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Quantitative Relationshipsin Chemical Reactions

• Coefficients in a reaction = quantity• Reactions occur in Mole multiples

– Moles are key reaction quantities– Mass to moles of reactants– Reaction product back to mass if need be

• Percent Yield– Actual / theoretical = yield

• Limiting Reactants– One reactant (almost) always in surplus

• Reaction energy– calories & kilocalories, Joules

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Stoichiometry ≡ Numerical Relationships

• Stoichiometry (reaction stoichiometry) is the calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions and chemical equations.

• The term stoichiometry is also often used for the molar proportions of elements in stoichiometric compounds. For example, the stoichiometry of hydrogen and oxygen in H2O is 2:1. In stoichiometric compounds, the molar proportions are whole numbers (what the law of multiple proportions is about).

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Simple Ratio Reaction

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Stoichiometry

• Stoichiometry rests upon the law of conservation of mass, and the law of definite proportions.

• chemical reactions combine in definite (usually simple) ratios of chemicals.

• chemical (non-nuclear) reactions can neither create nor destroy matter

• The amount of each element must be the same throughout the overall reaction. For example, the amount of element X on reactant side must equal the amount of X on the product side.

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Mass Balance Requirement

• Stoichiometry is used to balance chemical equations. For example, the two diatomic gases, hydrogen and oxygen, can combine to form liquid water, in an exothermic reaction described by the following equation

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• Stoichiometry is used not only to balance chemical equations but also is used in conversions — i.e. converting from grams to moles, or from grams to milliliters. For example, if there were 2.00 g of NaCl, to find the number of moles, one would do the following,

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Mole relationships

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Sequence of events

• Convert mass to moles (or molecules)– Cannot balance an equation with grams– Atoms weigh different amounts

• Balance equation of reactants + products– Mass Balance + Charge Balance– Ratio multipliers refer to moles or molecules

• Convert moles back to mass– Answers often desired in grams– We weigh in grams, calculate in moles

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Reactions amounts & mole ratios

• Stoichiometry is used to find the right amount of reactants to use in a chemical reaction. An example is thermite reaction,

To completely react with 85.0 grams of iron (III) oxide,

28.7 grams of aluminum are needed.

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Stepwise calculation

• GIVEN items– Reaction is Fe2O3 + 2 Al Fe + 2 Al2O3

– 85 grams of Fe2O3 available

• Need to go through Mole conversion– iron oxide = (2*55.85)+(3*16) = 159.7 g/mol– 85 gram / 159.7 gram/mole = 0.532 mole iron oxide

• Reaction requires 2 moles Al per Fe2O3

– Iron Oxide = 0.532 moles, so Al is 1.064 mole

• Convert answer in moles back to grams– Al at 1.064 mole * 27 gm/mole = 28.7

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Baby Steps vs One-Big-Calc• Baby steps more intuitive

– Get one item at a time right (e.g. gramsmoles)– Balance formulas/equations while in “mole mode”– Put dimensions in every step– Convert result to desired (e.g. molesgrams)– Apply yield issues

• One-Big-Calc simpler but can be hard to follow– Often simpler to break into a few pieces– Each piece makes sense on its own– Pieces combine to cancel unwanted units– Always put dimensions in every calculation !

• Helps prevent mistakes, easier to follow the logic

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Stoichiometry Conversions

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Yield• Lots of common usages

– Crop Yield– Investment return– Failure point in strength of materials– Energy of nuclear reaction (kilotons TNT)– Traffic sign

• Chemistry usage definition– Theoretical Yield (quantity)

• Output amount predicted by chemical reaction

– Practical or Actual yield (percentage)• What you really got versus theoretical amount• Due to various loss mechanisms

– Competing reactions, not 100% is desired product– Incomplete reactions– Errors, technical problems, accidents ….

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Yield Calculation• Example is Salt & Sand experiment

– Start with 6 grams of mixture (assumed)• Unknown ratio of salt and sand

– End result after separation• Separated Salt was (assumed) 2 grams• Separated Sand was (assumed) 3 grams• Total recovery was 2 + 3 = 5 grams

– Percentage of each constituent recovered• 2 / 5 = 40% Salt• 3 / 5 = 60% Sand• Recovery percentages should add to 100%

– Yield is recovery (output) versus input• 5 grams-out / 6 grams-in = 5 / 6 = 83 % yield• Loss is (6-5)/6 = 1 / 6 = 17%

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Percent into Mass

• Aluminum Ore example– Bauxite = Al2O3 is source for aluminum

– Aluminum molar mass is 27 gm/mol– Aluminum oxide molar mass is 102 gram/mol– % aluminum in the oxide is 27*2/102 = 53%– Every 100 grams oxide contains 53 gm Al

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Limiting Reactants

• Nature rarely provides 100% balance– Oxygen in air exceeds animal and fuel needs– Almost always an excess of all but 1 reactant– One which runs out first = “Limiting” reactant– Even with perfect balance of reactants

• Reaction may not end in timely fashion (or ever)

– A social analogy• Last single boy finding last single girl on our planet

– Earth has 6.6 billion people = 6.6*109

– Assume 50% are girls, 50% boys = 3.3*109 girls or boys

» Pairs form until last remaining singles are left

» Last couple’s chance of meeting = 1 in 3 billion

» 1 Mole=6*10E23, or 10E14 more than people on earth

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Limiting Reactants

– Limiting Reactant handled on moles basis• Find moles of each, and determine which runs out first

– A mole is a huge amount (6*10E23 molecules)

– Does not take much mole excess to dominate the reaction

• Social relationships have similarity to chemistry– 50/50 mix male/female changed by societal pressures

• Darwinism in reverse?– Chinese practice of favoring boy babies = no wives for many boys

– Wars eliminate mostly males = no husbands for many girls

– Which sex is today’s limiting reactant ?

