chapter 3 - stoichiometry
TRANSCRIPT
Chapter 4 — Intro—1 1
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CHAPTER 3 Stoichiometry and
Solution Concentration
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Topic Scopes:
• Molarity, molality, parts per million & percentage (w/w, w/v and v/v)
• Stoichiometry calculation
• Limiting reactant• Theoretical yield, actual yield and
percentage of yield
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Mole Concept
No. of Moles = Mass (g)molar mass (g/mol)
• 1 mole contains 1 Avogadro’s number (6.022 x 1023)
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Molarity ( M)
Molarity ( M) = Amount of solute (Mol)Volume of solution (L)
• Molarity (molar concentration) is the number of moles of a solute that is contained in 1 liter of solution
Chapter 4 — Intro—1 2
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Example: Saline Water Concentration
Typical seawaters contain sodium chloride, NaCl, as much as 2.7 g per 100 mL.
(a) What is the molarity of NaCl in the saline water?
(b) The MgCl2 content of the saline water is
0.054 M. Determine the weight (grams)
of MgCl2 in 50 mL of the saline water?
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Solution:
(a) Molecular weight of NaCl = 22.99 +35.54 = 58.44 g mol-1
Moles of NaCl in 100 mL of saline water= 2.7g /(58.44 g mol-1) = 0.046 mol
∴ Molarity of saline water = Mol/L
= 0.046 mol /(100/1000)L = 0.46 M
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Solution:(b) Molecular weight of MgCl2
= 24.30 + 2(35.45) = 95.20 g mol-1
Moles of MgCl2 =Molarity (M) x volume of solution (V)
∴Weight of MgCl2 in 50 mL of saline water= (M x V) x MW= 0.054 mol L-1 x (50/1000)L x 95.20 g mol-1
= 0.26 g8
Molality ( m)
• Molality is the number of moles of solute per kilogram (1000 g) of solvent
Molality ( m) = Amount of solute (Mol)Mass of solvent (kg)
Chapter 4 — Intro—1 3
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Percent Composition (Concentration In Percentage)
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Example:
A solution contains 118.5 g KI per liter of solution. Calculate the concentration in (a) % w/v & (b) % w/w. Given the density of the solution at 25°C is 1.078 g mL-1
Solution:
(a) % w/v = 118.5 g x 100%
1000 mL
= 11.85 % w/v
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Solution:
(b) % w/w = 118.5 g x 1 mL x 100%
1000 mL 1.078 g
= 10.99 % w/w
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Units of Low Concentration
• Parts per million, (ppm) is grams of solute per million grams of total solution/ mixture
• ppm = mass of solute
mass of sample
Unit of ppm: w/w = µg/g or mg/kg
w/v = µg/mL or mg/L
v/v = nL/mL or µL/L
x 106
Chapter 4 — Intro—1 4
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Units of Low Concentration
• Parts per billion, (ppb) is grams of solute per billion grams of total solution/ mixture
• ppb = mass of solute
mass of sample
Unit of ppb: w/w = ng/g or µg/kg
w/v = ng/mL or µg/L
v/v = nL/L
x 109
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Conversion of ppm to molarity
An aqueous solution (10.0 mL) contains
56 ppm SO2. Calculate the molarity of the solution.
Solution:
MW of SO2 = 32 + 2(16) = 64 g mol-1
56 ppm ≅ 56 mg/L
M10x8.75(1 L)molg64.0
g10x56 41-
3−
−=÷
=MMMM
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Stoichiometry
• The relationship between the quantities of chemical reactants and products
• Depend on the principle of the conservation of matter
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Reaction of Phosphorus with Cl 2Reaction of Phosphorus with Cl 2
Notice the stoichiometric coefficients and the physical states of the reactants and products
Chapter 4 — Intro—1 5
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Chemical Equations
• Depict the kind of reactants and products and their relative amounts in a reaction
4 Al(s) + 3 O2(g) → 2 Al2O3(s)
stoichiometric coefficients
• (s),(g),(l) – physical states of compounds
• (s) – solid, (g) – gas, (l) – liquid
(aq) – aqueous solution
reactants products
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Law of The Conservation of Matter
Lavoisier, 1788
• States that matter can be neither created nor destroyed
• An equation must be balanced
• It must have the same number of atoms of the same kind on both sides of the equation
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Law of The Conservation of Matter
• P4(s) + 6 Cl2(g) 4 PCl3(l)
• Total mass of reactants is 10g, must end up with 10g of products if the reaction completely converts reactants to products
12 Cl atoms 12 Cl atoms
4 P atoms 4 P atoms
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• Unbalanced equation:NH3(g) + O2(g) → NO(g) + H2O(g)
Balanced equation:2 NH3(g) + 5/2 O2(g) → 2 NO(g) + 3 H2O(g)
x each coefficient on both sides by 2:4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
Balanced Chemical Equation
Fraction
OR
Chapter 4 — Intro—1 6
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Theoretical Yield• The calculated maximum quantity of
product can be obtained form a chemical reaction / given quantities of reactants
Actual Yield
• The quantity of product that is actually obtained in laboratory / a chemical plant
• Actual yield < theoretical yield
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• Loss of product often occurs during isolation & purification steps
• Some reactions do not react completely to products of central interest (side reactions) but give > than 1 set of products (unintended products / by-products)
• If a reverse reaction occurs, some of the expected product may react to reform the reactants
? Actual Yield < Theoretical Yield
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Percentage (%) YieldPercentage = actual yield x 100%
Yield (%) theoretical yield
N2O = 42.00 g/mol
PROBLEM: If 454 g of NH4NO3 decomposes, how much N2O and H2O are formed? What is the theoretical yield of products?
