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PRINCIPLES OF CHEMISTRY I
CHEM 1211
CHAPTER 3
DR. AUGUSTINE OFORI AGYEMANAssistant professor of chemistryDepartment of natural sciences
Clayton state university
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CHAPTER 3
STOICHIOMETRY(CHEMICAL CALCULATIONS)
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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
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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
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- 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
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C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
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
CHEMICAL EQUATIONS
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- 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) + H2O(g)
2 C atoms 2 C atoms
Place the coefficient 2 in front of CO2 to balance C atoms
BALANCING CHEMICAL EQUATIONS
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C2H5OH(l) + O2(g) → 2CO2(g) + 3H2O(g)
(5+1)=6 H atoms 3(1x2)=6 H atoms
Place 3 in front of H2O to balance H atoms
- 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
BALANCING CHEMICAL EQUATIONS
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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
- 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
BALANCING CHEMICAL EQUATIONS
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C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
2 C atoms(5+1)=6 H atoms
1+(3x2)=7 O atoms
2 C atoms(3x2)=6 H atoms
(2x2)+3=7 O atoms
- Check to make sure equation is balanced- When the coefficient is 1, it is not written
- 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
BALANCING CHEMICAL EQUATIONS
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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)
BALANCING CHEMICAL EQUATIONS
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TYPES OF CHEMICAL REACTIONS
Five Types of Chemical Reactions
- Combination reaction
- Decomposition reaction
- Single-replacement reaction
- Double-replacement reaction
- Combustion reaction
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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)
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DECOMPOSITION REACTION
- Two or more products are formed from a single reactant
XY → X + Y
2H2O(l) → 2H2(g) + O2(g)
BaCO3(s) → BaO(s) + CO2(g)
2NaN3(s) → 2Na(s) + 3N2(g)
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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
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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)
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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
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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
ExamplesHCl(aq) → H+(aq) + Cl-(aq)
HNO3(aq) → H+(aq) + NO3-(aq)
Sum of charges on each side of equation must be equal
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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
ExamplesHCl, HNO3, H2SO4, HBr, H3PO4
ACIDS
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ACIDS
Weak Acids
- Transfer only a small percentage (< 5%) of theirprotons to H2O in aqueous solution
ExamplesOrganic acids: acetic acid, citric acid
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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)
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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
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Weak Bases
- produce small amounts of OH- ions in aqueous solution
Examplesmethylamine, 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
BASES
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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
ExampleHCl(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
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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)
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- Also called redox reactions- Involve transfer of electrons from one species to another
Oxidation - loss of electronsReduction - gain of electrons
Ionic solid sodium chloride (Na+ and Cl- ions) formed from solidsodium and chlorine gas
2Na(s) + Cl2(g) → 2NaCl(s)
The oxidation (rusting) of iron by reaction with moist air4Fe(s) + 3O2(g) → 2Fe2O3(s)
OXIDATION-REDUCTION REACTION
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- 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
NONREDOX REACTION
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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)
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OXIDATION NUMBERS
1. Oxidation number of uncombined elements = 0 Na(s), O2(g), H2(g), Hg(l)
2. Oxidation number of a monatomic ion = chargeNa+ = +1, Cl- = -1, Ca2+ = +2, Al3+ = +3
3. Oxygen is usually assigned -2H2O, CO2, SO2, SO3
Exceptions: H2O2 (oxygen = -1) OF2 (oxygen = +2)
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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 O Cl
6. The sum of oxidation numbers for - neutral compound = 0- polyatomic ion = chargeH2O = 0, CO3
2- = -2, NH4+ = +1
+1 -2 +1
OXIDATION NUMBERS
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CO2
The oxidation state of oxygen is -2 CO2 has no charge
