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Chapter 3. Chapter 3. StoichimetryStoichimetryCalculations with chemical formula and reactions are called Stoichiometry

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KEY CONCEPTSKEY CONCEPTSAtomic, molecular and formula massMole, Avogadro's number and Molar massMass percent of elementsMass % from analytical dataMass % from formulaEmpirical formula from Mass %Chemical formula from compositionChemical formula from mass % Chemical EquationsStoichiometric coefficients Balancing chemical equations Stoichimetry Stoichiometric coefficients and Limiting reactant Yields of chemical reactions

Actual yield and Thoretical yield Solutions concentration and dilution of solutionsSolution stoichiometry

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QuestionQuestionYou were given a chance to pick 100 kg of gold(I)periodate - AuIO4 or gold(I)nitrate- AuNO3

Which one you would pick?Rationalize your choice.

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GoldGold

1 oz = $ 4001 oz = 28.35 g1 kg = 35.27 oz1 kg Au = $ 14,109

Molecular mass vs. formula massMolecular mass vs. formula mass

Formula massFormula massAdd the masses of all the atoms in formula

- for molecular and ionic compounds.

Molecular massMolecular massCalculated the same as formula mass

- only valid for molecules.

Both have units of either u or grams/mole.

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Examples of M.W.Examples of M.W.H3PO4:

H = 1.01 g/mol,

P = 30.97 g/mol,

O = 16.00 g/mol 1 amu is equal to 1 g/mol

m.w. H3PO4 =

3 x 1.01+1 x 30.97+ 4 x 16.00

= 98.00 g/mol

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Examples of Examples of

F.W.F.W. K2CO3: K = 39.10 g/mol;

C = 12.01 g/mol;

O = 16.00 g/mol

f.w. K2CO3 =

2 x 39.10 + 1 x 12.01 + 3 x 16.00

= 138.2 g/mol

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Mole ConceptMole ConceptWe use masses weighed in grams in

chemical reaction. Need a conversion factor to convert grams to atoms and molecules.

atomic weight or molecular weight taken in grams contains

6.022 x 1023

atoms, molecules or particles.

The number 6.022 x 1023 is called Mole or Avagadro’s NumberAvagadro’s Number

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Can you guess thisCan you guess this??•How long is 1 mole of seconds? • 1.9096 x 1019 years • 2 x 1010 billion years •How deep is the layer of marbles, if 1mole of marbles are spread over the surface of earth?

• ~50 miles high

The Mole and Avogadro's numberThe Mole and Avogadro's number

Number of atoms in 12.000 grams of 12C

1 mol = 6.022 x 1023 atoms

1 g = 6.022 x 1023 u 1 u = 1 g/mol

Mole is the converssion factor between M.W and grams

1 mol = grams / molecular weight

mole = g/m.w. Or g/f.w.

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Conversion factors in Conversion factors in Stoichiometry?Stoichiometry?

6.022 x 1023 atoms = gram atomic weight6.022 x 1023 molecules = gram molecular weight 6.022 x 1023 atoms C = 12.01 grams of carbon

(C)

6.022 x 1023 molecules H2O = 18.02 g of H2O

6.022 x 1023 = 1 mol 1 g = 6.022 x 1023 amu (u) 1 amu = 1 g/mol

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Atomic MassesAtomic MassesAtomic mass in grams ?Mass of a copper atom in grams 63.55 g Cu 1 mole1 mole 6.022 x 1023

= 1.055 x 10-22 g Atomic mass in the periodic table divided

by Avagadro’s Number

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An atom weighs 7.47 x 10-23 g. What is the name of the element this atom belongs to?

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An atom weighs 7.47 x 10-23 g. What is the name of the element this atom belongs to?

Conversion factor: 1g = 6.022 x 1023 u7.47 x 10-23 g x6.022 x 1023 u

1g = 44.98 u or g/moleThe element is Sc.

Example - (NHExample - (NH44))22SOSO44

How many atoms are in 20.0 grams of ammonium sulfate?