Given molecular weight: NH4NO3 = 80.04 g/mol, H2O = 18.02 g/mol, N2O = 42.00 g/mol 24
STEP 1:Write the balanced chemical equation
NH4NO3 →→→→ N2O + 2 H2O
454 g of NH 4NO3 →→→→ N2O + 2 H2O454 g of NH 4NO3 →→→→ N2O + 2 H2O
STEP 2:Convert mass reactant of NH4NO3 (454 g) →moles
Moles of NH4NO3 :
454 g • 1 mol
80.04 g = 5.68 mol NH 4NO3
Chapter 4 — Intro—1 7
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STEP 3:
• Convert moles reactant → moles product
• 1 mol NH4NO3 → 2 mol H2O
• Express this relation as the STOICHIOMETRIC FACTOR
2 mol H 2O produced1 mol NH 4NO3 used
454 g of NH 4NO3 →→→→ N2O + 2 H2O454 g of NH 4NO3 →→→→ N2O + 2 H2O
26= 11.4 mol H2O produced
5.68 mol NH 4NO 3 • 2 mol H 2O produced1 mol NH 4NO 3 used
STEP 4:
Convert moles reactant (5.68 mol) → moles product
454 g of NH 4NO3 →→→→ N2O + 2 H2O454 g of NH 4NO3 →→→→ N2O + 2 H2O
Moles of H2O 2 mol H2OMoles of NH4NO3 1 mol NH4NO3
=
∴ Moles of H2O
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11.4 mol H 2O • 18.02 g1 mol
= 204 g H 2O
STEP 5:
Convert moles product (11.4 mol) → mass product
Called the THEORETICAL YIELD
ALWAYS FOLLOW THESE STEPS IN ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!
Theoretical Yield
Mass of H2O:
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STEP 6:
• How much N2O is formed?
• Total mass of reactants = total mass of
products
• 454 g NH4NO3 = ___ g N2O + 204 g H2O
• Mass of N2O = 250 g (Theoretical Yield)
454 g of NH 4NO3 →→→→ N2O + 2 H2O454 g of NH 4NO3 →→→→ N2O + 2 H2O
Chapter 4 — Intro—1 8
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• If you isolated only 131 g of N2O, what is
the percent yield?
• This compares the theoretical (250 g) and
actual (131 g) yields.
454 g of NH 4NO3 →→→→ N2O + 2 H2O454 g of NH 4NO3 →→→→ N2O + 2 H2O
Percentage (%) Yield
% yield = actual yield
theoretical yield • 100%
% yield = 131 g250. g
• 100% = 52.4%30
GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS
GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS
Mass Reactant (A)
StoichiometricfactorMoles
reactantMoles
product
Mass Product (B)
x (1 mol A/g A) x (g B/mol B)
x (y mol product B)
(x mol reactant A)
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PROBLEM:Using 5.00 g of H2O2, what mass of O2 and of H2O can be obtained?
PROBLEM:Using 5.00 g of H2O2, what mass of O2 and of H2O can be obtained?
• 2 H2O2(l) → 2 H2O(g) + O2(g)
• Reaction is catalyzed by MnO2
• Step 1: moles of H2O2
• Step 2: use STOICHIOMETRIC FACTOR to calculate moles of O2
• Step 3: mass of O232
Reactions Involving aLIMITING REACTANTReactions Involving aLIMITING REACTANT
• The reagent that is completely used up & it LIMITS the quantity of products formed in a reaction
Reactants Products
Chapter 4 — Intro—1 9
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STEP 1:Write the balanced chemical equation
2 Al + 3 Cl 2 →→→→ Al 2Cl6
PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form?
Molecular weight: Al = 27.00 g/mol, Cl = 35.45 g/mol
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Step 2: Calculate moles of each reactantMoles calculation based on limiting reactant
We have 5.40 g of Al and 8.10 g of Cl2 ,
Mole of Al = 5.40 g = 0.200 mol Al
27.00 g/mol
Mole of Cl2 = 8.10 g = 0.114 mol Cl2
70.90 g/mol
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• Reactants must be in the mole ratio:
Step 3: Determine limiting reactantCompare actual mole ratio of reactants to theoretical mole ratio
mol Cl 2mol Al
= 32
2 Al + 3 Cl 2 →→→→ Al2Cl6
Theoretical Theoretical mole ratiomole ratio
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Deciding on the Limiting Reactant
• If
• There is not enough Al to use up all the Cl2∴ Limiting reactant = Al
mol Cl2mol Al
> 32
• If
• There is not enough Cl2 to use up all the Al
∴ Limiting reactant = Cl2
mol Cl2mol Al
< 32
2 Al + 3 Cl 2 →→→→ Al2Cl6
Chapter 4 — Intro—1 10
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Step 4: Find mole ratio of reactants
This should be 3/2 or 1.5 if reactants are present in the exact stoichiometric ratio
Limiting reactant is Cl2
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0.57 = mol 0.200 mol 0.114
= AlmolCl mol 2 <
Actual Mole Ratio:
mol Cl2mol Al
= 32
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Limiting reactant = Cl2All calculations are based on Cl2Limiting reactant = Cl2All calculations are based on Cl2
moles Al2Cl6
massCl2
molesCl2
mass Al2Cl6
2
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Cl mol 3Cl Almol 1
PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form?
