 with excess oxygen.
• The 4.1 g SO
2 is the theoretical yield the maximum quantity of product that
– can be obtained from a chemical reaction, based on the amounts of starting materials.

• Limiting reactant : the reactant that is completely consumed when a chemical reaction occurs.

• In the below reaction of Cl
2
(green) and Na
(purple), Cl
2 is the limiting reactant . Once the limiting reactant is consumed the reaction stops and no more product forms.
The formed NaCl and the excess Na are present at the end of the reaction.

Example: Limiting Reactant
• Calculate the mass of the NH
3 product formed (theoretical yield) when 7.0 g of
N
2 reacts with 2.0 g of H
2
.

Mass of A
(reactant)
Molar mass of A
Mass of B
(reactant)
Molar mass of B
Moles of A
Moles of
Product
Moles of B
Coefficients in the equation
Choose smaller amount
Moles of
Product
Molar mass of product
Mass of product

Example: Limiting Reactant
• Calculate the mass of the NH
3 product formed (theoretical yield) when 7.0 g of
N
2 reacts with 2.0 g of H
2
.

•
•
As shown in the picture, in many reactions not all of the product formed can be isolated.
Actual yield : mass of product isolated in a reaction.
Yields are generally reported as a percent:
Percent yield

Actual yield
Theoretica l yield

100 %

Given the reaction,
PCl
3
+ Cl
2

PCl
5 what is the percent yield when 50.0 g of
PCl
3 reacts with 35.0 g Cl
2 and a chemist isolated 61.3 g of PCl
5
.?