Formula weight = 132.14 grams/molAtoms in formula = 15 atoms / unit

moles = 20.0 g x = 0.151 mol1 mol

132.14 g

atoms = 0.151 mol x 15 x 6.02 x1023 atomsunit

unitsmol

atoms = 1.36 x1024

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How do you convert grams to mole How do you convert grams to mole

and vice versa?and vice versa?Grams ---> molegrams /molecular (formula) weight

Moles ----> grams moles x molecular (formula) weight

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CalculationsCalculationsCalculate how many moles are in 100 g of sulfur-

S8

Calculate the grams of sugar in 3 moles of -glucose-C6H12O6

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% Element Composition in % Element Composition in CompoundsCompounds

n x Atomic weight % mass = --------------------- x 100 molecular weight

n = subscript of the element in the formula

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CalculationCalculation

Al2(SO4)3

calculate % Al, % S and % O

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ExamplesExamplesa) Al2(SO4)3: f.w. = 342.14 g/mol,

Atomic weight of oxygen = 16.00 g/mol, n = 12

12 x 16.00% O mass = ------------- x 100 = 56.12% oxygen.

342.14

c) CuSO4: f.w. = 159.61 g/mol

Atomic weight of oxygen = 16.00 g/mol n = 4 4 x 16.00% O mass =-------------- = 40.01% oxygen. 159.61

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Calculate % composition Calculate % composition from analysisfrom analysis

A 12.5 g sample of a compound that contains only phosphorus and sulfur was analyzed and found to contain 7.04 g of phosphorus and 5.46 g of sulfur.

Calculate the percent composition of elements.

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ExamplesExamplesGlucose has a molecular formula

of C6H12O6 (M.W. 180.16 g/mol).

a) How many grams of C, H and O are available in 1 mole of glucose?

b) Calculate mass percents of elements C, H and O in glucose.

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1 mol C6H12O6= 6 mol C = 6 x 12.01 =72.06g C

1 mol C6H12O6=12 mol H= 12 x 1.01

= 12.12g H

1 mol C6H12O6=6 mol O =6 x 16.00

= 96.00g O

How many grams of C, H and O are available in 1 mole of glucose?

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6 x 12%C = --------- x 100 = 40.00% C

180.16 12 x 1.01% H = ------------- x 100 = 6.73% H 180.16 6 x 16.00 %O = ------------ x 100 = 53.29% O

180.16 ----------- 100.02%

Mass percent of element. C6H12O6 = 180.16 g/mol

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What is Empirical FormulaWhat is Empirical Formula

Simple whole number ratio of each atom expressed in the subscript of the formula.

Molecular Formula = C6H12O6 of glucose

Empirical Formula = CH2O

Emiprical formula is calculated from % composition

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How do you get Empirical Formula How do you get Empirical Formula

from % composition and vice from % composition and vice

versa?versa?

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% composition to Empirical formula% composition to Empirical formula

•Divide percent composition by atomic weight.

•Divide answer by lowest number to get simple ratio of moles or atoms.

•Multiply by a factor to get whole number ratio.

•Write a formula with whole number ratio as subscripts to get empirical formula.

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ExamplesExamples

•A 12.5 g sample of a compound that contains only phosphorus and sulfur was analyzed and found to contain 7.04 g of phosphorus and 5.46 g of sulfur.

•a) Calculate the percent composition of elements.

•b) The empirical formula of the compound

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CalculationCalculation mass of element% Element = ------------------- x 100 mass of sample 7.04 % P = --------- x 100 = 56.3% P

12.5

5.46% S = ------- x 100 = 43.7% S 12.5 100.0

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CalculationCalculation% of P and S: 56.3 43.7Moles of P and S:56.3g 43.7g------- = 1.82 mol P ------ = 1.36

mol S30.97 32.06Atom ratio: 1.82 atoms P 1.36 atoms

S 1.33 atoms P 1.00 atoms

S

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Calculation..continueCalculation..continuedd

1.82 atoms P 1.36 atoms S Divide by lowest number: 1.33 atoms P 1.00 atoms Ssimple atom ratio: 1.33 atom P 1 atom S to get a simple whole number ratio.

Multiply both numbers by 2, 3, and 4 etc. until 1.33 becomes close to a whole number.

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Calculation..continueCalculation..continueddTo get simple whole number ratio:

P S1.33 atom 1 atom 2 x 1.33 = 2.66 2 x 1 2.66 23 x 1.33= 3.99 3 x 1 4.00 3.0Therefore, Formula should contain 4 P

atoms and 3 S atoms.Therefore, the empirical Formula of the

compound is: P4S3

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More ExampleMore ExampleGlucose contains the elements C, H,

and O in 39.99% C, 6.71% H and 53.28% O, respectively, by mass.

a) Calculate the empirical formula of glucose.

b) Molecular weight determination in solution for glucose has shown a molecular weight close to 180 g/mol. What is the molecular formula of glucose?