The sum of oxidation states of carbon and oxygen = 01 carbon atom and 2 oxygen atoms
1(x) + 2(-2) = 0x = +4
CO2
x -2 for each oxygen
OXIDATION NUMBERS
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CH4
x +1
1(x) + 4(+1) = 0x = -4
OXIDATION NUMBERS
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NO3-
x -2
1(x) + 3(-2) = -1x = +5
OXIDATION NUMBERS
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CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
+1-4 +4 +10 -2 -2
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
-4
+1 +1
+48e- loss
-2 -20
8e- gain
OXIDATION NUMBERS
Carbonlosses
8 electrons
EachOxygen
atom gains 2 electrons
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Oxidation
Loss of electronsIncrease in oxidation number
Reducing Agent (electron donor)
Reduction
Gain of electronsDecrease in oxidation number
Oxidizing Agent (electron acceptor)
MnemonicOIL RIG
Oxidation Involves Loss; Reduction Involves Gain
Redox reactions are characterized by transfer of electrons
OXIDATION NUMBERS
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• Oxidation Number Method
• Half Reaction Method
- Acidic Solutions- Acidic Solutions
- Basic Solutions- Basic Solutions
BALANCING REDOX EQUATIONS
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MnO4-(aq) + 2Br-(aq) → Mn2+(aq) + Br2(aq)
OXIDATION NUMBER METHOD
Step 1: Balance the net ionic equation for all atoms other than H and O
Step 2: Assign oxidation numbers to all atoms
+7 -1 +2 0
MnO4-(aq) + Br-(aq) → Mn2+(aq) + Br2(aq)
MnO4-(aq) + 2Br-(aq) → Mn2+(aq) + Br2(aq)
-2
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Step 3: Determine which atoms have changed oxidation numbers and by how much
Step 4: Multiply net gain and net loss of electrons by suitable factors to balance
+7 5 e- gain
2 e- loss -1
Net gain: 5 x 2 = 10Net loss: 2 x 5 = 10
+2
0
MnO4-(aq) + 2Br-(aq) → Mn2+(aq) + Br2(aq)
OXIDATION NUMBER METHOD
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Step 5: Multiply the coefficients by respective factors
Net gain: 5 x 2 = 10 Net loss: 2 x 5 = 10
2[MnO4-(aq)] + 5[2Br-(aq)] → 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
OXIDATION NUMBER METHOD
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Cl-(aq) → Cl2(aq)
Cr2O72-(aq) → Cr3+(aq)
HALF REACTION METHOD
Cr2O72-(aq) + Cl-(aq) → Cr3+(aq) + Cl2(aq) (In Acid)
Oxidation half-reaction
Reduction half-reaction
+6 -2 -1 +3 0
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Step 2: - Balance all the elements except hydrogen and oxygen
2Cl-(aq) → Cl2(aq)
Oxidation half-reaction
Reduction half-reaction
Cr2O72-(aq) → 2Cr3+(aq)
HALF REACTION METHOD
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2Cl-(aq) → Cl2(aq)
Step 2: - Balance all the elements except hydrogen and oxygen- Balance oxygen using H2O
Oxidation half-reaction
Reduction half-reaction
Cr2O72-(aq) → 2Cr3+(aq) + 7H2O(l)
HALF REACTION METHOD
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Cr2O72-(aq) + 14H+(aq) → 2Cr3+(aq) + 7H2O(l)
2Cl-(aq) → Cl2(aq)
Step 2: - Balance all the elements except hydrogen and oxygen- Balance oxygen using H2O- Balance hydrogen using H+
Oxidation half-reaction
Reduction half-reaction
HALF REACTION METHOD
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Step 2: - Balance all the elements except hydrogen and oxygen - Balance oxygen using H2O- Balance hydrogen using H+
- Balance charge using electrons (e-)
2Cl-(aq) → Cl2(aq) + 2e-
Oxidation half-reaction
Reduction half-reaction
Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)
HALF REACTION METHOD
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3 x [2Cl-(aq) → Cl2(aq) + 2e-]
Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)
Step 3: Multiply both balanced half-reactions by suitable factors to equalize electron count
6Cl-(aq) → 3Cl2(aq) + 6e-
Oxidation half-reaction
Reduction half-reaction
HALF REACTION METHOD
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Step 4: - Add the half-reactions and cancel identical species - Check that the elements and charges are balanced
6Cl-(aq) → 3Cl2(aq) + 6e-
Cr2O72-(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
HALF REACTION METHOD
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2MnO4-(aq) + 3C2O4
2-(aq) + 2H2O(l) →
2MnO2(s) + 6CO32-(aq) + 4H+(aq)
MnO4-(aq) + C2O4
2-(aq) → MnO2(s) + CO32-(aq) (In Base)
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
BASIC SOLUTIONS
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2MnO4-(aq) + 3C2O4
2-(aq) + 2H2O(l) + 4OH-(aq)
→ 2MnO2(s) + 6CO3
2-(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) + C2O4
2-(aq) → MnO2(s) + CO32-(aq) (In Base)
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2MnO4-(aq) + 3C2O4
2-(aq) + 2H2O(l) + 4OH-(aq)
→ 2MnO2(s) + 6CO3
2-(aq) + 4H+(aq) + 4OH-(aq)
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
4H2O
MnO4-(aq) + C2O4
2-(aq) → MnO2(s) + CO32-(aq) (In Base)
2
BASIC SOLUTIONS
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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) + C2O4
2-(aq) → MnO2(s) + CO32-(aq) (In Base)
2MnO4-(aq) + 3C2O4
2-(aq) + 4OH-(aq) →
2MnO2(s) + 6CO32-(aq) + 2H2O(l)
BASIC SOLUTIONS
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THE MOLE
The amount of substance of a system, which contains as manyelementary 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
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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
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THE MOLE
How many atoms are there in 0.40 mole nitrogen?
