Chapter 3 - Stoichiometry

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


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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)


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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


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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


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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


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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


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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


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


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)



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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.


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


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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


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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:

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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:


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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


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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


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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

Date post:	24-Oct-2014
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Chapter 3 - Stoichiometry

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