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CalculationCalculationEmpirical formulaEmpirical formulafrom % elemental composition.% composition: 39.99% C, 6.71% H, 53.28% O ii) mole ratio: C H O39.99 6.71 53.28------- = 3.33; -------- = 6.64; -------- = 3.3312.01 1.01 16.003.33 mol C: 6.64 mol H: 3.33 mol O

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Calculation..continuedCalculation..continuedAtom ratio3.33 mol C: 6.64 mol H: 3.33 mol OSimple atom: 3.33 6.64 3.33 ------- ------- ------- 3.33 3.33 3.33

1 1.99 1Simple whole number 1 2 1number ratio of atomThe empirical formulaempirical formula of the

compound is

CH2O

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Calculation..continueCalculation..continuedd

Molecular Formula = n x empirical Formula

Molecular weight 180

n = -------------------------------------- = ------ = 6

Empirical Formula Weight 30

Molecular Formula = (CH2O)n =

(CH2O)6

Molecular Formula = C6H12O6 of glucose

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QuestionQuestion

A molecular compound contains 92.3% carbon and 7.7% hydrogen by weight. What is its empirical formula?

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QuestionQuestion

A molecular compound has empirical formula CH. If 0.050 mol of the compound weighs 3.90 g, what is its molecular formula?

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Chemical EquationChemical Equation P4O10 (s) + 6H2O (l) = 4 H3PO4(l)

reactantsreactants enter into a reaction. productsproducts are formed by the reaction. Parantheses Parantheses represent physical state stoichiometric coefficientsstoichiometric coefficients are numbers in front of chemical formula formula

gives the amounts (moles) of each substance used and each substance produced.

Equation Must be balanced!

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Chemical ReactionChemical ReactionCould be described in words

Chemical equationChemical equation:

Reactants?

Products?

reaction conditions?

=, ---> , <==> or ?

stoichiometric coefficients?

Number in front of substances representing moles, atoms, molecules

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Steps in Stoichiometric Steps in Stoichiometric CalculationsCalculations

•Check whether chemical equation is balanced

•get the moles from grams of materials

•find the limiting reactant•calculate moles of products from the

limiting reactant•convert moles of the products to

grams•find the actual yield of the reaction•calculate % yield of the reaction

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QuestionQuestionHow much hydrogen gas is produced when 1 kg of sodium reacts with water?

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ExamplesExamples

Calculate the following using the chemical equation given below:

4 NH3(g) + 5 O2(g) ----> 4 NO(g) + 6 H2O(g)

a) moles of NO(g) from 2 moles of NH3(g) and excess O2(g).

b) moles of H2O(g) from 3 moles of O2(g) and excess NH3(g).

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ExamplesExamples

How many moles of H2O will

be produced by 0.80 mole of

O2 with excess H2 according

to the equation?

2H2(g) + O2(g) = 2

H2O(l)

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QuestionQuestion

2Al(s) + 6HCl(aq)--> 2AlCl3(aq) +3H2(g) According to the equation above, how many grams of aluminum are needed to react with 0.582 mol of hydrochloric acid?

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5.23 g Al

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What is the limiting What is the limiting reagent?reagent?

Limiting reagent is the reactant, which is used up first.

To find the limiting reactant you have to compare the amounts of reactants in moles.

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ExamplesExamples

A 300.0 g sample of phosphorus

( M.W. 123.88 g/mol) burns in 500.0 g of oxygen (M.W. 32.00 g/mol) according to following equation:

P4(s) + 5O2(g) = P4O10(s)

What is the limiting reagent?

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Theoretical yieldTheoretical yield

Theoretical yield is the amount (grams) of products formed according to chemical equation.

Use the limiting reagent to calculate the moles of the product and then convert moles to grams.

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Actual YieldActual YieldActual yield is the grams of the product obtained by an experiment.

Actual yield should be less than the theoretical yield if the experiment was carried out meticulously. If the products are contaminated with impurities or the formula of product was wrong the actual yield could be higher.

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% Yield% Yield

actual yield % yield = ------------------------ x 100 theoretical yield

If the products are contaminated with impurities or the formula of product was wrong the % yield could be higher than 100%.

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QuestionQuestion

Sulfur trioxide, SO3 , is made from the oxidation of SO2 and the reaction is represented by the equation

2SO2 + O2 -----> 2SO3

A 16.0-g sample of SO2 gives 18.0 g of SO3. The percent yield of SO3 is

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ExamplesExamples

• A 300.0 g sample of phosphorus ( M.W. 123.88 g/mol) burns in 500.0 g of oxygen (M.W. 32.00 g/mol) according to following equation:

• P4(s) + 5O2(g) = P4O10(s) • a) What is the limiting reagent?