= 2.4 x 1023 nitrogen atoms
How many molecules are there in 1.2 moles water?
= 7.2 x 1023 water molecules
nitrogenmol1
atomsnitrogen10x6.022xnitrogenmol0.40
23
watermol1
moleculeswater10x6.022xwatermol1.2
23
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How many H atoms are there in 1.2 moles water?
= 1.4 x 1024 H atoms
molecule)water(1
atoms)H(2x
water)mol(1
molecules)water10x(6.022xwatermol1.2
23
THE MOLE
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MOLAR MASS
- The mass of a substance in grams that is numerically equal tothe 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
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MOLAR MASS
Consider the following
Sodium (Na) has an atomic mass of 22.99 uThis 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 uThe mass of 1 mole of NaCl = 58.44 g
Molar mass of NaCl = 58.88 g/mol
Formula mass of CaCO3 = 100.09 uThe mass of 1 mole of CaCO3 = 100.09 g
Molar mass of CaCO3 = 100.09 g/mol
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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
= 204 g NaNO3
= 2.0 x 102 g NaNO3
3
333 NaNOmol1
NaNOg85.00xNaNOmol2.4NaNOg
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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
= 0.028 mol NaNO3
= 2.8 x 10-2 mol NaNO3
3
333 NaNOg85.00
NaNOmol 1xNaNOg2.4NaNOmol
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PERCENTAGE COMPOSITION
- Percentage by mass contributed by individual elements in a compound
100%xcompoundofmass
element of masselement%
100%xcompoundofmassformula
element)ofatomsofumberelement)(nofmass(atomicelement%
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PERCENTAGE COMPOSITION
Calculate the percentage of carbon, hydrogen, and oxygen, inethanol (C2H5OH)
% 13.13100%xu 46.07
u)(6) (1.01H%
% 73.34100%xu 46.07
u)(1) (16.00O%
% 52.14100%xu 46.07
u)(2) (12.01C%
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PERCENTAGE COMPOSITION
Calculate the percent composition by mass of each elementin the following compounds
C9H8O4
(NH4)2PtCl4
C2H2F4
C8H10N4O2
Pt(NH3)2Cl2
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COMBUSTION ANALYSIS
- Used for the determination of mass percentages and empiricalformula of compounds
- A combustion train is usually used for analysis of compoundscontaining 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
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COMBUSTION ANALYSIS
furnace
CaCl2 NaOHO2
sample
H2O trap CO2 trap
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COMBUSTION ANALYSIS
Combustion of a 0.2000-g sample of a compound made up ofonly carbon, hydrogen, and oxygen yields 0.200 g H2O and0.4880 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
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COMBUSTION ANALYSIS
OHmol0.0111g18.02
mol1xg0.200OHmol 22
Hmol0.0222OHmol1
Hmol2xOHmol0.0111Hmol
22
H0.0224Hmol1
H01.1xHmol0.0222Hmass g
g
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COMBUSTION ANALYSIS
22 COmol0.01109g44.01
mol1xg0.4880COmol
Cmol0.01109COmol1
Cmol1xCOmol0.01109Cmol
22
Cg0.1332Cmol1
Cg12.01xCmol0.01109Cmass
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COMBUSTION ANALYSIS
Mass O = 0.2000 g sample – (0.0224 g + 0.1332 g)
= 0.0444 g O
% .6066100%x0.2000
g 0.1332C%
% .211100%x0.2000
g 0.0224H%
% .222100%x0.2000
g 0.0444O%
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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)
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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
Clmol2.021Clg35.45
Clmol1xClg71.65
Cmol2.021Cg12.01
Cmol1xCg24.27
Hmol04.4 Hg 1.01
Hmol1xHg07.4
71.65 g Cl
24.27 g C
4.07 g H
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EMPIRICAL FORMULA
1.0002.021
2.021:Cl
1.0002.021
2.021:C
- Calculate mol ratios
99.12.021
4.04:H
- Round to nearest integers and write empirical formula
Cl: 1, C: 1, H: 2 giving CH2Cl
