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PRINCIPLES OF CHEMISTRY I
CHEM 1211
CHAPTER 3
DR. AUGUSTINE OFORI AGYEMAN
Assistant professor of chemistry
Department of natural sciences
Clayton state university
CHAPTER 3
STOICHIOMETRY
(CHEMICAL CALCULATIONS)
STOICHIOMETRY
- The area of study involved with calculation of the quantities of
substances consumed or produced in a chemical reaction
- Chemical reactions are represented by chemical equations
- Reactants are substances that are consumed
- Products are substances that are formed
CHEMICAL EQUATIONS
- Reactants are written on the left side of a chemical
equation and products on the right side
- An arrow pointing towards the products, is used to
separate the reactants from the products
- The plus sign (+) is used to separate different reactants
or different products
CHEMICAL EQUATIONS
- Chemical equations must be consistent with
experimental facts
(reactants and products in a reaction that actually takes place)
- Chemical equations must be balanced
(equal numbers of atoms of each kind on both sides)
(Daltons atomic theory)
CHEMICAL EQUATIONS
States of reactants and products
Physical states of reactants and products are represented by:
(g): gas
(l): liquid
(s): solid
(aq): aqueous or water solution
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
BALANCING CHEMICAL EQUATIONS
- Whole numbers are placed on the left side of the formula (called
coefficients) to balance the equation (subscripts remain unchanged)
- The coefficients in a chemical equation are the smallest set
of whole numbers that balance the equation
C2H5OH(l) + O2(g)
2 C atoms
→ 2CO2(g) + H2O(g)
2 C atoms
Place the coefficient 2 in front of CO2 to balance C atoms
BALANCING CHEMICAL EQUATIONS
- Whole numbers are placed on the left side of the formula (called
coefficients) to balance the equation (subscripts remain unchanged)
- The coefficients in a chemical equation are the smallest set
of whole numbers that balance the equation
C2H5OH(l) + O2(g) → 2CO2(g) + 3H2O(g)
3(1x2)=6 H atoms
(5+1)=6 H atoms
Place 3 in front of H2O to balance H atoms
BALANCING CHEMICAL EQUATIONS
- Whole numbers are placed on the left side of the formula (called
coefficients) to balance the equation (subscripts remain unchanged)
- The coefficients in a chemical equation are the smallest set
of whole numbers that balance the equation
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
1+(3x2)=7 O atoms
(2x2)+3=7 O atoms
Place 3 in front of O2 to balance O atoms
BALANCING CHEMICAL EQUATIONS
- Whole numbers are placed on the left side of the formula (called
coefficients) to balance the equation (subscripts remain unchanged)
- The coefficients in a chemical equation are the smallest set
of whole numbers that balance the equation
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
2 C atoms
2 C atoms
(5+1)=6 H atoms
(3x2)=6 H atoms
1+(3x2)=7 O atoms
(2x2)+3=7 O atoms
- Check to make sure equation is balanced
- When the coefficient is 1, it is not written
BALANCING CHEMICAL EQUATIONS
Balance the following chemical equations
Fe(s) + O2(g) → Fe2O3(s)
C12H22O11(s) + O2(g) → CO2(g) + H2O(g)
(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + H2O(g)
TYPES OF CHEMICAL REACTIONS
Five Types of Chemical Reactions
- Combination reaction
- Decomposition reaction
- Single-replacement reaction
- Double-replacement reaction
- Combustion reaction
COMBINATION REACTION
- Addition or synthesis reaction
- Two or more reactants produce a single product
