The Mole Concept and Stoichiometry From Chapters 8 and 9

CHEM 1305:Introductory Chemistry

The Mole Concept and Stoichiometry

From Chapters 8 and 9

Textbook “Introductory Chemistry: Concepts and Critical Thinking” Seventh Edition by Charles H. Corwin

© 2014 Pearson Education, Inc.

Avogadro’s Number

• When describing chemical reactions, it is very important to consider the number of atoms present in the reactants/products

• For each element, there is a standard number of atoms related to that element’s atomic mass

• For example, how many atoms are in 12.01 g carbon (atomic mass of carbon = 12.01 amu)?

• There are 6.02 × 1023 atoms of carbon in 12.01 g of carbon

• 6.02 × 1023 is referred to as Avogadro’s number (symbol N)

2


Atomic Mass and Avogadro’s Number

• State the mass of Avogadro’s number of atoms for each of the following:• copper

6.02 × 1023 atoms of Cu = 63.55 g

• mercury

6.02 × 1023 atoms of Hg = 200.59 g

• sulfur

6.02 × 1023 atoms of S = 32.07 g

• helium

6.02 × 1023 atoms of He = 4.00 g

• State the mass for each of the following:• 1 atom of Au 196.67 amu • 6.02 × 1023 atoms of Au 196.67 g

3


Mole Calculations I

• The mole (symbol mol) is a unit of measure for an amount of a chemical substance – defined as the amount of a substance that contains Avogadro’s number of particles (6.02 × 1023 particles)

1 mol = Avogadro’s number (N) = 6.02 × 1023 atoms (or molecules or units)

• The mole relationship can be used to convert between the number of particles and the mass of substance – we will again use the unit analysis method from Chapter 2

• Example – how many molecules are in 0.250 mol chlorine gas?

4


0.250 mol Cl2 × = 1.51 × 1023 molecules Cl26.02 × 1023 molecules Cl2

1 mol Cl2

Mole Calculations I

• Calculate the number of sodium atoms in 0.120 mol Na

• Calculate the number of moles of potassium in 1.25 × 1021 atoms K

5


0.120 mol Na × = 7.22 × 1022 atoms Na6.02 × 1023 atoms Na

1 mol Na

1.25 × 1021 atoms K × = 0.00208 mol K1 mol K

6.02 × 1023 atoms K

Visualizing the Mole Concept

1 mol = 6.02 × 1023 atoms (or molecules, or units of anything)

6


1 mol of softballs

(6.02 × 1023 softballs) occupy

the same volume as Earth

1 mol of shotput balls have

the same mass as Earth

Molar Mass

• The atomic mass of any substance expressed in g corresponds to 1 mol of the substance

• 1 mol of carbon has a mass of 12.01 g; 1 mol of lithium has a mass of 6.94 g

• The atomic mass of a substance expressed in g is known as the molar mass (symbol MM) with units of g/mol

• The molar mass of iron is 55.85 g/mol; the molar mass of boron is 10.81 g/mol

• The molar mass of oxygen gas (O2) is 32.00 g/mol (2 × 16.00 g/mol)

• The molar mass of a compound is calculated by adding the molar masses of each element

• Fe2O3: (2 × 55.85 g/mol Fe) + (3 × 16.00 g/mol O) = 159.70 g/mol

• Note that 16.00 g/mol should be used for O – and not 32.00 g/mol which is for O2

7


Molar Mass Calculations

• Calculate the molar mass for each of the following substances:• silver metal, Ag

107.87 g/mol• ammonia gas, NH3

14.01 g/mol + (3 × 1.01 g/mol) = 17.04 g/mol• magnesium nitrate, Mg(NO3)2

24.31 g/mol + (2 × 14.01 g/mol) + (6 × 16.00 g/mol) = 148.33 g/mol• manganese metal, Mn

54.94 g/mol• sulfur hexafluoride, SF6

32.07 g/mol + (6 × 19.00 g/mol) = 146.07 g/mol• strontium acetate, Sr(C2H3O2)2

87.62 g/mol + (4 × 12.01 g/mol) + (6 × 1.01 g/mol) + (4 × 16.00 g/mol) = 205.72 g/mol

8


Mole Calculations II

• The molar mass of a substance can be used as a unit factor• Molar mass of lead, Pb = 207.21 g/mol; 207.21 g Pb = 1 mol Pb

• Remember also, 6.02 × 1023 particles (atoms or molecules) = 1 mol

• Use this to calculate the mass of 2.55 × 1023 atoms of Pb

9


2.55 × 1023 atoms Pb × = 0.424 mol Pb1 mol Pb

6.02 × 1023 atoms Pb

0.424 mol Pb × = 87.86 g Pb207.21 g Pb

1 mol Pb

Mole Calculations II

• What is the mass of 7.75 × 1022 formula units of lead(II) sulfide, PbS?

• How many molecules are found in 0.175 g of fluorine gas, F2?

10


1 mol PbS

6.02 × 1023 molecules PbS7.75 × 1022 molecules PbS × × = 30.80 g PbS

239.28 g PbS

1 mol PbS

6.02 × 1023 molecules F2

1 mol F2

0.175 g F2 × × = 2.77 × 1021 molecules F2

1 mol F2

38 g F2

Percent Composition

• The percent composition of a compound lists the mass percent of each element

• The percent composition of water, H2O is 11% H and 89% O – this is always true, a drop, a mL, and a pool full of water all contain 11% H and 89% O (law of definite composition, Chapter 3)

• To calculate percent composition – assume 1 mol is present• 1 mol H2O contains 2 mol H and 1 mol O

(2 × molar mass H) + (1 × molar mass O) = (1 × molar mass H2O)

(2 × 1.01 g H) + 16.00 g O = 18.02 g H2O

11


2.02 g H

18.02 g H2O× 100% = 11.2% H

16.00 g O

18.02 g H2O× 100% = 88.79% O

Percent Composition of a Substance

• Calculate the percent composition of trinitrotoluene (TNT), C7H5(NO2)3

• Percent composition: 37.01% C, 2.22% H, 18.50% N, 42.26% O

12


(7 × 12.01 g C) + (5 × 1.01 g H) + (3 × 14.01 g N) + (6 × 16.00 g O) = g C7H5(NO2)3

84.07 g C + 5.05 g H + 42.03 g N + 96.00 g O = 227.15 g C7H5(NO2)3

84.07 g C

227.15 g C7H5(NO2)3× 100% = 37.01% C

5.05 g H

227.15 g C7H5(NO2)3× 100% = 2.22% H

42.03 g N

227.15 g C7H5(NO2)3× 100% = 18.50% N

96.00 g O

227.15 g C7H5(NO2)3× 100% = 42.26% O

Empirical Formula

• The empirical formula of a compound corresponds to the simplest whole number ratio of atoms of each element in a compound

• A 1.640 g sample of radium was heated with oxygen to produce 1.755 g of radium oxide, calculate g oxygen:

(1.755 g radium oxide – 1.640 g radium) = 0.115 g oxygen

• Calculate moles of each reactant:

• Mole ratio is Ra0.00726O0.00719 simplify by dividing by the smaller number:

13


1.640 g Ra × = 0.00726 mol Ra1 mol Ra

226.03 g Ra0.115 g O × = 0.00719 mol O

1 mol O

16.00 g O

Ra O = Ra1.01O1 RaO0.00726

0.00719

0.00719

0.00719

Empirical Formula from Mass Composition

• If 0.558 g iron metal reacts with chlorine gas to form 1.621 g iron chloride, is FeCl2 or FeCl3 the correct formula for the product?

14


0.558 g Fe × = 0.00100 mol Fe1 mol Fe

55.85 g Fe

1.063 g Cl × = 0.00300 mol Cl1 mol Cl

35.45 g Cl

Fe Cl = Fe1Cl3 FeCl30.00100

0.00100

0.00300

0.00100

Empirical Formulas from Percent Composition

• Calculate the empirical formula for benzene, 92.2% C and 7.83% H

• Assume 100 g of sample – 92.2 g C and 7.83 g H

15


92.2 g C × = 7.68 mol C1 mol C

12.01 g C

7.83 g H × = 7.75 mol H1 mol H

1.01 g H

C H = C1H1.01 CH7.68

7.68

7.75

7.68

Molecular Formula

• The empirical formula for benzene was found to be CH, but the actual formula is C6H6

• The actual formula, or molecular formula, is a multiple of the empirical formula – represented as (CH)n, where n is 6 for benzene

• Acetylene and styrene also have the empirical formula CH – the molar mass for each compound can be used to determine molecular formula

16


Benzene: =(CH)n

CH

78 g/mol

13 g/molAcetylene: =

(CH)n

CH

26 g/mol

13 g/molStyrene: =

(CH)n

CH

104 g/mol

13 g/mol

n = 6

(CH)6

C6H6

n = 2

(CH)2

C2H2

n = 8

(CH)8

C8H8

Molecular Formula from Empirical Formula

• The empirical formula for fructose is CH2O, find the molecular formula given that the molar mass of fructose is 180 g/mol

• The empirical formula for ethylene dibromide is CH2Br, find the molecular formula given that the molar mass is 190 g/mol

17


Fructose: =(CH2O)n

CH2O

180 g/mol

30 g/mol

n = 6

(CH2O)6

C6H12O6

Ethylene dibromide: =(CH2Br)n

CH2Br

190 g/mol

94 g/mol

n = 2

(CH2Br)2

C2H4Br2

The Mole Interpretation of Equation Coefficients

• Look at the balanced chemical equation for nitrogen monoxide gas reacting with oxygen in UV light to form nitrogen dioxide (smog)

2 NO(g) + O2(g) 2 NO2(g)

• 2 molecules of NO react with one molecule of O2 to form 2 molecules of NO2 – these coefficients are ratios, any number can be used

• Using Avogadro’s number as the coefficient gives the mole ratio of reactants and products

2 mol NO(g) + 1 mol O2(g) 2 mol NO2(g)

18


UV

UV

The Volume Interpretation of Equation Coefficients

• According to Avogadro’s theory, there are equal numbers of molecules in equal volumes of gas at STP conditions

• Therefore, twice the number of molecules occupies twice the volume of gas – the previous chemical reaction can be written as:

2 L NO(g) + 1 L O2(g) 2 L NO2(g)

• The following table summarizes the information we can obtain from interpreting the coefficients of a balanced chemical equation

19


UV

Interpreting Chemical Equation Coefficients

• Methane gas burns with oxygen to form carbon dioxide and water vapor

CH4(g) + O2(g) CO2(g) + H2O(g)

• Balance the chemical equation to find coefficients:

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

• Mole interpretation of the reaction:

1 mol CH4(g) + 2 mol O2(g) 1 mol CO2(g) + 2 mol H2O(g)

• Volume interpretation of the reaction:

1 L CH4(g) + 2 L O2(g) 1 L CO2(g) + 2 L H2O(g)

• Try the same for: C3H8(g) + O2(g) CO2(g) + H2O(g)

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spark

spark

spark

spark

spark

Verifying the Conservation of Mass Law

• Law of conservation of mass – mass is neither created or destroyed during a chemical reaction; combined mass of reactants must equal combined mass of products

• Let’s verify this using the previous reaction example:

2 NO(g) + O2(g) 2 NO2(g)

(2 mol NO(g) + 1 mol O2(g) 2 mol NO2(g))

• Molar mass of NO is 30.01 g/mol; O2 is 32.00 g/mol; NO2 is 46.01 g/mol

(2 mol × 30.01 g/mol) + (1 mol × 32.00 g/mol) (2 mol × 46.01 g/mol)

60.02 g + 32.00 g 92.02 g

92.02 g 92.02 g

21


UV

UV

Mole-Mole Relationships

N2(g) + O2(g) 2 NO(g)

• 1 mole nitrogen and 1 mole oxygen form 2 moles nitrogen monoxide

• The following mole ratios (unit factors) can be written for this equation:

• How many moles of oxygen react with 2.25 mol of nitrogen?