• b) How many moles of P4O10 are produced theoretically?

• c) If 612 g of P4O10 is actually produced in this reaction, calculate the percent yield.

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How do you calculate moles How do you calculate moles of substances in solutionsof substances in solutions

•Use concentration of solution to convert L or mL of solution in to moles

•What is concentration of a solution?•The relative amounts of solute and solvent

•There are so many ways to show amount: g, mole, equivalents,volume

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Concentration UnitsConcentration Units• a) Molarity (M) • b) Normality (N) • c) Molality (m)

• f) Mole fraction (a) • g) Mass percent (% weight) • h) Volume percent (%

volume) • i) "Proof" • j) ppm and ppb

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MolarityMolarity

M = moles solute molliters of solution L

=

• [ ] - special symbol which means molar ( mol/L )

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MolarityMolarity

What’s the molarity of a solution that has 18.23 g HCl in 2.0 liters?

First, you need the FM of HCl.First, you need the FM of HCl.FMHCl = 1.008 x 1 H + 35.45 x 1 Cl

= 36.46 g/mol

Next, find the number of moles.Next, find the number of moles.molesHCl = 18.23 gHCl / 36.46 g/mol

= 0.50 mol

Finally, divide by the volume.Finally, divide by the volume.MHCl = 0.50 mol / 2.0 L

= 0.25 M

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ExamplesExamplesCalculate the molarity of a solution

prepared by dissolving 200.0 g of K2SO4 in enough water to make 500.0 mL solution.

moles of soluteMolarity(M) = ---------------------- Liters of solution

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solute = K2SO4; F.W. = 174.27 g/mol; mass= 200g

moles of K2SO4 = ?

200 g /174.27 g K2SO4

= 1.148 mol K2SO4

500.0 mL = ? Liters of solution = 0.5 LMolarity?

1.148 mol K2SO4

Molarity of K2SO4 sol. = ------------------------

0.5 Liters of solution

= 2.30 mole/Liter = 2.30 M (M = moles/liters)

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Normality (N)Normality (N) Equivalents of solute Normality(N) = ---------------------------- Liters of solution

moles Equivalents = ----------------------- n n is # of H, OH in or e- in acid or base N(normality) = Molarity x n (# of H, OH

or e-)

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ExamplesExamples

How many grams of KNO3 are contained in 500 mL of a 0.500 M solution of potassium nitrate?

How many mL of 2.00 M solution of HNO3 are required with water to make a 250 mL of 1.50 M nitric acid solution?

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Mole fraction (Mole fraction (aa))

moles of solute (substance)

a = -------------------------------------

moles of solute + solvent

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Calculate the mole fraction Calculate the mole fraction

of benzene in a of benzene in a

benzene(Cbenzene(C66HH66)-)-

chloroform(CHClchloroform(CHCl33) solution ) solution

which contains 60 g of which contains 60 g of

benzene and 30 g of benzene and 30 g of

chloroform.chloroform.

M.W. = 78.12M.W. = 78.12 ( (CC66HH66))

M.W. = 119.37 M.W. = 119.37 ( (CHClCHCl33))

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Weight/Weight %Weight/Weight %

Weight/Weight % = Mass SoluteTotal Mass

x 100

If a ham contained 5 grams of fat in 200 gof ham, what is the % wt/wt?

5 g / 200g * 100 = 2.5 wt/wt%

On the label, it would say 97.5 % fat free.

Use the same units for bothUse the same units for both

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Volume/Volume %Volume/Volume %

Volume/Volume % = Volume SoluteTotal Volume

x 100

If 10 ml of alcohol is dissolved in water to make 200 ml of solution, what is the concentration?

10 ml / 200 ml * 100 = 5 V/V%

Alcohol in wine is measured as a V/V%.

Use the same units for bothUse the same units for both

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Weight/Volume %Weight/Volume %

Weight/Volume % = Mass solute Total Volume

x 100

If 5 grams of NaCl is dissolved in water to make 200 ml of solution, what is the concentration?

5 g / 200 ml * 100 = 2.5 wt/v%

Saline is a 0.9 wt/v% solution of NaCl in water.

use g and mluse g and ml

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Very low concentrationsVery low concentrations

Pollutants in air and water are typically found at very low concentrations. Two common units are used to express these trace amounts.