X + Y → XY
N2(g) + 3H2(g) → 2NH3(g)
2Mg(s) + O2(g) → 2MgO(s)
SO3(g) + H2O(l) → H2SO4(aq)
DECOMPOSITION REACTION
- Two or more products are formed from a single reactant
XY → X + Y
2H2O(l) → 2H2(g) + O2(g)
2NaN3(s) → 2Na(s) + 3N2(g)
BaCO3(s) → BaO(s) + CO2(g)
SINGLE-REPLACEMENT REACTION
- Substitution or Displacement reaction
- An atom or molecule replaces another atom or molecule
A + BY → B + AY
Fe(s) + CuSO4(aq)
→ Cu(s) + FeSO4(aq)
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
Cl2(g) + 2NaBr(aq) → 2NaCl(aq) + Br2(g)
- Metal replaces metal and nonmetal replaces nonmetal
- Cation replaces cation and anion replaces anion
DOUPLE-REPLACEMENT REACTION
- Exchange or metathesis (transpose) reaction
- Parts of two compounds switch places to form two new compounds
AX + BY → AY + BX
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
CaCl2(aq) + Na2CO3(aq) → CaCO3(s) + 2NaCl(aq)
COMBUSTION REACTION
- Reaction between a substance and oxygen (air) accompanied
by the production of heat and light
- A common synonym for combustion is burn
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) + heat
2CH3OH + 3O2(g) → 2CO2(g) + 4H2O(g) + heat
2Mg(s) + O2(g) → 2MgO(s) + heat
Hydrocarbons are the most common type of compounds that undergo
combustion producing CO2 and H2O
ACIDS
- Ionize in aqueous solutions to form hydrogen (H+) ions
- Proton (H+) donors
- H+ and H3O+ are used interchangeably
Ionize
- Dissolving in solution (water) to form ions
Examples
HCl(aq) → H+(aq) + Cl-(aq)
HNO3(aq) → H+(aq) + NO3-(aq)
Sum of charges on each side of equation must be equal
ACIDS
Strong Acids
- Transfer 100% (or very nearly 100%) of their protons
to H2O in aqueous solution
- Completely or nearly completely ionize in aqueous solution
- Strong electrolytes
Examples
HCl, HNO3, H2SO4, HBr, H3PO4
ACIDS
Weak Acids
- Transfer only a small percentage (< 5%) of their
protons to H2O in aqueous solution
Examples
Organic acids: acetic acid, citric acid
BASES
- Proton (H+) acceptors
- Produce hydroxide ion (OH-) when dissolved in water
Examples
NaOH → Na+(aq) + OH-(aq)
Ca(OH)2 → Ca2+(aq) + 2OH-(aq)
BASES
Strong Bases
- Completely or nearly completely ionize in aqueous solution
-Strong electrolytes
Examples
- Hydroxides of Groups 1A and 2A are strong bases
LiOH, CsOH, Ba(OH)2, Ca(OH)2
most common in lab: NaOH and KOH
BASES
Weak Bases
- produce small amounts of OH- ions in aqueous solution
Examples
methylamine, cocaine, morphine
most common: NH3
- Small amounts of NH4+ and OH- ions are
produced in aqueous solution
- The name aqueous ammonia is preferred over
ammonium hydroxide
ACID-BASE REACTION
- Also referred to as Neutralization Reaction
- Occurs when solutions of an acid and a base are mixed
- Products are salt and water when the base is a metal hydroxide
Example
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l)
A cation from a base combines with an anion from an acid
to form a salt
ACID-BASE REACTION
Gas Formation
- Carbonates and bicarbonates react with acids to form CO2 gas
HCl(aq) + NaHCO3(aq) → NaCl(aq) + H2CO3(aq)
2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2CO3(aq)
- H2CO3 (carbonic acid) is unstable and decomposes to produce CO2
H2CO3(aq) → H2O(l) + CO2(g)
- Hydrogen sulfide (H2S) is produced when Na2S reacts with an acid
2HCl(aq) + Na2S(aq) → 2NaCl(aq) + H2S(g)
OXIDATION-REDUCTION REACTION
- Also called redox reactions
- Involve transfer of electrons from one species to another
Oxidation - loss of electrons
Reduction - gain of electrons