2.25 mol N2 × = 2.25 mol O2

• How many moles of nitrogen monoxide are produced?

2.25 mol N2 × = 4.50 mol NO

22


∆

1 mol N2

1 mol O2

1 mol O2

1 mol N2

1 mol N2

2 mol NO

2 mol NO

1 mol N2

1 mol O2

2 mol NO

2 mol NO

1 mol O2

1 mol O2

1 mol N2

2 mol NO

1 mol N2

Mole-Mole Relationships

• Carbon monoxide is produced by passing oxygen gas over hot coal. How many moles of oxygen react with 2.50 mol of carbon?

2 C(s) + O2(g) 2 CO(g)

2.50 mol C × = 1.25 mol O2

• Iron is produced by passing carbon monoxide through iron(III) oxide. How many moles of iron are produced from 2.50 mol of Fe2O3?

Fe2O3(l) + 3 CO(g) 2 Fe(l) + 3 CO2(g)

2.50 mol Fe2O3 × = 5.00 mol Fe

23


∆

1 mol O2

2 mol C

∆

2 mol Fe

1 mol Fe2O3

Types of Stoichiometry Problems

• Stoichiometry refers to the relationship between quantities in a chemical reaction according to the balanced chemical equation

• There are three types of stoichiometry problems

a A + b B c C + d D

• Mass-mass problem: ex., given grams of A, calculate grams of C• How many grams of Zn metal react with hydrochloric acid to give 0.500 g ZnCl2?

• Mass-volume problem: ex., given grams of A, calculate liters of C• How many kilograms of Fe react with H2SO4(aq) to produce 50.0 mL H2 gas?

• Volume-volume problem: ex., given liters of A, calculate liters of C• How many liters of H2 gas react with Cl2 gas to give 50.0 cm3 of HCl gas?

24


Mass-Mass Problems

• Under high temperature, 14.4 g of iron(II) oxide is converted to elemental iron using aluminum metal

3 FeO(l) + 2 Al(l) 3 Fe(l) + Al2O3(l)

• How many grams of aluminum metal are necessary for the reaction?

• 2 moles Al react with 3 moles of FeO... what next?

• Molar mass relates grams to moles!

• Calculate molar mass of each compound and use unit analysis method

14.4 g FeO × × × = 3.60 g Al

25


1 mol FeO

71.85 g FeO

2 mol Al

3 mol FeO

26.98 g Al

1 mol Al

Mass-Mass Stoichiometry

• Calculate the mass of solid potassium iodide required to yield 1.78 g of mercury(II) iodide precipitate

2 KI(s) + Hg(NO3)2(aq) HgI2(s) + 2 KNO3(aq)

• Molar mass of KI is 166.00 g/mol, molar mass of HgI2 is 454.39 g/mol

1.78 g HgI2 × × × = 1.30 g KI

• Calculate the mass of iron filings required to produce 0.455 g Ag metal

Fe(s) + AgNO3(aq) Fe(NO3)3(aq) + Ag(s)

0.455 g Ag × × × = 0.0785 g Fe

26


1 mol HgI2454.39 g HgI2

2 mol KI

1 mol HgI2

166.00 g KI

1 mol KI

Fe(s) + 3 AgNO3(aq) Fe(NO3)3(aq) + 3 Ag(s)

1 mol Ag

107.87 g Ag

1 mol Fe

3 mol Ag

55.85 g Fe

1 mol Fe

Limiting Reactant Concept

• We want to build the following block tower:

