Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

Chapter 3Chapter 3Stoichiometry:Stoichiometry:Calculations with Calculations with

Chemical Formulas Chemical Formulas and Equationsand Equations

Law of Conservation of Law of Conservation of MassMass

““We may lay it down as an We may lay it down as an incontestable axiom that, in incontestable axiom that, in all the operations of art and all the operations of art and nature, nothing is created; nature, nothing is created; an equal amount of matter an equal amount of matter

exists both before and after exists both before and after the experiment. Upon this the experiment. Upon this principle, the whole art of principle, the whole art of

performing chemical performing chemical experiments depends.”experiments depends.”

--Antoine Lavoisier, 1789--Antoine Lavoisier, 1789

Chemical EquationsChemical Equations

Concise representations of Concise representations of chemical reactionschemical reactions

Anatomy of a Chemical Anatomy of a Chemical EquationEquation

CHCH4 (4 (gg)) + 2 O + 2 O2 (2 (gg) ) CO CO2 (2 (gg)) + 2 H + 2 H22OO ( (gg))


ReactantsReactants appear on appear on the left side of the the left side of the

equation.equation.

CHCH44 ( (gg)) + 2 + 2 OO22 ( (gg) ) CO CO2 (2 (gg)) + 2 H + 2 H22OO ( (gg))


ProductsProducts appear on appear on the right side of the the right side of the

equation.equation.

CHCH4 (4 (gg)) + 2 O + 2 O2 (2 (gg) ) COCO22 ((gg)) + 2 + 2 HH22OO ((gg))


The The statesstates of the reactants and of the reactants and products are written in parentheses products are written in parentheses to the right of each compound.to the right of each compound.

CHCH4 4 ((gg)) + 2 O + 2 O2 2 ((gg)) CO CO22 ((gg)) + 2 H + 2 H22OO ((gg))


CHCH4 (4 (gg)) + + 22 O O2 (2 (gg) ) CO CO22 ((gg)) + + 22 H H22OO ((gg))

CoefficientsCoefficients are inserted to balance are inserted to balance the equation.the equation.

Subscripts and Subscripts and Coefficients Give Coefficients Give

Different InformationDifferent Information

SubscriptsSubscripts tell the number of atoms of tell the number of atoms of each element in a moleculeeach element in a molecule

Subscripts and Subscripts and Coefficients Give Coefficients Give

Different InformationDifferent Information

Subscripts tell the number of atoms of Subscripts tell the number of atoms of each element in a moleculeeach element in a molecule

CoefficientsCoefficients tell the number of tell the number of moleculesmolecules

ReactioReaction Typesn Types

Combination ReactionsCombination Reactions

Examples:Examples:NN2 (2 (gg)) + 3 H + 3 H2 (2 (gg)) 2 NH 2 NH3 (3 (gg))

CC33HH6 (6 (gg)) + Br + Br2 (2 (ll)) C C33HH66BrBr2 (2 (ll))

2 Mg2 Mg ( (ss)) + O + O2 (2 (gg)) 2 MgO 2 MgO ( (ss))

Two or Two or more more substances substances react to react to form one form one productproduct

2 Mg2 Mg ( (ss)) + O + O2 (2 (gg)) 2 MgO 2 MgO

((ss))

Decomposition ReactionsDecomposition Reactions

Examples:Examples:CaCOCaCO3 (3 (ss)) CaO CaO ( (ss)) + CO + CO2 (2 (gg))

2 KClO2 KClO3 (3 (ss)) 2 KCl 2 KCl ( (ss)) + O + O2 (2 (gg))

2 NaN2 NaN3 (3 (ss)) 2 Na 2 Na ( (ss)) + 3 N + 3 N2 (2 (gg))

One substance One substance breaks down into breaks down into two or more two or more substancessubstances

Combustion ReactionsCombustion Reactions

Examples:Examples:CHCH4 (4 (gg)) + 2 O + 2 O2 (2 (gg)) CO CO2 (2 (gg)) + 2 H + 2 H22OO ( (gg))

CC33HH8 (8 (gg)) + 5 O + 5 O2 (2 (gg)) 3 CO 3 CO2 (2 (gg)) + 4 H + 4 H22OO ( (gg))

Rapid reactions Rapid reactions that produce a that produce a flameflame

Most often involve Most often involve hydrocarbons hydrocarbons reacting with reacting with oxygen in the airoxygen in the air

FormulFormula a

WeightsWeights

Formula Weight (FW)Formula Weight (FW) Sum of the atomic weights for the Sum of the atomic weights for the

atoms in a chemical formulaatoms in a chemical formula So, the formula weight of calcium So, the formula weight of calcium

chloride, CaClchloride, CaCl22, would be, would be Ca: 1(40.1 amu)Ca: 1(40.1 amu) + Cl: 2(35.5 amu)+ Cl: 2(35.5 amu)

111.1 amu111.1 amu These are generally reported for These are generally reported for

ionic compoundsionic compounds

Molecular Weight (MW)Molecular Weight (MW)

Sum of the atomic weights of the Sum of the atomic weights of the atoms in a moleculeatoms in a molecule

For the molecule ethane, For the molecule ethane, CC22HH66, the , the molecular weight would bemolecular weight would be

C: 2(12.0 amu)+ H: 6(1.0 amu)

30.0 amu

Percent CompositionPercent Composition

One can find the percentage of the One can find the percentage of the mass of a compound that comes mass of a compound that comes from each of the elements in the from each of the elements in the compound by using this equation:compound by using this equation:

% element =(number of atoms)(atomic weight)