Ionic solid sodium chloride (Na+ and Cl- ions) formed from solid
sodium and chlorine gas
2Na(s) + Cl2(g) → 2NaCl(s)
The oxidation (rusting) of iron by reaction with moist air
4Fe(s) + 3O2(g) → 2Fe2O3(s)
NONREDOX REACTION
- There is no transfer of electrons from one reactant
to another reactant
BaCO3(s) → BaO(s) + CO2(g)
- Double-replacement reactions
- Most reactions we have already come across
OXIDATION NUMBER (STATE)
The concept of oxidation number
- provides a way to keep track of electrons in redox reactions
- not necessarily ionic charges
Conventionally
- actual charges on ions are written as n+ or n- oxidation numbers are written as +n or -n
Oxidation - increase in oxidation number (loss of electrons)
Reduction - decrease in oxidation number (gain of electrons)
OXIDATION NUMBERS
1. Oxidation number of uncombined elements = 0
Na(s), O2(g), H2(g), Hg(l)
2. Oxidation number of a monatomic ion = charge
Na+ = +1, Cl- = -1, Ca2+ = +2, Al3+ = +3
3. Oxygen is usually assigned -2
H2O, CO2, SO2, SO3
Exceptions: H2O2 (oxygen = -1)
OF2 (oxygen = +2)
OXIDATION NUMBERS
4. Hydrogen is usually assigned +1 (-1 when bonded to metals)
+1: HCl, NH3, H2O
-1: CaH2, NaH
5. Halogens are usually assigned -1 (F, Cl, Br, I)
Exceptions: when Cl, Br, and I are bonded to oxygen
Cl2O: Cl
+1
O
-2
Cl
+1
6. The sum of oxidation numbers for
- neutral compound = 0
- polyatomic ion = charge
H2O = 0, CO32- = -2, NH4+ = +1
OXIDATION NUMBERS
CO2
The oxidation state of oxygen is -2
CO2 has no charge
The sum of oxidation states of carbon and oxygen = 0
1 carbon atom and 2 oxygen atoms
1(x) + 2(-2) = 0
x = +4
CO2
x
-2 for each oxygen
OXIDATION NUMBERS
CH4
x
+1
1(x) + 4(+1) = 0
x = -4
OXIDATION NUMBERS
NO3x
-2
1(x) + 3(-2) = -1
x = +5
OXIDATION NUMBERS
CH4(g) + 2O2(g)
-4
+1
→
0
CO2(g) + 2H2O(g)
+4
-2
+1
-2
Carbon
losses
8 electrons
8e- gain
0
CH4(g) + 2O2(g)
-4
-2
→
-2
CO2(g) + 2H2O(g)
+4
8e- loss
+1
+1
Each
Oxygen
atom gains
2 electrons
OXIDATION NUMBERS
Redox reactions are characterized by transfer of electrons
Oxidation
Reduction
Loss of electrons
Increase in oxidation number
Gain of electrons
Decrease in oxidation number
Reducing Agent (electron donor) Oxidizing Agent (electron acceptor)
Mnemonic
OIL RIG
Oxidation Involves Loss; Reduction Involves Gain
BALANCING REDOX EQUATIONS
• Oxidation Number Method
• Half Reaction Method
- Acidic Solutions
- Basic Solutions
OXIDATION NUMBER METHOD
Step 1: Balance the net ionic equation for all atoms other than
H and O
MnO4-(aq) + Br-(aq) → Mn2+(aq) + Br2(aq)
MnO4-(aq) + 2Br-(aq) → Mn2+(aq) + Br2(aq)
Step 2: Assign oxidation numbers to all atoms
MnO4-(aq) + 2Br-(aq) → Mn2+(aq) + Br2(aq)
+7 -2
-1
+2
0
OXIDATION NUMBER METHOD
Step 3: Determine which atoms have changed oxidation numbers
and by how much
MnO4-(aq) + 2Br-(aq) → Mn2+(aq) + Br2(aq)
+7
-1
5 e- gain
0
2 e- loss
+2
Step 4: Multiply net gain and net loss of electrons by suitable
factors to balance
Net gain: 5 x 2 = 10
Net loss: 2 x 5 = 10
OXIDATION NUMBER METHOD
Step 5: Multiply the coefficients by respective factors
Net gain: 5 x 2 = 10
2[MnO4-(aq)] + 5[2Br-(aq)]
Net loss: 2 x 5 = 10
→
2Mn2+(aq) + 5Br2(aq)
2MnO4-(aq) + 10Br-(aq) → 2Mn2+(aq) + 5Br2(aq)
Step 6: Balance the equation for O by adding H2O and for H
by the adding H+
2MnO4-(aq) + 10Br-(aq) + 16H+(aq)
→
2Mn2+(aq) + 5Br2(aq) + 8H2O(l)
Net charge: (2 x -1) + (-10) + (+16) = +4
Net charge: (2 x +2) = +4
HALF REACTION METHOD
Step 1: Assign oxidation numbers and write separate equations
for the oxidation and reduction half reactions
