Stoichiometry! The heart of chemistry. The Mole The mole is the SI unit chemists use to represent an amount of substance. 1 mole of any substance = 6.02.

Stoichiometry!The heart of chemistry

The Mole

•The mole is the SI unit chemists use to represent an amount of substance.

•1 mole of any substance = 6.02 x 1023 representative particles. This is known as Avogadro’s number.

Types of Representative Particles

1. The representative particle of most elements is the atom. Ex. 1 mol Ca = 6.02 x 1023 atoms Ca

2. The representative particle of covalent compounds (including diatomic elements) is the molecule. Ex. 1 mol SO3 = 6.02 x 1023 molecules SO3.

3. The representative particle of ionic compound is the formula unit. Ex. 1 mol NaCl = 6.02 x 1023 formula units NaCl.

Remember

•A mole represents number of particles, not mass.

•Each element has a unique atomic mass.•Substances with the same number of

moles can have different masses.

Molar Mass•Molar mass

represents how much mass there is in one mole of a substance.

•Molar mass is expressed in units of g/mol.

atom

ic

mass

Ex) Molar mass of Sr

87.62 g/mol

How do I calculate molar mass?1. Molar mass of an element = atomic mass

of an element (periodic table)Ex. 1 mol He = 4.00 g He. The molar mass of

He is 4.00 g/mol.1 mol S = 32.07 g S. The molar mass of S is 32.07 g/mol.

Calculating molar mass cont.

2. Molar mass of a compound = sum of the masses of the elements in the compound

a) What is the molar mass of water? H2OH: 2 x 1.01 = 2.02O: 1 x 16.00 = 16.00

= 18.02 g/mol b) What is the molar mass of oxygen gas? O2

O: 2 x 16.00 = 32.00 g/mol

Target Check• What is the molar mass of lithium?

6.94 g/mol• What is the molar mass of carbon?

12.01 g/mol• What is the molar mass of calcium nitrate?

Ca(NO3)2

Ca: 1 x 40.08 = 40.08N: 2 x 14.01 = 28.02O: 6 x 16.00 = 96.00= 164.10 g/mol

What if I have more/less than one mole of a substance?!1. Calculate the number of moles of NaCl in 175.5 g

of the salt.

2. How many atoms are in 2.5 moles of sodium metal?

3. A sample of nitrogen gas contains 1.20 x 1025 molecules. Determine the number of grams of nitrogen in the sample.

NaCl mol 3.003 =NaCl) g (58.44

NaCl) mol (1 x NaCl g 175.5

22

2

223

22

25 N g 559 N mol 1

N g 28.02 x

N molecules10 x 6.02

N mol 1 x N molecules10 x 1.20

Na atoms10 x 1.5 Na mol 1

Na atoms 10 x 6.02 x Na mol 5.2 24

23

Target Check

1. Calculate the mass of 0.800 moles of H2SO4.

2. How many atoms are in a 10.0 g sample of calcium metal?

4242

4242 SOH g 78.5

SOH mol 1

SOH g 98.09 x SOH mol 800.0

Ca atoms 10 x 1.50 Ca mol 1

Ca atoms 10 x 6.02 x

Ca g 40.08

Ca mol 1 x Ca g 0.10 23

23

Molar Volume

•The volume of a gas is usually measured at a standard temperature and pressure (STP). ▫Standard temperature is 0°C.▫Standard pressure is 1 atm.

•A mole of any gas at STP occupies a volume of 22.4 L. This is the molar volume of a gas.

Molar Volume Examples

•Determine the volume, in liters, of 0.600 moles of sulfur dioxide gas at STP.

•Determine the number of molecules in 33.6 L of hydrogen gas at STP.

22

22 SO L 13.4

SO mol 1

SO L 22.4 x SO mol 600.0

223

2

223

2

22 H molecules 10 x 9.03

H mol 1

H molecules 10 x 6.02 x

H L 4.22

H mol 1 x H L 6.33

Target Check

•Determine the number of moles of 58.6 L of CO2 gas at STP.

•Determine the number of liters of 76.0 g of oxygen gas at STP.

22

22 CO mol 2.62

CO L 4.22

CO mol 1 x CO L 6.58

22

2

2

22 O L 53.2

O mol 1

O L 22.4 x

O g 0.32

O mol 1 x O g 0.76

Percent Composition

•Percentage composition is the percent by mass of each element in a compound.

% of element in compound =

100 x compound of massmolar

compound mol 1in element of mass

Percentage Composition ExampleFind the percentage composition of

aluminum sulfate.Al2(SO4)3 molar mass = 342.17 g/mol

Al: 2 x 26.98 g = 53.96 g Al % Al = (53.96 g/342.17 g) x 100 = 15.8%S: 3 x 32.07 g = 96.21 g S% S = (96.21 g/342.17 g) x 100 = 28.1%O: 12 x 16.00 g = 192.00 g O% O = (192.00 g/342.17 g) x 100 = 56.1%

Target Check

1. Find the percent of carbon in carbon dioxide.

CO2 molar mass = 44.01 g/mol

C: 1 x 12.01 g = 12.01 g C% C = (12.01 g/44.01 g) x 100 = 27.3%

Empirical Formulas

An empirical formula consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number mole ratio of the different atoms in the compound.

Examples: CH4, NH3, CO, LiClO3

Non-examples: N2H4 or C2H6

Empirical Formulas Example Problems

•What is the empirical formula of a compound that is 79.8% carbon and 20.2% hydrogen?

79.8 g C x (1 mol C/12.01 g C) = 6.64 mol C6.64 mol C/6.64 = 120.2 g H x (1 mol H/1.01 g H) = 20.0 mol H20.0 mol H/6.64 = 3The empirical formula is CH3.

Empirical Formulas Example Problems

•What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen?

25.9 g N x (1 mol N/14.01 g N) = 1.85 mol N

1.85 mol N/1.85 = 1 x 2 = 274.1 g O x (1 mol O/16.00 g O) = 4.63 mol

O4.63 mol O/1.85 = 2.5 x 2 = 5The empirical formula is N2O5.

Target CheckA compound is found to contain 46.0 g sodium,

52.0 g chromium, and 64.0 g oxygen. What is the empirical formula of the compound?

46.0 g Na x (1 mol Na/22.99 g Na) = 2.00 mol Na2.00 mol Na/1.00 = 252.0 g Cr x (1 mol Cr/52.00 g Cr) = 1.00 mol Cr1.00 mol Cr/1.00 = 164.0 g O x (1 mol O/16.00 g O) = 4.00 mol O4.00 mol O/1.00 = 4The empirical formula is Na2CrO4.

Molecular Formulas

The molecular formula of a compound is the actual formula of molecular compound. It is either the same as the experimentally determined empirical formula, or it is some whole-number multiple of it. In order to find the molecular formula of a compound, you must know the empirical formula first.

Molecular Formulas Example Problem

The molar mass of a compound is 92 g/mol. Analysis of a sample of the compound diciates that it contains 0.606 g N and 1.390 g O. Find its molecular formula.

0.606 g N x (1 mol N/14.01 g N) = 0.0433 mol N 0.0433 mol N/0.0433 = 11.390 g O x (1 mol O/16.00 g O) = 0.0869 mol O0.0869 mol O/0.0433 = 2The empirical formula is NO2 (46.01 g/mol).

92 g/mol/46.01 g/mol = 2The molecular formula is N2O4.

Target Check 1. The percent composition of a compound is 40.0% C, 6.7%

H, and 53.3 % O. The molar mass of the compound is 90.0 g/mol. What is the molecular formula of the compound?

40.0 g C x (1 mol C/12.01 g C) = 3.33 mol C3.33 mol C/3.33 = 16.7 g H x (1 mol H/1.01 g H) = 6.63 mol H6.63 mol H/3.33 = 253.3 g O x (1 mol O/16.00 g O) = 3.33 mol O3.33 mol O/3.33 = 1The empirical formula is CH2O (30.0 g/mol).

90.0 g/mol / 30.0 g/mol = 3The molecular formula is C3H6O2.

Target Check

2. Calculate the molecular formula of CH4N. The molar mass is 120.0 g/mol.

CH4N = 30.0 g/mol

120 g/mol / 30.0 g/mol = 4The molecular formula is C4H16N4.

Stoichiometry

Stoichiometry is the study of the amount of substances produced and consumed in chemical reactions.

Calculations are carried out step-by-step using dimensional analysis.

Molecular Relationships in Balanced EquationsA balanced equation shows the smallest

whole number of moles of each component involved in the reaction.

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

1 molecule 2 molecules 1 molecules 2 molecules

1 mole 2 moles 1 mole 2 moles

Molecular Relationships Example ProblemsC3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

1. For everyone one mole of C3H8 that completely reacts, how many moles of CO2 are produced?

3 moles CO2

2. For every 3 moles of CO2 produced by the reaction, how many moles of H2O will be produced?

4 moles H2O

3. What is the ratio of moles of O2 to moles of CO2? 5:3

Mole A Mole B Calculations

In a mole-mole problem, the number of moles of one substance in a given reaction is given and you are asked to calculate the number of moles of another substance.

Steps in solving mole-mole calculations.▫Write a balanced chemical equation.▫Use the mole ratio to convert moles of

given to moles of unknown.

Mole A Mole B Calculations Example

How many moles of oxygen gas would result from the decomposition of 13.4 moles of water?

2H2O 2H2 + O2

22

22 O mol 6.70

OH mol 2

O mol 1 x OH mol 4.13

Target Check

How many moles of nitrogen gas are required to completely react with 4.5 moles of hydrogen gas to produce ammonia?

N2 + 3H2 2NH3

22

22 N mol 1.5

H mol 3

N mol 1 x H mol 5.4

Mole A Mass B Calculations• In a mole-mass problem, the number of moles

of one substance in a given reaction is given and you are asked to calculate the number of grams of another substance, or vice versa.

• Information needed to solve a mole-mass calculation:▫A balanced chemical equation.▫Mole ratio.▫Molar mass of substance being converted to or

from grams.

Mole A Mass B Calculations Example ProblemHow many grams of zinc chloride would be

produced in the reaction of 4.5 moles of hydrochloric acid with excess zinc?

Zn + 2HCl ZnCl2 + H2

22

22 ZnClg 307 ZnClmol 1

ZnClg 136.92 x

HCl mol 2

ZnClmol 1 x HCl mol 5.4

Mass A Mass B Calculations• In a mass-mass problem, the mass of one

substance in a reaction is given and you are asked to calculate the mass of one or more of the other substances involved in the reaction.

• Steps in solving mass-mass calculations:▫Write a balanced equation.▫Convert the mass of the given substance to moles.▫Use the mole ratio to calculate the moles of the

second substance needed to react with the given substance.

▫Convert the moles of the second substance to grams.

Mass A Mass B Calculations Example ProblemIf 30.0 g of lithium react with an excess of

water, how many grams of lithium hydroxide will be produced?

LiOH g 104 LiOH mol 1

LiOH g 23.95 x

Li mol 2

LiOH mol 2 x

Li g 6.94

Li mol 1 x Li g 0.30

Target Check

1. Sodium chloride is prepared by the reaction of sodium metal with chlorine gas. How many moles of sodium are required to produce 40.0 g of sodium chloride?

2Na + Cl2 2NaClNa mol 0.684

NaCl mol 2

Na mol 2 x

NaCl g 44.58

NaCl mol 1 x NaCl g 40

Target Check

2. How many grams of aluminum will be produced by the decomposition of 25.0 g of aluminum oxide?2Al2O3 2Al + 3O2

Al g 13.2 Al mol 1

Al g 26.98 x

OAl mol 2

Al mol 4 x

OAl g 101.96

OAl mol 1 x OAl g 0.25

3232

3232

Mass A Volume B Calculations•In mass-volume problems, either the

given or the unknown quantity is the volume of a gas at STP.

Mass A Volume B Calculations Example ProblemOxygen is prepared in the lab by the

decomposition of potassium chlorate.2KClO3 2KCl + 3O2

a. Calculate the volume of oxygen in liters, measured at STP that would be obtained from 183.9 g of potassium chlorate.

22

2

3

2

3

33 O L 50.42

O mol 1

O L 22.4 x

KClO mol 2

O mol 3 x

KClO g 122.55

KClO mol 1 x KClO g 9.183

Mass A Volume B Calculations Example ProblemOxygen is prepared in the lab by the

decomposition of potassium chlorate.2KClO3 2KCl + 3O2

b. Calculate the grams of potassium chlorate required to produce 2.80 L of oxygen at STP.

33

3

2

3

2

22 KClO g 10.2

KClO mol 1

KClO g 122.55 x

O mol 3

KClO mol 2 x

O L 22.4

O mol 1 x O L 80.2

Volume A Volume B Calculations•In volume-volume problems, both the

substance sought and the substance given are gases. Under the same conditions of temperature and pressure, the volume of reacting gases are proportional to the number of moles of the gases in the balanced equation.

Volume A Volume B Calculations Sample Problem2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g)

Calculate the volume of CO2 results from the combustion of 6.0 L of C2H6.

Shortcut:

22

2

62

2

62

6262 CO L 12

CO mol 1

CO L 22.4 x

HC mol 2

CO mol 4 x

HC L 4.22

HC mol 1 x HC L 0.6

262

262 CO L 12

HC L 2

CO L 4 x HC L 0.6

Target Check

1. Aluminum reacts with hydrochloric acid to produce aluminum chloride and hydrogen gas. If 150L of hydrogen gas is collected at STP, how many grams of aluminum were used?

2Al + 6HCl 2AlCl3 + 3H2Al g 120

Al mol 1

Al g 26.98 x

H mol 3

Al mol 2 x

H L 22.4

H mol 1 x H L 150

22

22

Target Check

2. Calculate the volume of water vapor resulting from the combustion of sufficient C2H6 to require 2.0 L of O2.

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g)

OH L 1.7 O L 7

OH L 6 x O L 0.2 2

2

22

Limiting Reagents

•The limiting reagent in a chemical reaction limits the amounts of the other reactants that can combine—and the amount of product that can form—in a chemical reaction.

•In other words, it is the reactant that runs out first.

•The reaction will stop once the limiting reagent is used up.

Excess Reagent

•The excess reagent in a chemical reaction is the substance that is not used up completely in a chemical reaction.

•In other words, it is the reactant that you have leftovers of.

Limiting Reagent Sample ProblemNitrogen gas and hydrogen gas react to

form ammonia. What will occur will 6.70 moles of N2 reacts with 3.20 moles of H2?

N2 + 3H2 2NH3

a. What is the limiting reagent? Step 1: Convert both of the reactants to the products.

32

32 NH mol 13.4

N mol 1

NH mol 2 x N mol 70.6

32

32 NH mol 2.13

H mol 3

NH mol 2 x H mol 20.3

Step 2: The reactant that produced the least amount of product ran out first, and thus, is the limiting reagent.

The hydrogen gas is the limiting reactant.b. What is the maximum number of moles

of NH3 that can be produced?

Based on the answer to part a, only 2.13 mol of NH3 can be produced.

Target Check

A student reacts 80.0 g of Copper metal with 25.0 g of Sulfur to form copper (I) sulfide.

2Cu + S Cu2S

a. What is the limiting reagent?

Cu is the limiting reagent!

SCu g 100. SCu mol 1

SCu g 159.17 x

Cu mol 2

SCu mol 1 x

Cu g 63.55

Cu mol 1Cu x g 0.80 2

2

22

SCu g 124 SCu mol 1

SCu g 159.17 x

S mol 1

SCu mol 1 x

S g 32.07

S mol 1 x S g 0.25 2

2

22

Target Check cont.

b. What is the maximum number of grams of copper(I) sulfide that could be produced?

Based on the answer from part a, only 100. g of Cu2S can be produced.

Percent Yield

When a chemical reaction is used to calculate the amount of product that will form during a reaction, then a value for the theoretical yield is obtained. This is the maximum amount of product that could be formed.

The amount of product that forms when the reaction is carried out in the laboratory is called the actual yield. It is often less than the theoretical yield.

Percent Yield

The percent yield is the ratio of the actual yield to the theoretical yield.

% yield = 100 x yield ltheoretica

yield actual

Percent Yield

There are many factors that cause percent yield to be less than 100%.

•Reactions do not always go to completion.•Impure reactants and competing side

reactions may cause other products to be formed.

•Some of the products may be lost during purification.

Percent Yield Example Problem

A student decomposes 24.8 g of calcium carbonate in the lab.

CaCO3 CaO + CO2

a. Calculate the theoretical yield of calcium oxide.

CaO g 13.9 CaO mol 1

CaO g 56.08 x

CaCO mol 1

CaO mol 1 x

CaCO g 09.100

CaCO mol 1 x CaCO g 8.24

33

33

Percent Yield Example Problem

b. What is the percent yield if the student only produced 13.1 g of calcium oxide in the laboratory?

94.2% 100 x g 13.9

g 13.1 yield %

Target Check

When 50 g of silicon dioxide is heated with an excess of carbon, 32.2 g of silicon carbide is produced.

SiO2(s) + 3C(s) SiC(s) + 2CO(g)

a. First, calculate the theoretical yield.

SiC g 33.4 SiC mol 1

SiC g 40.10 x

SiO mol 1

SiC mol 1 x

SiO g 09.60

SiO mol 1 x SiO g 0.50

22

22

Target Check

b. Second, calculate the percent yield.% yield = 96.4% 100 x

g 33.4

g 2.32