PRINCIPLES OF CHEMISTRY I CHEM 1211 CHAPTER 3 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of...
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Transcript of PRINCIPLES OF CHEMISTRY I CHEM 1211 CHAPTER 3 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of...
PRINCIPLES OF CHEMISTRY I
CHEM 1211
CHAPTER 3
DR. AUGUSTINE OFORI AGYEMANAssistant professor of chemistryDepartment 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 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
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
- 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
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
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
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
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
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)
BaCO3(s) → BaO(s) + CO2(g)
2NaN3(s) → 2Na(s) + 3N2(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
ExamplesHCl(aq) → H+(aq) + Cl-(aq)
HNO3(aq) → H+(aq) + NO3-(aq)
Sum of charges on each side of equation must be equal
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
ACIDS
Weak Acids
- Transfer only a small percentage (< 5%) of theirprotons to H2O in aqueous solution
ExamplesOrganic 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)
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
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
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
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)
- 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
- 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
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 = chargeNa+ = +1, Cl- = -1, Ca2+ = +2, Al3+ = +3
3. Oxygen is usually assigned -2H2O, CO2, SO2, SO3
Exceptions: H2O2 (oxygen = -1) OF2 (oxygen = +2)
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
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
CH4
x +1
1(x) + 4(+1) = 0x = -4
OXIDATION NUMBERS
NO3-
x -2
1(x) + 3(-2) = -1x = +5
OXIDATION NUMBERS
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
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
• Oxidation Number Method
• Half Reaction Method
- Acidic Solutions- Acidic Solutions
- Basic Solutions- Basic Solutions
BALANCING REDOX EQUATIONS
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
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
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
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
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
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
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
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
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
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
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
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)
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
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
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
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?
= 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
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
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
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
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
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
PERCENTAGE COMPOSITION
- Percentage by mass contributed by individual elements in a compound
100%xcompoundofmass
element of masselement%
100%xcompoundofmassformula
element)ofatomsofumberelement)(nofmass(atomicelement%
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%
PERCENTAGE COMPOSITION
Calculate the percent composition by mass of each elementin the following compounds
C9H8O4
(NH4)2PtCl4
C2H2F4
C8H10N4O2
Pt(NH3)2Cl2
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
COMBUSTION ANALYSIS
furnace
CaCl2 NaOHO2
sample
H2O trap CO2 trap
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
COMBUSTION ANALYSIS
OHmol0.0111g18.02
mol1xg0.200OHmol 22
Hmol0.0222OHmol1
Hmol2xOHmol0.0111Hmol
22
H0.0224Hmol1
H01.1xHmol0.0222Hmass g
g
COMBUSTION ANALYSIS
22 COmol0.01109g44.01
mol1xg0.4880COmol
Cmol0.01109COmol1
Cmol1xCOmol0.01109Cmol
22
Cg0.1332Cmol1
Cg12.01xCmol0.01109Cmass
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%
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
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
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
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 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
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
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
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
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
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
- 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?
- 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?
- 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:
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?
- 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
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)
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
- 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