Cr2O72-(aq) + Cl-(aq) → Cr3+(aq) + Cl2(aq) (In Acid)
+6 -2
-1
+3
0
chloride oxidized from -1 to 0, chromium reduced from +6 to +3
Oxidation half-reaction
Cl-(aq) → Cl2(aq)
Reduction half-reaction
Cr2O72-(aq) → Cr3+(aq)
HALF REACTION METHOD
Step 2: - Balance all the elements except hydrogen and oxygen
Oxidation half-reaction
2Cl-(aq) → Cl2(aq)
Reduction half-reaction
Cr2O72-(aq) → 2Cr3+(aq)
HALF REACTION METHOD
Step 2: - Balance all the elements except hydrogen and oxygen
- Balance oxygen using H2O
Oxidation half-reaction
2Cl-(aq) → Cl2(aq)
Reduction half-reaction
Cr2O72-(aq) → 2Cr3+(aq) + 7H2O(l)
HALF REACTION METHOD
Step 2: - Balance all the elements except hydrogen and oxygen
- Balance oxygen using H2O
- Balance hydrogen using H+
Oxidation half-reaction
2Cl-(aq) → Cl2(aq)
Reduction half-reaction
Cr2O72-(aq) + 14H+(aq) → 2Cr3+(aq) + 7H2O(l)
HALF REACTION METHOD
Step 2: - Balance all the elements except hydrogen and oxygen
- Balance oxygen using H2O
- Balance hydrogen using H+
- Balance charge using electrons (e-)
Oxidation half-reaction
2Cl-(aq) → Cl2(aq) + 2eReduction half-reaction
Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)
HALF REACTION METHOD
Step 3: Multiply both balanced half-reactions by suitable
factors to equalize electron count
Oxidation half-reaction
3 x [2Cl-(aq) → Cl2(aq) + 2e-]
6Cl-(aq) → 3Cl2(aq) + 6eReduction half-reaction
Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)
HALF REACTION METHOD
Step 4: - Add the half-reactions and cancel identical species
- Check that the elements and charges are balanced
6Cl-(aq) → 3Cl2(aq) + 6eCr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)
Cr2O72-(aq) + 14H+(aq) + 6Cl-(aq)
→
3Cl2(aq) + 2Cr3+(aq) + 7H2O(l)
Net Charge: (-2) + (+14) + (6 x -1) = +6
Net Charge: (2 x +3) = +6
BASIC SOLUTIONS
Balance equation as if it were acidic
Note H+ ions and add same number of OH- ions to both sides
Cancel H+ and OH- (=H2O) with H2O on other side
MnO4-(aq) + C2O42-(aq) → MnO2(s) + CO32-(aq) (In Base)
2MnO4-(aq) + 3C2O42-(aq) + 2H2O(l)
→
2MnO2(s) + 6CO32-(aq) + 4H+(aq)
BASIC SOLUTIONS
Balance equation as if it were acidic
Note H+ ions and add same number of OH- ions to both sides
Cancel H+ and OH- (=H2O) with H2O on other side
MnO4-(aq) + C2O42-(aq) → MnO2(s) + CO32-(aq) (In Base)
2MnO4-(aq) + 3C2O42-(aq) + 2H2O(l) + 4OH-(aq)
→
2MnO2(s) + 6CO32-(aq) + 4H+(aq) + 4OH-(aq)
BASIC SOLUTIONS
Balance equation as if it were acidic
Note H+ ions and add same number of OH- ions to both sides
Cancel H+ and OH- (=H2O) with H2O on other side
MnO4-(aq) + C2O42-(aq) → MnO2(s) + CO32-(aq) (In Base)
2MnO4-(aq) + 3C2O42-(aq) + 2H2O(l) + 4OH-(aq)
→
2MnO2(s) + 6CO32-(aq) + 4H+(aq) + 4OH-(aq)
4H2O
2
BASIC SOLUTIONS
Balance equation as if it were acidic
Note H+ ions and add same number of OH- ions to both sides
Cancel H+ and OH- (=H2O) with H2O on other side
MnO4-(aq) + C2O42-(aq) → MnO2(s) + CO32-(aq) (In Base)
2MnO4-(aq) + 3C2O42-(aq) + 4OH-(aq)
→
2MnO2(s) + 6CO32-(aq) + 2H2O(l)
THE MOLE
The amount of substance of a system, which contains as many
elementary entities as there are atoms in 12 grams of carbon-12
- abbreviated mol
1 mole (mol) = 6.02214179 x 1023 entities
- known as the Avogadro’s number (after Amedeo Avogadro)
- usually rounded to 6.022 x 1023
THE MOLE
The number of entities (or objects) can be atoms or molecules
1 mol C = 6.022 x 1023 atoms C
1 mol CO2 = 6.022 x 1023 molecules CO2
2 conversion factors can be derived from each
THE MOLE
How many atoms are there in 0.40 mole nitrogen?