• How many identical towers can we build with the following pieces? 2

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• Let’s apply this to a chemical reaction:

Fe(s) + S(s) FeS(s)

• There is an excess of sulfur – iron is called the limiting reactant and it controls the amount of product which is formed

• Let’s say 2.50 mol of Fe are heated with 3.00 mol S

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∆

Experiment Mol Fe Mol S Mol FeS

FeS

Fe FeS

S

S

S FeS

Fe FeS

S

S

S

+

+ -

-

+

+ -

-

+

+ -

-

∆


• 1.00 mol of iron(II) oxide is heated with 1.00 mol of aluminum metal to form molten iron and solid aluminum oxide as the products

3 FeO(l) + 2 Al(l) 3 Fe(l) + Al2O3(s)

• Identify the limiting reactant and calculate the moles of iron produced

• First, determine how many moles of Fe are produced by each reactant

1.00 mol FeO × = 1.00 mol Fe

1.00 mol Al × = 1.50 mol Fe

• FeO is the limiting reactant and only produces 1.00 mol Fe metal

29


∆

3 mol Fe

3 mol FeO

3 mol Fe

2 mol Al

Limiting Reactant Problems

3 FeO(l) + 2 Al(l) 3 Fe(l) + Al2O3(s)

• How much Fe is produced from 25.0 g of FeO reacting with 25.0 g of Al?

• The number of moles must first be determined

25.0 g FeO × × = 0.348 mol Fe

25.0 g Al × × = 1.39 mol Fe

• FeO is the limiting reactant and will determine mass of Fe produced

0.348 mol Fe × = 19.4 g Fe

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∆

3 mol Fe

3 mol FeO

1 mol FeO

71.85 g FeO

1 mol Al

26.98 g Al

3 mol Fe

2 mol Al

55.85 g Fe

1 mol Fe

Mass-Mass Limiting Reactant

• 50.0 g manganese(IV) oxide reacts with 25.0 g of aluminum to form manganese metal and aluminum oxide. What is the limiting reactant?

3 MnO2(l) + 4 Al(l) 3 Mn(l) + 2 Al2O3(s)

• What mass of Mn metal is produced?

31


∆

50.0 g MnO2 × × × = 31.6 g Mn1 mol MnO2

86.94 g MnO2

3 mol Mn

3 mol MnO2

54.94 g Mn

1 mol Mn

25.0 g Al × × × = 38.2 g Mn1 mol Al

26.98 g Al

3 mol Mn

4 mol Al

54.94 g Mn

1 mol Mn

MnO2 is the limiting reactant, 31.6 g of Mn is produced

Percent Yield

• Laboratory experiment: a student weighs a 1.00-g sample of CdCl2 and dissolves it in water, next aqueous Na2S solution is added to obtain a yellow CdS precipitate and an aqueous NaCl solution

CdCl2(aq) + Na2S(aq) CdS(s) + 2 NaCl(aq)

• The student collects the CdS on some filter paper, and finds it has a mass of 0.775 g – this is called the actual yield

• How much CdS should have been produced? Calculated as 0.788 g, known as theoretical yield

• Percent yield can be calculated as the ratio below:

32


actual yield

theoretical yield × 100% = percent yield

0.775 g

0.788 g× 100% = 98.4%

Percent Yield

• A student dissolves 1.50 g of copper(II) nitrate in water. After adding aqueous sodium carbonate solution, the student obtains 0.875 g of CuCO3 precipitate. If the theoretical yield is 0.988 g, what is the percent yield?

Cu(NO3)2(aq) + Na2CO3(aq) CuCO3(s) + 2 NaNO3(aq)

• 15.0 kg ammonia and nitric acid react to produce ammonium nitrate with an actual yield of 71.5 kg – what is the percent yield?

101%

33


× 100% = 88.6%0.875 g

0.988 g

Documents

The Mole Concept and Stoichiometry From Chapters 8 and 9