stoichiometry and the mole chapter 8 what is stoichiometry? quantitative aspects of chemistry ...

Stoichiometry and the Stoichiometry and the molemole

Chapter 8Chapter 8

What is stoichiometry?What is stoichiometry?

Quantitative aspects of chemistryQuantitative aspects of chemistry Stoicheon Greek root (element)Stoicheon Greek root (element) Metron Greek root( to measure)Metron Greek root( to measure) Calculate how much of a reactant Calculate how much of a reactant

needed to produce a product or how needed to produce a product or how much product could be expectedmuch product could be expected

StoichiometryStoichiometry

How much do I want?How much do I want? How much do I have?How much do I have? How much will I get?How much will I get?

The Mole•Defined as the number of carbon atoms in exactly 12 grams of carbon-12. •1 mole is 6.02 x 1023 particles. •Treat it like a very large dozen •6.02 x 1023 is called Avogadro's number.

The MoleThe Mole

One mole = 6.022 x 10One mole = 6.022 x 1023 23 (Avogadro’s (Avogadro’s number)number)

Avogadro’s hypothesisAvogadro’s hypothesis

When all of the following are the same:When all of the following are the same: Volumes of containersVolumes of containers Temperature of the gasesTemperature of the gases The pressure exerted by and on each The pressure exerted by and on each

gasgas

Then, the number of molecules in each Then, the number of molecules in each container will be the same, too. container will be the same, too. However, the masses are not equal.However, the masses are not equal.

NN2(g)2(g) + 3H + 3H2(g)2(g)2NH2NH3(g)3(g)

11 33 22

1 dozen1 dozen

(12)(12)3 dozen3 dozen

(36)(36)2 dozen2 dozen

(24)(24)

1 gross1 gross

(144)(144)3 gross3 gross

(432)(432)2 gross2 gross

(288)(288)

1mole1mole(6.02x10(6.02x102323))

3 moles3 moles(18.06x10(18.06x102323))

2 moles2 moles(12.04 x10(12.04 x102323))

Representative particles

•The smallest pieces of a substance.

•For an element it is an atom. –Unless it is diatomic

•For a molecular compound it is a molecule.

•For an ionic compound it is a formula unit.

Molar MassMolar Mass

A substance’s A substance’s molar mass molar mass (molecular weight) is the mass in (molecular weight) is the mass in grams of one mole of the compound. grams of one mole of the compound.

COCO22 = 44.01 grams per mole = 44.01 grams per mole

HH22O = 18.02 grams per moleO = 18.02 grams per mole

Ca(OH)Ca(OH)22 = 74.10 grams per mole = 74.10 grams per mole

Chemical EquationsChemical Equations

Chemical change involves a Chemical change involves a reorganization of reorganization of

the atoms in one or more substances.the atoms in one or more substances.CC22HH55OH + 3OOH + 3O22 2CO2CO22 + 3H + 3H22OO

reactantsreactants productsproducts

11 mole of ethanol mole of ethanol reacts with reacts with 33 moles of moles of oxygen oxygen

to produce to produce 22 moles of carbon dioxide moles of carbon dioxide and and 33 moles of watermoles of water

When the equation is balanced it has When the equation is balanced it has quantitative significance:quantitative significance:

Mole Mole RelationsRelations

Calculating Masses of Reactants Calculating Masses of Reactants and Productsand Products

1.1. Balance the equation. Balance the equation.

2.2. Convert mass to moles. Convert mass to moles.

3.3. Set up mole ratios. Set up mole ratios.

4.4. Use mole ratios to calculate moles Use mole ratios to calculate moles of desired substance. of desired substance.

5.5. Convert moles to grams, if Convert moles to grams, if necessary.necessary.

Working a Stoichiometry Working a Stoichiometry ProblemProblem

6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.

1. Identify reactants and products and write the balanced equation.

Al + O2 Al2O3

b. What are the reactants?

a. Every reaction needs a yield sign!

c. What are the products?

d. What are the balanced coefficients?

4 3 2

How to solve stoichiometry How to solve stoichiometry problems?problems?

Write the balanced chemical equationWrite the balanced chemical equation Convert to moles information on Convert to moles information on

reactants and productsreactants and products Use mole ratios from equation to Use mole ratios from equation to

determine number of moles of determine number of moles of unknownunknown

Convert moles to unit desiredConvert moles to unit desired g xg xmoles xmoles xmoles ymoles yg yg y

Working a Stoichiometry Working a Stoichiometry ProblemProblem

6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?

4 Al + 3 O2 2Al2O3

=6.50 g Al

? g Al2O3

1 mol Al

26.98 g Al 4 mol Al

2 mol Al2O3

1 mol Al2O3

101.96 g Al2O3

6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 =

12.3 g Al2O3

Percent yieldPercent yield

The % yield is calculated from:The % yield is calculated from:

(actual yield / theoretical yield) x 100(actual yield / theoretical yield) x 100 The theoretical yield is how much The theoretical yield is how much

product is predicted from balanced product is predicted from balanced chemical equation.chemical equation.

The actual yield is how much is The actual yield is how much is recovered when actual experiment is recovered when actual experiment is conducted.conducted.

Limiting Limiting ReactantReactant

The The limiting reactantlimiting reactant is the reactant is the reactant

that is that is consumed firstconsumed first,, limiting the amounts of limiting the amounts of products formed.products formed.