2 Al + 3 Cl 2 →→→→ Al2Cl6
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Calculation of mass of Al 2Cl6 expectedCalculation of mass of Al 2Cl6 expected
Step 1: Calculate mole of Al2Cl6 expectedbased on limiting reactant
622
622 Cl Almol 0.0380 =
Cl mol 3Cl Almol 1
• Cl mol 0.114
Mole of Al2Cl6 1 mol Al2Cl6=
Mole of Cl2 3 mol Cl2
Mole of Al2Cl6:
40
Step 2: Calculate mass of Al2Cl6 expected based on limiting reactant
6262
62 Cl Alg 10.1 = mol
Cl Alg 266.4 • Cl Almol 0.0380
Mass of Al2Cl6
= Mole of Al2Cl6 x molar mass of Al2Cl6
Mass of Al2Cl6:
Chapter 4 — Intro—1 11
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• Cl2 was the limiting reactant.
Therefore, Al was present in
excess. But how much?
• First find how much Al was required.
• Then find how much Al is in excess..
Problem:How much of which reactant will remain when reaction is complete?
Problem:How much of which reactant will remain when reaction is complete?
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2 Al + 3 Cl2 products
0.200 mol0.200 mol 0.114 mol = LR0.114 mol = LR
Calculating Excess AlCalculating Excess Al
Excess Al = Al available - Al required
= 0.200 mol - 0.0760 mol
= 0.124 mol Al in excess (convert to mass)
required Almol 0.0760 = Cl mol 3 Almol 2
• Cl mol 0.1142
2
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Chemical Analysis
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Chemical Analysis
• An impure sample of the mineral thenardite contains Na2SO4.
• Mass of mineral sample = 0.123 g
• The Na2SO4 in the sample is converted to insoluble BaSO4
• The mass of BaSO4 = 0.177 g
• What is the mass percent of Na2SO4 in the mineral?
Given molecular weight:
BaSO4 = 233.4 g/mol, Na2SO4 = 142.0 g/mol
Chapter 4 — Intro—1 12
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• Mole of BaSO4
= 0.177 g BaSO4 /(233.4 g mol-1) = 7.58 x 10-4 mol
• Using stoichiometric factor,
Moles of Na2SO4 1 mol Na2SO4
7.58 x 10-4 mol BaSO4 1 mol BaSO4
• Moles of Na2SO4 = 7.58 x 10-4 mol
Na2SO4(aq) + BaCl2(aq) → 2 NaCl(aq) + BaSO4(s)
=
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• Mass of Na2SO4
= 7.58 x 10-4 mol Na2SO4 (142.0 g/mol) = 0.108 g Na2SO4
• Mass percent of Na2SO4 in the mineral,
= (0.108 g Na2SO4/0.123 g sample)100% = 87.6% Na2SO4
Na2SO4(aq) + BaCl2(aq) → 2 NaCl(aq) + BaSO4(s)
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Using Stoichiometry to Determine a Formula
Burn 0.115 g of a hydrocarbon, CxHy and produce 0.379 g of CO2 and 0.1035 g of H2O.
CxHy + O2 → 0.379 g CO2 + 0.1035 g H2O
What is the empirical formula of CxHy?
Problem:
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First, recognize that all C in CO2 and all H in H2O is from CxHy.
1. Calculate amount of C in CO2
8.61 x 10-3 mol CO2 → 8.61 x 10-3 mol C
2. Calculate amount of H in H2O
5.744 x 10-3 mol H2O→ 1.149 x 10-2 mol H
CCxxHHy y + + O2 →→ 0.379 g CO0.379 g CO22 + 0.1035 g H+ 0.1035 g H22OO
Chapter 4 — Intro—1 13
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Now find ratio of [mol H : mol C] to find values of x and y in CxHy.
Mole ratio [ mol H : mol C]
= 1.149 x 10 -2 mol H : 8.61 x 10-3 mol C
= 1.33 mol H : 1.00 mol C
= 4 mol H : 3 mol C
Empirical formula = C3H4
CCxxHHy y + + O2 →→ 0.379 g CO0.379 g CO22 + 0.1035 g H+ 0.1035 g H22OO
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SummaryAfter this lecture, you should be able to understand and calculate:
• The mole concept
• Molarity, parts per million & percentage
• Stoichiometry factor
• Theoretical yield, actual yield & % yield
• Limiting reactant & excess reactant