Parts per million - ppmParts per million - ppm

Parts per billion - ppbParts per billion - ppb

Both are modifications of the % system which could be viewed as parts per hundred - pph.

Both mass and volume % systems are used.

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Low concentrations in airLow concentrations in air

Trace amounts in are are expressed as volume/volume ratios.

ppm = x 106

ppb = x 109

Example.Example. One cm3 of SO2 in one m3 of air would be expressed as 1 ppm or 1000 ppb.

volume solutevolume solution

volume solutevolume solution

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Low concentrations in waterLow concentrations in water

Mass percentages are used for water pollutants.

ppm = x 106

ppb = x 109

Example.Example. One ppm of a toxin in water is the same as 1 mg / liter since one liter of water has a mass of approximately 106 mg.

mass solutemass solution

mass solutemass solution

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Solution preparationSolution preparation

Solutions are typically prepared by:

Dissolving the proper amount of solute and diluting to volume.

Dilution of a concentrated solution.

Lets look at an example of the calculations required to prepare known molar solutions using both approaches.

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Dilution ProblemsDilution ProblemsWhy we dilute solutions?

Preparing solutions by adding water to concentrated solutions

moles before = moles after

MiVi = MfVf

Mi = initial molarity

Vi = initial volume

Mf = final molarity

Vf = final molarity

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ExamplesExamplesHow many mL of 2.00 M solution of HNO3 are

required with water to make a 250 mL of 1.50 M nitric acid solution?

MiVi = MfVf

Mi = 2.00

Vi = ?

Mf = 1.50

Vf = 250 mL

MfVf 1.50 x 250

Vi = --------------- = ---------------- = 187.5 mL Mi 2.00

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Stoichiometric calculations Stoichiometric calculations of solutions reactionsof solutions reactions•Check whether chemical equation

is balanced•get the moles from volume of from volume of

solutionssolutions•find the limiting reactant•calculate moles of products from

the limiting reactant•convert moles of the products to

grams•find the actual yield of the

reaction•calculate % yield of the reaction

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Solution stoichiometry exampleSolution stoichiometry example

Determine the volume of 0.100 M HCl that must be added to completely react with 250 ml of 2.50 M NaOH

Balanced chemical equationBalanced chemical equation

HCl(aq) + NaOH(aq) NaCl(aq) + H2O (l)

The first step is to determine how many moles of NaOH we have.

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Solution stoichiometry exampleSolution stoichiometry example

We have 250 ml of a 2.50 M solution.

molNaOH = 0.250 L x 2.50 mol/L

= 0.625 molNaOH

From the balanced chemical equation, we know that we need one mole of HCl for each mole of NaOH.

That means we need 0.625 molHCl.

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Solution stoichiometry exampleSolution stoichiometry example

Now we can determine what volume of our 0.100 M HCl solution is required.

L = molHCl / MHCl

= 0.625 mol

= 6.26 L

1 L0.100 mol( )

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ExamplesExamplesHow many mLs of 0.100 M BaCl2 are

required to react completely with 25 mL of 0.200 M Fe2(SO4)3?

3 BaCl2(aq) + Fe2(SO4)3(aq)---> 3 BaSO4(s) + 2 Fe Cl3(aq)

3 BaCl2 = 1 Fe2(SO4)3

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Calculate the moles of Fe2(SO4)3 :

moles = Molarity x Liters of solution

0.200 M x 0.025 L = 0.005 mole Fe2(SO4)3

Then convert Fe2(SO4)3 to BaCl2 mole, BaCl2 moles

to liters and liters to mL.

0.005 mole Fe2(SO4)3 ----> BaCl2 moles

0.005 mol Fe2(SO4)3 x 3 = 0.015 BaCl2 moles

moles = Molarity x Liters of solution

0.015 = 0.100 x Liters

Liters BaCl2 = 0.15L

= 150 mL of BaCl2

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ExamplesExamplesHow many mLs of 0.300 M NaOH are

required to react with 500 mL of 0.170 M H2SO4 completely?

Acid Base Reactions . Na xVa = Nb x Vb .

Na = Normality of acid,

Nb = Normality of base V= volumeN(normality) = Molarity x n (# of H or OH)

Na (0.170 x 2) xVa (500) = Nb (0.300x 1)x Vb(?) .

Vb= 0.170 x 2 x 500/ 0.300 = 566.7 mL NaOH