Limiting reactantLimiting reactant

Reactant present in short supplyReactant present in short supply

Excess reactant•Reactant in excess relative to limiting reactant

FormulasFormulas

molecular formula = (empirical molecular formula = (empirical formula)formula)nn [ [nn = integer] = integer]

molecular formula = Cmolecular formula = C66HH66 = = (CH)(CH)6 6

empirical formula = CHempirical formula = CH

Empirical formula: the lowest whole number ratio of atoms in a compound.

Molecular formula: the true number of atoms of each element in the formula of a compound.

FormulasFormulas (continued)(continued)

Formulas for Formulas for ionic compoundsionic compounds are are ALWAYSALWAYS empirical (lowest whole empirical (lowest whole number ratio).number ratio).Examples:Examples:

NaCl MgCl2 Al2(SO4)3 K2CO3

FormulasFormulas (continued)(continued)

Formulas for Formulas for molecular compoundsmolecular compounds MIGHTMIGHT be empirical (lowest whole be empirical (lowest whole number ratio).number ratio).

Molecular:Molecular:

H2O

C6H12O6 C12H22O11

Empirical:

H2O

CH2O C12H22O11

Empirical Formula Empirical Formula DeterminationDetermination

1.1. Base calculation on 100 grams of Base calculation on 100 grams of compound. compound.

2.2. Determine moles of each element in 100 Determine moles of each element in 100 grams of compound. grams of compound.

3.3. Divide each value of moles by the Divide each value of moles by the smallest of the values. smallest of the values.

4.4. Multiply each number by an integer to Multiply each number by an integer to obtain all whole numbers.obtain all whole numbers.

Empirical Formula Empirical Formula DeterminationDetermination

Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

49.32 14.107

12.01

g C mol Cmol C

g C

6.85 16.78

1.01

g H mol Hmol H

g H

43.84 12.74

16.00

g O mol Omol O

g O

Empirical Formula Empirical Formula DeterminationDetermination

(part 2)(part 2)

4.1071.50

2.74

mol C

mol O

6.782.47

2.74

mol H

mol O

2.741.00

2.74

mol O

mol O

Divide each value of moles by the smallest Divide each value of moles by the smallest of the values.of the values.

Carbon:Carbon:

Hydrogen:Hydrogen:

Oxygen:Oxygen:

Empirical Formula Empirical Formula DeterminationDetermination

(part 3)(part 3)Multiply each number by an integer to Multiply each number by an integer to obtain all whole numbers.obtain all whole numbers.

Carbon: 1.50Carbon: 1.50 Hydrogen: 2.50Hydrogen: 2.50 Oxygen: 1.00Oxygen: 1.00x 2 x 2 x 2

33 55 22

Empirical formula:C3H5O

2

Finding the Molecular Finding the Molecular FormulaFormula

The empirical formula for adipic acid The empirical formula for adipic acid is Cis C33HH55OO22. The molecular mass of . The molecular mass of adipic acid is 146 g/mol. What is the adipic acid is 146 g/mol. What is the molecular formula of adipic acid?molecular formula of adipic acid?

1. Find the formula mass of 1. Find the formula mass of CC33HH55OO22

3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g

Finding the Molecular Finding the Molecular FormulaFormula

The empirical formula for adipic acid The empirical formula for adipic acid is Cis C33HH55OO22. The molecular mass of . The molecular mass of adipic acid is 146 g/mol. What is the adipic acid is 146 g/mol. What is the molecular formula of adipic acid?molecular formula of adipic acid?

3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g

2. Divide the molecular mass by 2. Divide the molecular mass by the mass given by the emipirical the mass given by the emipirical formula.formula.

1462

73

Finding the Molecular Finding the Molecular FormulaFormula

The empirical formula for adipic acid The empirical formula for adipic acid is Cis C33HH55OO22. The molecular mass of . The molecular mass of adipic acid is 146 g/mol. What is the adipic acid is 146 g/mol. What is the molecular formula of adipic acid?molecular formula of adipic acid?

3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g146

273

3. Multiply the empirical formula by 3. Multiply the empirical formula by this number to get the molecular this number to get the molecular formula.formula.

(C(C33HH55OO22) x 2 ) x 2 ==

CC66HH1010OO44

Combustion AnalysisCombustion Analysis Technique that requires burning of an unknown substance and trap Technique that requires burning of an unknown substance and trap

the gases from burning.the gases from burning.

Used to determine molecular Used to determine molecular formula of an unknownformula of an unknown

Combustion= burningCombustion= burning Using oxygen Using oxygen

Points about combustionPoints about combustion Element that makes unknown almost always contain carbon Element that makes unknown almost always contain carbon

and hydrogen. Oxygen is often involved and nitrogen is and hydrogen. Oxygen is often involved and nitrogen is involved sometimes.involved sometimes.

Must know mass of the unknown substance before burning Must know mass of the unknown substance before burning itit

Unknown will be burnt in pure oxygen, present in excessUnknown will be burnt in pure oxygen, present in excess Carbon dioxide and water are the productsCarbon dioxide and water are the products All the carbon winds up as carbon dioxide and all the All the carbon winds up as carbon dioxide and all the

hydrogen winds up as waterhydrogen winds up as water The end result will be to determine the empirical formula of The end result will be to determine the empirical formula of

the substancethe substance