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Limiting Reactant Gram Calculations

• Grams versus grams is another method– Calculate moles of product from each reactant– Apply stoichiometry ratios– Minimum amount of product defines limiting reactant

• Example: 100 grams each of Hydrogen, Oxygen– Reaction Stoichiometry: H2+ ½ O2 H2O– 100 gm H 100/1.008 99.2/2= 49.6 moles water– 100 gm O 100/16 = 6.25 Moles water (lesser amt)– Reaction limited by smaller amount = oxygen– Net result is 6.25 moles of water can be formed

• 6.25 moles H2O * 18 grams water / mole = 113 grams water• 99.2 – (2*6.25) = 86 moles of Hydrogen left over = 87 grams• Still have 200 grams total, so mass is conserved

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Today’s Experiment

• Making oxygen from household chemicals– Hydrogen peroxide H2O2+ Bleach NaOCl

• Constant bleach, variable peroxide– Several trials, see when O2 is constant

• Constant peroxide, variable bleach– Several trials, see when O2 is constant

• Oxygen levels out due to limited reagent– That which runs out first dominates reaction

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Today’s Experiment

• How to determine quantities in an equation– We will use empirical data to decide– Find limiting quantities of two ingredients– Convert quantities to moles– Take ratios of moles = equation coefficients– Apply ratios to balanced chemical equation

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Calculations

Part A, constant Bleach, Sodium hypochlorite

• NaOCl, 5.25% ≡ 5.25 grams/100mL– 5.25 gm/100ml = .0525 gm/ml bleach– We use constant 4.0 ml bleach in part A

• 4.0 mL*.0525 gm/mL = 0.210 gm NaOCL• 0.210gm / (23+16+35.5) g/mol = 0.00282 moles

• All part A trials have same moles NaOCl– now to determine influence of H2O2

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CalculationsPart A, Bleach = NaOCl (hypochlorite)• At what point is all NaOCl consumed?

– Keep adding H2O2 until no more O2 gas

– Plot gas evolution versus added H2O2

Oxygen Generation

0.0

50.0

100.0

0 1 2 3 4 5 6 7 8

milliliters of Peroxide

mill

iter

s of

O

xyge

n

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Part A calculations• NaOCl consumed after 3.5mL H2O2 added

– How many moles in 3.5 mL H2O2 ?– H2O2 concentration is 3.0%, weight/volume

• Same as 3.0 grams in 100ml• Same as 0.030 grams H2O2 per milliliter

– 3.5 mL * 0.030 gm/mL = 0.105 gm H2O2

• 0.105 gm H2O2 /(32+2)gm/mole = 0.0031 moles

– 0.0031 moles H2O2 0.0028 moles NaOCl• Ratio is 0.0031/0.0028 = 1.11• So approximately EQUAL moles involved

– Equation must include 1 NaOCl + 1 H2O2

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Calculations

Part B, constant Hydrogen Peroxide

• H2O2, 3.0% ≡ 3.0 grams/100mL– 3.0 gm/100ml = .030 gm/ml bleach– We use constant 4.0 ml peroxide in part B

• 4.0 mL*.030 gm/mL = 0.120 gm H2O2

• 0.120gm / (1+1+16+16) g/mol = 0.00353 moles

• All part B trials equal moles of H2O2

– now to determine influence of NaOCl

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CalculationsPart B, constant H2O2 Hydrogen Peroxide

• At what point is all peroxide consumed?– Keep adding NaOCl until no more O2 gas

– Plot gas evolution versus added NaOCl

Oxygen Generation

0.050.0

100.0

0 1 2 3 4 5 6 7 8

milliliters of Bleach

Mill

ilite

rs o

f O

xyge

n

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Part B calculations• H2O2 consumed after 5.5mL NaOCl added

– How many moles were in 5.5 mL NaOCl ?– NaOCl concentration is 5.25%, weight/volume

• Same as 5.25 grams in 100ml• Same as 0.0525 grams NaOCl per milliliter

– 5.5 mL * 0.0525 gm/mL = 0.289 gm NaOCl• 0. 289 gm H2O2 /74.4 gm/mole = 0.00388 moles

– 0.00353 moles H2O2 = 0.00388 moles NaOCl• Ratio is 0.00388/0.00353 = 1.09• So approximately EQUAL moles involved

– Equation must include 1 NaOCl + 1 H2O2

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Calculation Conclusion

• Equal moles of reactants 1:1

• We know equal moles, + O2 generation– 1 H2O2 + 1 NaOCl O2 + ??

• Now to finish & balance the equation– Water is typical reaction product– Leaves NaCl to balance materials

– H2O2 + NaOCl O2 + NaCl + H2O

• Thus we have derived the equation !