(FW of the compound)x 100

Percent CompositionPercent Composition

So the percentage of carbon in So the percentage of carbon in ethane is…ethane is…

%C =(2)(12.0 amu)

(30.0 amu)

24.0 amu

30.0 amu= x 100

= 80.0%

MolesMoles

Avogadro’s NumberAvogadro’s Number

6.02 x 106.02 x 102323

1 mole of 1 mole of 1212C has C has a mass of 12 ga mass of 12 g

Molar MassMolar Mass

By definition, these are the mass By definition, these are the mass of 1 mol of a substance (i.e., of 1 mol of a substance (i.e., g/mol)g/mol) The molar mass of an element is the The molar mass of an element is the

mass number for the element that mass number for the element that we find on the periodic tablewe find on the periodic table

The formula weight (in amu’s) will The formula weight (in amu’s) will be the same number as the molar be the same number as the molar mass (in g/mol)mass (in g/mol)

Using MolesUsing Moles

Moles provide a bridge from the Moles provide a bridge from the molecular scale to the real-world molecular scale to the real-world scalescale

Mole RelationshipsMole Relationships

One mole of atoms, ions, or molecules contains One mole of atoms, ions, or molecules contains Avogadro’s number of those particlesAvogadro’s number of those particles

One mole of molecules or formula units One mole of molecules or formula units contains Avogadro’s number times the number contains Avogadro’s number times the number of atoms or ions of each element in the of atoms or ions of each element in the compoundcompound

Finding Finding Empirical Empirical FormulasFormulas

Calculating Empirical Calculating Empirical FormulasFormulas

One can calculate the empirical formula One can calculate the empirical formula from the percent compositionfrom the percent composition


The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.


Assuming 100.00 g of para-aminobenzoic acid,

C: 61.31 g x = 5.105 mol C

H: 5.14 g x = 5.09 mol H

N: 10.21 g x = 0.7288 mol N

O: 23.33 g x = 1.456 mol O

1 mol12.01 g

1 mol14.01 g

1 mol1.01 g

1 mol16.00 g


Calculate the mole ratio by dividing by the smallest number of moles:

C: = 7.005 7

H: = 6.984 7

N: = 1.000

O: = 2.001 2

5.105 mol0.7288 mol

5.09 mol0.7288 mol

0.7288 mol0.7288 mol

1.458 mol0.7288 mol


These are the subscripts for the empirical formula:

C7H7NO2

Combustion AnalysisCombustion Analysis

Compounds containing C, H and O are routinely Compounds containing C, H and O are routinely analyzed through combustion in a chamber like analyzed through combustion in a chamber like thisthis C is determined from the mass of COC is determined from the mass of CO22 produced produced H is determined from the mass of HH is determined from the mass of H22O producedO produced O is determined by difference after the C and H have O is determined by difference after the C and H have

been determinedbeen determined

Elemental AnalysesElemental Analyses

Compounds Compounds containing other containing other elements are elements are analyzed using analyzed using methods methods analogous to analogous to those used for C, those used for C, H and OH and O

Stoichiometric Stoichiometric CalculationsCalculations

The coefficients in the balanced The coefficients in the balanced equation give the ratio of equation give the ratio of molesmoles of reactants and productsof reactants and products


From the mass of From the mass of Substance A you Substance A you can use the ratio can use the ratio of the coefficients of the coefficients of A and B to of A and B to calculate the calculate the mass of Substance mass of Substance B formed (if it’s a B formed (if it’s a product) or used product) or used (if it’s a reactant)(if it’s a reactant)


Starting with 1.00 g of CStarting with 1.00 g of C66HH1212OO66… …

we calculate the moles of Cwe calculate the moles of C66HH1212OO66……

use the coefficients to find the moles of Huse the coefficients to find the moles of H22O…O…

and then turn the moles of water to gramsand then turn the moles of water to grams

C6H12O6 + 6 O2 6 CO2 + 6 H2O

LimitinLimiting g

ReactanReactantsts

How Many Cookies Can I How Many Cookies Can I Make?Make?

You can make You can make cookies until you run cookies until you run out of one of the out of one of the ingredientsingredients

Once this family runs Once this family runs out of sugar, they out of sugar, they will stop making will stop making cookies (at least any cookies (at least any cookies you would cookies you would want to eat)want to eat)

How Many Cookies Can I How Many Cookies Can I Make?Make?

In this example In this example the sugar would the sugar would be the be the limiting limiting reactantreactant, because , because it will limit the it will limit the amount of cookies amount of cookies you can makeyou can make

Limiting ReactantsLimiting Reactants

The limiting The limiting reactant is the reactant is the reactant reactant present in the present in the smallest smallest stoichiometric stoichiometric amountamount


The limiting reactant is the reactant The limiting reactant is the reactant present in the smallest stoichiometric present in the smallest stoichiometric amountamount In other words, it’s the reactant you’ll run In other words, it’s the reactant you’ll run

out of first (in this case, the Hout of first (in this case, the H22))


In the example below, the OIn the example below, the O22 would be the would be the excess reagentexcess reagent

Theoretical YieldTheoretical Yield

The The theoretical yieldtheoretical yield is the is the amount of product that can be amount of product that can be mademade In other words it’s the amount of In other words it’s the amount of

product possible as calculated product possible as calculated through the stoichiometry problemthrough the stoichiometry problem

This is different from the This is different from the actual actual yieldyield, the amount one actually , the amount one actually produces and measuresproduces and measures

Percent YieldPercent Yield

A comparison of the amount A comparison of the amount actually obtained to the amount actually obtained to the amount it was possible to makeit was possible to make

Actual YieldTheoretical YieldPercent Yield = x 100

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Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations