The Mole and Stoichiometry

The Mole and The Mole and StoichiometryStoichiometry

Chemistry gets Real….Chemistry gets Real….

Tough that isTough that is

The “mole”The “mole”

A term for a certain number of A term for a certain number of something.something.

Brainstorm other counting words!Brainstorm other counting words! Dozen = Dozen = Pair = Pair = Gross = Gross =

A “mole” of something is A “mole” of something is 6.02 x 106.02 x 102323 of something of something..

602, 000, 000, 000, 000, 000, 000, 000602, 000, 000, 000, 000, 000, 000, 000

12

2

144

Molecular WeightMolecular Weight

M.W. = M.W. = the the weightweight (in grams) of a (in grams) of a molemole of of substancesubstance

On your periodic tablesOn your periodic tables Round to the nearest tenthRound to the nearest tenth

Hydrogen is 1.00797 Hydrogen is 1.00797 1.0 g/mol 1.0 g/mol

Mole Weight is an INTENSIVE propertyMole Weight is an INTENSIVE property—doesn’t depend on amount—doesn’t depend on amount

Try these MW’sTry these MW’s

CaCa 40.140.1

HH22

2 (1.0) = 2.02 (1.0) = 2.0

BaFBaF22

[137.3 + 2(19.0)] = 175.3[137.3 + 2(19.0)] = 175.3

MW of 2BaFMW of 2BaF22 is still 175.3 is still 175.3

mol

g

mol

g

mol

g

1 mole of Ca weighs

40.1 grams

1 mole of H2 weighs 2.0 grams

mol

g

1 penny = 2.681 penny = 2.68

6 pennies =6 pennies = g g

6 pennies = 6 pennies =

coin

g

coin

g2.68

2.68 X 6 = 16.08

Avogadro’s NumberAvogadro’s Number

NNAA = 6.02 x 10 = 6.02 x 102323 of of anythinganything

Avogadro (1776-1856)

1 mole K = 1 mole K = gramsgrams

MW of potassium = MW of potassium =

1 dozen K = 1 dozen K = atoms atoms 1 mole K = atoms 1 mole K = atoms

mol

g39.1

39.10

12

6.02 x 1023

MW of COMW of CO22 = = 3 moles of CO3 moles of CO22 = =

3 moles of CO3 moles of CO22 = ____ g/mol = ____ g/mol

MW of nitrogen gas = MW of nitrogen gas = NN22 (g) = (g) =

mol

g44.0

2 ( 14.0) = 28.0 mol

g

gramsmol

gmoles 00.1320.443

mol

g44.0

Which elements exist as diatomic Which elements exist as diatomic molecules?molecules?

HH2 2 NN22 OO22 F F22 ClCl22 BrBr22 II22

How Big Is The Mole?How Big Is The Mole? One mole of marbles would cover the entire One mole of marbles would cover the entire

Earth to a depth of fifty milesEarth to a depth of fifty miles One mole of hockey pucks would equal the One mole of hockey pucks would equal the

mass of the moon.mass of the moon. One mole of rice grains is more than the One mole of rice grains is more than the

number of grains of all crops grown since the number of grains of all crops grown since the beginning of time.beginning of time.

If one mole of pennies was divided up equally If one mole of pennies was divided up equally between all the people on Earth, you would between all the people on Earth, you would have enough money to spend a million have enough money to spend a million dollars every hour, 24 hours a day, for your dollars every hour, 24 hours a day, for your entire life. When you died, you would have entire life. When you died, you would have spent less than half of your riches.spent less than half of your riches.

Lab 11Lab 11

FormulaFormula Mass (g)Mass (g) MW MW MolesMoles MoleculMoleculeses

AtomsAtoms

AgNOAgNO33 15.0015.00 169.9169.9 0.0880.088 5.3 x 5.3 x 10102222

2.7 x 2.7 x 10102323

KOHKOH 18.9818.98 56.156.1 0.340.34 2.0 x 2.0 x 10102323

6.0 x 6.0 x 10102323

COCO 47.0047.00 28.028.0 1.681.68 1.0 x 1.0 x 10102424

2.0 x 2.0 x 10102424

NaNa22SOSO44 185.00185.00 142.1142.1 1.301.30 7.8x 7.8x 10102323

5.4 x 5.4 x 10102424

HH22 1.001.00 2.02.0 0.500.50 3.0 x 3.0 x 10102323

6 x 106 x 102323

NaNa22SS 13.7613.76 78.178.1 0.180.18 1.1 x 1.1 x 10102323

3.3 x 3.3 x 10102323

NaOHNaOH 214.75214.75 40.040.0 5.375.37 3.2 x 3.2 x 10102424

9.6 x 9.6 x 10102424

PbPb 6.506.50 207.2207.2 0.0310.031 1.8 x 1.8 x 10102222

1.8 x 1.8 x 10102222

PbPb33(PO(PO44))

22

18.7518.75 811.6811.6 0.0230.023 1.4 x 1.4 x 10102222

1.8 x 1.8 x 10102323

NaClNaCl 62.2562.25 58.558.5 1.061.06 6.6 x 6.6 x 10102323

1.3 x 1.3 x 10102424

mol

g

÷MW X NA X#atoms

Percent CompositionPercent Composition

A. Determined from Formulas (“Accepted A. Determined from Formulas (“Accepted Value”)Value”)

Is NaCl 50.0% Na by weight?Is NaCl 50.0% Na by weight?

NoNo, Na is 23.0 g/mole and Cl is 35.5 g/mole , Na is 23.0 g/mole and Cl is 35.5 g/mole

To Prove, % Na =To Prove, % Na =

%3.39100 X )235.35(

23

100 NaCl

Na X

MW

MW

Percent CompositionPercent Compositionfrom Formulafrom Formula

% oxygen in CaCO% oxygen in CaCO33??

Grams of Mg in 4.00 grams of MgO?Grams of Mg in 4.00 grams of MgO?

O%0.48%9520.47 100 X (100.1)

(48)

)]16(30.121.40[

)16(3

Mg%3.60%2977.60)163.24(

3.24

gMgg 41.2)00.4(603.0

First: Calculate % Mg in MgO

Second: Calculate g Mg in 4.00 g MgO

B. Experimental Percent CompositionB. Experimental Percent Composition FROM DATA (“Experimental Value”)FROM DATA (“Experimental Value”) 4.00g of Ag4.00g of Ag22O is O is decomposeddecomposed to yield 3.65g Ag. to yield 3.65g Ag.

The Experimental % = ? The Experimental % = ?

The Accepted % = ?The Accepted % = ?

Your Experimental Error?Your Experimental Error?

Experimental % AgExperimental % Ag

Equation:Equation:

AgAg22O O

Experimental % Ag:Experimental % Ag:

4.00 g 3.65 g

%3.91%25.9100.4

65.3

g

g

Ag + O22 4

True (Accepted) %AgTrue (Accepted) %Ag

%9.1%9333.1100 X 1.93

1.933.91 100 X

valueaccepted

value)accepted - valuealexperiment(

%1.93%097.93 100 X 231.8

(215.8)

]16)9.107(2[

)9.107(2

100 Ag

Ag

2

XOMW

MW

Percent Error?

Empirical FormulaEmpirical Formula

Definition:Definition:The The simplestsimplest formula indicating the formula indicating the

mole ratiomole ratio of elements in a compound of elements in a compound Examples:Examples:

HH22OO22 HO HO CC66HH66CHCH NN22OO44?? COCO22 ? ?

NO2

CO2

Empirical Formula STEPSEmpirical Formula STEPS

1.1. Change grams to molesChange grams to moles

2.2. Divide by the least # moles for a Divide by the least # moles for a RATIORATIO

3.3. Apply ratio to the formulaApply ratio to the formula

Solving Empirical Formula—Solving Empirical Formula—determined from gram compositiondetermined from gram composition

A compound contains 0.90 g Ca and 1.60 A compound contains 0.90 g Ca and 1.60 g Clg Cl

Camoles 022.01.40

90.0

2Ca 022.0

Cl 045.0

moles

molesCaCl2

Clmoles 045.05.35

60.1

Try This…Try This…

0.556 g Carbon and 0.0933 g Hydrogen0.556 g Carbon and 0.0933 g Hydrogen

molesC046.00.12

0556.0 molesH0933.0

1

0933.0

2C 046.0

H 0933.0

moles

molesCH2

Why is the Empirical Formula a Why is the Empirical Formula a ratio of small WHOLE numbers?ratio of small WHOLE numbers?

Can’t have half of an atomCan’t have half of an atom Atoms combine as Atoms combine as wholewhole units units Shows the simplest way that atoms Shows the simplest way that atoms

can paircan pair

What formula would this ratio give?What formula would this ratio give?

K = 0.26 molesK = 0.26 moles N = 0.25 molesN = 0.25 moles O = 0.78 molesO = 0.78 moles Yields an Empirical Formula of…Yields an Empirical Formula of…

KNOKNO33

Try This: 70.5 % Fe and 29.5 % OTry This: 70.5 % Fe and 29.5 % O

___ moles Fe, ___ moles O___ moles Fe, ___ moles O

Assume 100g of substance:Assume 100g of substance:

Femolesiron

molgIrongof

26.19.55

5.70

1.26

1.84

Fe2O3

Omolesoxygen

molgOxygengof

84.10.16

5.29

1.26 moles 1.26 moles

FeO1.5X 2

Try This: 40.0% CTry This: 40.0% C 6.7% H 53.3% O6.7% H 53.3% O

___ moles C, ___ moles H, ___ moles O___ moles C, ___ moles H, ___ moles O

Assume 100g of substance:Assume 100g of substance:

molesCcarbon

molggofcarbon

3.30.12

00.40

3.3 6.7 3.3

After dividing by the least moles yields:

CH2

O

Why doesn’t the ratio of the % give Why doesn’t the ratio of the % give the empirical formula?the empirical formula?

Must account for differing Must account for differing massesmasses of of elements.elements.

Why does the ratio of the moles give Why does the ratio of the moles give the empirical formula?the empirical formula? The ratio of the # of atoms normalizes The ratio of the # of atoms normalizes

for mass differences.for mass differences.

Molecular FormulasMolecular Formulas

Definiton: Definiton: Formula of an actual Formula of an actual compoundcompound

as it exists in molecules.as it exists in molecules. Benzene exists as CBenzene exists as C66HH6 6 not CH. not CH. Hydrogen Peroxide exists as HHydrogen Peroxide exists as H22OO2 2

not HO.not HO.

Empirical Empirical FormulaFormula

MW MW EmpiricalEmpirical

MW MW MolecularMolecular

Molecular Molecular FormulaFormula

HOHO 34 g/mol34 g/mol

CHOClCHOCl22 200 200 g/molg/mol

CClNCClN22 226.5 226.5 g/molg/mol

H2O217 g/mol17 g/mol

X 2

100 g/mol100 g/mol

75.5 g/mol75.5 g/mol

CC22HH22OO22ClCl44

CC33ClCl33NN66

X 2

Why is the M.W. needed to Why is the M.W. needed to determine the molecular formula?determine the molecular formula? Need M.W. of actual compound to find Need M.W. of actual compound to find

how many each type of atom is in a how many each type of atom is in a molecule.molecule.

Stoichiometry – The Big LeaguesStoichiometry – The Big Leagues

A. Define:A. Define:Problem Solving involving mass-mass Problem Solving involving mass-mass relationships in chemical changesrelationships in chemical changes Ex. How many grams of rust are formed Ex. How many grams of rust are formed

when 12.00 g of Fe reacts with oxygen.when 12.00 g of Fe reacts with oxygen.

B. Must use B. Must use balancedbalanced equations for equations for the correct mole ratiosthe correct mole ratios

C. CoefficientsC. Coefficients yield the mole yield the mole ratio!!!ratio!!!

22 H H22 + O + O22 22 H H22OO

2 : 12 : 1 : 2 : 2

D. ExampleD. Example

4 Fe + 3 O4 Fe + 3 O22 2 Fe 2 Fe22OO33

4 : 3 : 24 : 3 : 2

If 4 moles of iron rust, If 4 moles of iron rust, moles of Fe moles of Fe22OO33 will form will form

If 8 moles of iron rust,If 8 moles of iron rust, moles of Femoles of Fe22OO33 will form will form4

2

Solving Solving Mass-Mass Mass-Mass ProblemsProblems

1) Determine the mole ratio (from the balanced equation)

2) Convert grams to moles

3) Apply the mole ratio

4) Convert moles to grams

28.00 g of iron yields 28.00 g of iron yields ? ? g of rust? g of rust?

4 Fe + 3 O4 Fe + 3 O22 2 Fe 2 Fe22OO33

28.00 g 28.00 g FeFe

0.50 moles0.50 moles

molgg

8.55

00.28

2:1

molesmol

g25.00.160


40.00 g 40.00 g FeFe22OO33

? g Fe2O3

36.00 g of water resulted from 36.00 g of water resulted from ? ? g of methane? g of methane?

CHCH44 + O + O22 CO CO22 + H + H22OO

36.00 g 36.00 g HH22OO


gmol

gmole 00.160.161

1:2

moles

molgg

20.18

00.36


16.00g of 16.00g of methanemethane

2 2

Balance

? g CH4

Variation:Variation:12 moles of oxygen combusting will yield 12 moles of oxygen combusting will yield

how many grams of COhow many grams of CO22??

12 moles12 moles 2:1

molesmol

g60.44

6 moles6 moles

264.00 g 264.00 g COCO22

CHCH44 + O + O22 CO CO22 + H + H22OO2 2

? g CO2

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The Mole and Stoichiometry