6.022 x 10 23 nitrogen atoms
0.40 mol nitrogen x
1 mol nitrogen
= 2.4 x 1023 nitrogen atoms
How many molecules are there in 1.2 moles water?
6.022 x 10 23 water molecules
1.2 mol water x
1 mol water
= 7.2 x 1023 water molecules
THE MOLE
How many H atoms are there in 1.2 moles water?
(6.022 x 1023 water molecules)
(2 H atoms)
1.2 mol water x
x
(1 mol water)
(1 water molecule)
= 1.4 x 1024 H atoms
MOLAR MASS
- The mass of a substance in grams that is numerically equal to
the formula mass of that substance
- Add atomic masses to get the formula mass (in u)
= molar mass (in g/mol)
- The mass, in grams, of 1 mole of the substance
MOLAR MASS
Consider the following
Sodium (Na) has an atomic mass of 22.99 u
This implies that the mass of 1 mole of Na = 22.99 g
Molar mass of Na = 22.99 g/mol
Formula mass of NaCl = 58.44 u
The mass of 1 mole of NaCl = 58.44 g
Molar mass of NaCl = 58.88 g/mol
Formula mass of CaCO3 = 100.09 u
The mass of 1 mole of CaCO3 = 100.09 g
Molar mass of CaCO3 = 100.09 g/mol
MOLAR MASS
Calculate the mass of 2.4 moles of NaNO3
Molar mass NaNO3 = 22.99 + 14.01 + 3(16.00)
= 85.00 g /mol NaNO3
85.00 g NaNO3
g NaNO3  2.4 mol NaNO3 x
1 mol NaNO3
= 204 g NaNO3
= 2.0 x 102 g NaNO3
MOLAR MASS
How many moles are present in 2.4 g NaNO3
Molar mass NaNO3 = 22.99 + 14.01 + 3(16.00)
= 85.00 g /mol NaNO3
1 mol NaNO3
mol NaNO3  2.4 g NaNO3 x
85.00 g NaNO3
= 0.028 mol NaNO3
= 2.8 x 10-2 mol NaNO3
PERCENTAGE COMPOSITION
- Percentage by mass contributed by individual
elements in a compound
% element 
% element 
mass of element
x 100%
mass of compound
(atomic mass of element)(n umber of atoms of element)
x 100%
formula mass of compound
PERCENTAGE COMPOSITION
Calculate the percentage of carbon, hydrogen, and oxygen, in
ethanol (C2H5OH)
%C 
(12.01 u)(2)
x 100%  52.14 %
46.07 u
(1.01 u)(6)
%H 
x 100%  13.13 %
46.07 u
(16.00 u)(1)
%O 
x 100%  34.73 %
46.07 u
PERCENTAGE COMPOSITION
Calculate the percent composition by mass of each element
in the following compounds
C9H8O4
(NH4)2PtCl4
C2H2F4
C8H10N4O2
Pt(NH3)2Cl2
COMBUSTION ANALYSIS
- Used for the determination of mass percentages and empirical
formula of compounds
- A combustion train is usually used for analysis of compounds
containing only carbon, hydrogen, and oxygen
- A first compartment with CaCl2 traps H2O
- A second compartment with NaOH traps CO2
- Masses of trapped H2O and CO2 are then determined
COMBUSTION ANALYSIS
O2
sample
furnace
CaCl2
NaOH
H2O trap
CO2 trap
COMBUSTION ANALYSIS
Combustion of a 0.2000-g sample of a compound made up of
only carbon, hydrogen, and oxygen yields 0.200 g H2O and
0.4880 g CO2. Calculate the mass and mass percentage of
each element present in the 0.2000-g sample.