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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
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MOLECULAR FORMULA
From previous example, calculate the molecular formula if themolar 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
2g/mol49.48
g/mol98.96
MassFormulaEmpirical
massMolar
Molecular formula = (CH2Cl)2 = C2H4Cl2
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CHEMICAL FORMULA
Subscripts represent both atomic and molar amounts
Consider Na2S2O3:
- Two atoms of sodium, two atoms of sulfur, and three atoms ofoxygen are present in one molecule of Na2S2O3
- Two moles of sodium, two moles of sulfur, and three moles ofoxygen are present in one mole of Na2S2O3
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CHEMICAL FORMULA
How many moles of sodium atoms, sulfur atoms, and oxygenatoms 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
Namol3.6OSNamol1
Namol2xOSNamol1.8Namol
322322
Smol3.6OSNamol1
Smol2xOSNamol1.8Smol
322322
Omol5.4OSNamol1
Omol3xOSNamol1.8Omol
322322
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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
= 7.5 x 1020 molecules (NH2)2CO
CO)(NHmole1
CO)NH(molecules10x6.022x
CO)(NHg60.07
CO)(NHmole1xCO)(NHg0.075
22
2223
22
2222
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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
= 0.0511 g carbon
Cmol1
Cg12.01x
OHCmol1
Cmol6x
OHCg176.14
OHCmol1xOHCg0.125
686686
686686
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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
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- 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
= 349 g O2
CHEMICAL EQUATIONS(STOICHIOMETRIC CALCULATIONS)
2
2
83
2
83
8383 Omol1
Og32.00x
HCmol1
Omol5x
HCg44.11
HCmol1xHCg96.1
Given: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
What mass of oxygen will react with 96.1 g of propane?
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- 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
= 288 g CO2
CHEMICAL EQUATIONS(STOICHIOMETRIC CALCULATIONS)
2
2
83
2
83
8383 COmol1
COg44.01x
HCmol1
COmol3x
HCg44.11
HCmol1xHCg96.1
Given: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
What mass of CO2 will be produced from 96.1 g of propane?
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- 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
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Consider the following reaction for producing nitrogen gas from gaseous ammonia and solid copper(II) oxide:
LIMITING REACTANT
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?
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- Make sure the equation is balanced- Calculate moles of desired product from each reactant
LIMITING REACTANT
2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(g)
23
2
3
33 Nmol0.530
NHmol2
Nmol1x
NH g 17.03
NHmol1xNHg18.1
22 Nmol0.380
CuOmol3
Nmol1x
CuO g 79.55
CuOmol1xCuOg90.4
CuO is limiting since it produces smaller amount of N2
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PERCENT YIELD
%100xyieldltheoretica
yieldactualyieldPercent
Theoretical YieldThe calculated quantity of product formed,
assuming all of the limiting reactant is used up
Actual YieldThe amount of product actually obtained in a
reaction (always less than the theoretical yield)
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PERCENT YIELD
Given actual yield:
- Determine the limiting reactant
- Calculate the theoretical yield from the limiting reactant
- Calculate percent yield
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PERCENT YIELD
Calculate the percent yield of N2 from the previous example if 9.04 g of N2 is actually produced
- CuO is the limiting reactant
- Calculate the theoretical yield
2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(g)
22
22 Ng10.6Nmol1
Ng28.02x
CuOmol3
Nmol1x
CuO g 79.55
CuOmol1xCuOg90.4
%85.3%100xNg10.6
Ng9.04%100x
yieldltheoretica
yieldactualyieldPercent
2
2