- Convert mass H2O/CO2 to moles using molar mass
- Determine moles H/C from number of atoms and moles H2O/CO2
- Convert moles H/C to mass H/C using molar mass
- Determine mass O by subtracting total mass H and C
from mass sample
- Calculate percentages as discussed earlier
COMBUSTION ANALYSIS
1 mol
mol H 2 O  0.200 g x
 0.0111 mol H 2 O
18.02 g
mol H  0.0111 mol H 2 O x
2 mol H
 0.0222 mol H
1 mol H 2 O
mass H  0.0222 mol H x
1.01 g H
 0.0224 g H
1 mol H
COMBUSTION ANALYSIS
mol CO 2  0.4880 g x
1 mol
 0.01109 mol CO 2
44.01g
mol C  0.01109 mol CO 2 x
1 mol C
 0.01109 mol C
1 mol CO 2
mass C  0.01109 mol C x
12.01g C
 0.1332 g C
1 mol C
COMBUSTION ANALYSIS
Mass O = 0.2000 g sample – (0.0224 g + 0.1332 g)
= 0.0444 g O
%H 
0.0224 g
x 100%  11.2 %
0.2000
0.1332 g
%C 
x 100%  66.60 %
0.2000
0.0444 g
%O 
x 100%  22.2 %
0.2000
EMPIRICAL FORMULA
Given mass % elements:
- Convert to g elements assuming 100.0 g sample
- Convert to mole elements using molar mass
- Calculate mole ratio
(divide each by the smallest number of moles)
- Round each to the nearest integer
- Multiply through by a suitable factor if necessary
( __.5 x 2
or
__.33 x 3
or
__ .25 x 4)
EMPIRICAL FORMULA
Determine the empirical formula for a compound that gives the
following percentages upon analysis (in mass percents):
71.65 % Cl
24.27 % C
4.07 % H
- Assume 100.0 g of sample and convert grams to moles
71.65 g Cl
71.65 g Cl x
1 mol Cl
 2.021 mol Cl
35.45 g Cl
24.27 g C
24.27 g C x
1 mol C
 2.021 mol C
12.01 g C
4.07 g H
4.07 g H x
1 mol H
 4.04 mol H
1.01 g H
EMPIRICAL FORMULA
- Calculate mol ratios
Cl :
2.021
 1.000
2.021
C:
2.021
 1.000
2.021
4.04
H:
 1.99
2.021
- Round to nearest integers and write empirical formula
Cl: 1, C: 1, H: 2 giving CH2Cl
MOLECULAR FORMULA
Given the molar mass:
- Determine the empirical formula
- Calculate the empirical formula mass
- Divide the given molar mass by the empirical formula mass to
obtain a whole-number multiple
- Multiply all subscripts in the empirical formula by the multiple
MOLECULAR FORMULA
From previous example, calculate the molecular formula if the
molar mass is known to be 98.96 g/mol
Empirical formula = ClCH2
Empirical formula mass = 35.45 + 12.01 + 2(1.01) = 49.48 g/mol
Molar mass
98.96 g/mol

2
Empirical Formula Mass 49.48 g/mol
Molecular formula = (CH2Cl)2 = C2H4Cl2
CHEMICAL FORMULA
Subscripts represent both atomic and molar amounts
Consider Na2S2O3:
- Two atoms of sodium, two atoms of sulfur, and three atoms of
oxygen are present in one molecule of Na2S2O3
- Two moles of sodium, two moles of sulfur, and three moles of
oxygen are present in one mole of Na2S2O3
CHEMICAL FORMULA
How many moles of sodium atoms, sulfur atoms, and oxygen
atoms are present in 1.8 moles of a sample of Na2S2O3?
I mol Na2S2O3 contains 2 mol Na, 2 mol S, and 3 mol O
mol Na 1.8 mol Na 2S2 O3 x
2 mol Na
 3.6 mol Na
1 mol Na 2S2 O3
mol S 1.8 mol Na 2S2 O 3 x
2 mol S
 3.6 mol S
1 mol Na 2S2 O 3
mol O 1.8 mol Na 2S2 O 3 x
3 mol O
 5.4 mol O
1 mol Na 2S2 O 3
CHEMICAL CALCULATIONS
Calculate the number of molecules present in 0.075 g of urea,
(NH2)2CO
Given mass of urea:
- Convert to moles of urea using molar mass
- Convert to molecules of urea using Avogadro’s number
1 mole (NH 2 ) 2 CO 6.022 x 1023 molecules ( NH2 ) 2 CO
0.075 g (NH 2 ) 2 CO x
x
60.07 g (NH 2 ) 2 CO
1 mole (NH 2 ) 2 CO
= 7.5 x 1020 molecules (NH2)2CO
CHEMICAL CALCULATIONS
How many grams of carbon are present in a 0.125 g of vitamin C,
C6H8O6?
Given mass of vitamin C:
- Convert to moles of vitamin C using molar mass
- Convert to moles of C (1 mole C6H8O6 contains 6 moles C)
- Convert moles carbon to g carbon using molar mass
1 mol C 6 H 8O 6
6 mol C
12.01 g C
0.125 g C 6 H 8O 6 x
x
x
176.14 g C 6 H 8O 6 1 mol C 6 H 8O 6 1 mol C
= 0.0511 g carbon
CHEMICAL EQUATIONS
(STOICHIOMETRIC CALCULATIONS)
Coefficients represent both molecular and molar amounts
Given:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
- 1 molecule of C3H8 reacts with 5 molecules of O2 to produce
3 molecules of CO2 and 4 molecules of H2O
- 1 mole of C3H8 reacts with 5 moles of O2 to produce
3 moles of CO2 and 4 moles of H2O
CHEMICAL EQUATIONS
(STOICHIOMETRIC CALCULATIONS)
Given:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
What mass of oxygen will react with 96.1 g of propane?
- make sure the equation is balanced
- calculate moles of propane from given mass and molar mass
- determine moles of oxygen from mole ratio (stoichiometry)
- calculate mass of oxygen
1 mol C 3 H 8
5 mol O 2
32.00 g O 2
96.1 g C 3 H 8 x
x
x
44.11 g C 3 H 8 1 mol C 3 H 8
1 mol O 2
= 349 g O2
CHEMICAL EQUATIONS
(STOICHIOMETRIC CALCULATIONS)
Given:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
What mass of CO2 will be produced from 96.1 g of propane?
- make sure the equation is balanced
- calculate moles of propane from given mass and molar mass
- determine moles of CO2 from mole ratio (stoichiometry)
- calculate mass of CO2
1 mol C 3 H 8
3 mol CO2 44.01 g CO2
96.1 g C 3 H 8 x
x
x
44.11 g C 3 H 8 1 mol C 3 H 8 1 mol CO2
= 288 g CO2
LIMITING REACTANT
- Also called limiting reagent
- The reactant that is completely consumed in a reaction
- The reactant(s) with leftovers is (are) known as the excess
reactant(s) or excess reagent(s)
To determine the limiting reactant:
- Write and balance the equation for the reaction
- Use given amount of each reactant to determine amount of desired product
- The reactant that gives the smallest amount of product is the limiting
LIMITING REACTANT
Consider the following reaction for producing nitrogen gas
from gaseous ammonia and solid copper(II) oxide:
2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(g)
If a sample containing 18.1 g of NH3 is reacted with 90.4 g of
CuO, which is the limiting reactant?
LIMITING REACTANT
2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(g)
- Make sure the equation is balanced
- Calculate moles of desired product from each reactant
1 mol NH3
1 mol N 2
18.1 g NH3 x
x
 0.530 mol N 2
17.03 g NH3 2 mol NH3
90.4 g CuO x
1 mol CuO
1 mol N 2
x
 0.380 mol N 2
79.55 g CuO 3 mol CuO
CuO is limiting since it produces smaller amount of N2
PERCENT YIELD
actual yield
Percent yield 
x 100 %
theoretica l yield
Theoretical Yield
The calculated quantity of product formed,
assuming all of the limiting reactant is used up
Actual Yield
The amount of product actually obtained in a
reaction (always less than the theoretical yield)
PERCENT YIELD
Given actual yield:
- Determine the limiting reactant
- Calculate the theoretical yield from the limiting reactant
- Calculate percent yield
PERCENT YIELD
Calculate the percent yield of N2 from the previous example
if 9.04 g of N2 is actually produced
2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(g)
- CuO is the limiting reactant
- Calculate the theoretical yield
90.4 g CuO x
1 mol CuO
1 mol N 2
28.02 g N 2
x
x
 10.6 g N 2
79.55 g CuO 3 mol CuO
1 mol N 2
Percent yield 
actual yield
9.04 g N 2
x 100 % 
x 100 %  85.3 %
theoretica l yield
10.6 g N 2