Chapter 3
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Chapter 3. Stoichiometry. Stoichiometry. The study of quantities of materials consumed and produced in chemical reactions. 3.1 Chemical equations. The '+' is read as 'reacts with' and the arrow '' means 'produces'.

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Chapter 3

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Chapter 3



  • The study of quantities of materials consumed and produced in chemical reactions.

3.1 Chemical equations

  • The '+' is read as 'reacts with' and the arrow '' means 'produces'

Because atoms are neither created nor destroyed in a reaction, a chemical equation must have an equal number of atoms of each element on each side of the arrow (i.e. the equation is said to be 'balanced').

Balancing Equations

  • Write 'un-balanced' equation using formulas of reactants and products

  • Write 'balanced' equation by determining coefficients that provide equal numbers of each type of atom on each side of the equation (generally, whole number values)

  • Note! Subscripts should never be changed when trying to balance a chemical equation. Changing a subscript changes the actual identity of a product or reactant. Balancing a chemical equation only involves changing the relative amounts of each product or reactant.


We seem to be o.k. with our number of carbon atoms in both the reactants and products, but we have only half the hydrogen in our products as in our reactants. We can fix this by doubling the relative number of water molecules in the list of products:

Note that while this has balanced our carbon and hydrogen atoms, we now have 4 oxygen atoms in our products, and only have 2 in our reactants. We can balance our oxygen atoms by doubling the number of oxygen atoms in our reactants:

  • The physical state of each chemical can be indicated by using the symbols (g), (l), (s), and (aq) (for gas, liquid, solid, and aqueous respectively):

    Na (s) + H2O (l)  NaOH (aq) + H2(g)

Balance the following equations

C2H5OH (aq) + O2 (g)  CO2 (g) + H2O (g)

Fe (s) + O2 (g)  FeO2 (s)

C2H4 (g)+ O2 (g)  CO2 (g) + H2O (l)


2Na (s) + 2H2O (l)  2NaOH (aq) + H2(g)

Fe (s) + O2 (g)  FeO2 (s)

C2H4 (g)+ 3O2 (g)  2CO2 (g) + 2H2O (l)

Which box represents the reaction between NO and O2 to produce NO2.


Balancing / Writing Rxn wks

Classifying reactions movie

Chemical Reactivity

Combination/Synthesis Reaction:

2 or more substances react to form one new product

A + B  C

+ 

solid magnesium and oxygen gas react to produce solid magnesium oxide

2Mg (s) + O2(g)  2MgO (s)

Metal nonmetal ionic compound Diatomic

2+ 2-

Decomposition Rxn

  • One substance undergoes a reaction to produce two or more substances.

  • Typically occurs when things are heated.

    AX  A + X

     +

Solid calcium carbonate reacts to produce solid calcium oxide and carbon dioxide gas

CaCO3 (s)  CaO (s) + CO2 (g)

2+ (2-) 2+ 2- 4+ 2(2-)

Single displacement

  • One element replaces a similar element in a compound

    A + BX  AX + B

    BX + Y  BY + X

    +  +

Solid copper is dissolved in aqueous silver nitrate to produce solid silver and aqueous copper II nitrate.

Cu(s) + AgNO3 (aq) Ag(s) + Cu(NO3)2 (aq)

Write the sentence for this reaction:

Fe (s) + Cu(NO3)2 (aq)Fe(NO3)2 (aq)+ Cu (s)

Activity Series

  • We need to know what metals are most likely to oxidize others.

  • Example: We can’t store nickel nitrate in an iron container because the solution would eat through the container.

Activity Series

  • A list of metals arranged in order of decreasing ease of oxidation.

  • Page 139 table

Using activity series

  • Any metal on the list can be oxidized by the metal below it.

  • Give: FeCl2 + Mg

  • Find: will iron oxidize

    Magnesium metal?

  • I finger on Fe

  • 1 finger on Mg

  • Is the bound chemical below

  • Yes Fe is below Mg.

  • Then complete the reaction

  • Give: NaCl + Mg

  • Find: will sodium oxidize

    Magnesium metal?

  • I finger on Na

  • 1 finger on Mg

  • Is the bound chemical below

  • no

  • Then the reaction is not possible

What if you don’t have an Activity Series table?

  • Down Group 1 (I) the "Alkali Metals" the activity increases Cs > Rb > K > Na > Li

  • Down Group 2 (II) the activity increases e.g. Ca > Mg

  • On the same period, the Group 1 metal is more reactive than the group 2 metal

  • the group 2 metal is more reactive than the Group 3 metal,

  • All three are more reactive than the "Transition Metals". e.g. Na > Mg > Al (on Period 3) and K > Ca > Ga > Fe/Cu/Zn etc. (on Period 4)

Double Replacement Rxn/ Metathesis

  • The ions of two compounds exchange places in an aqueous solution to form two new compounds.

    AX + BY  AY + BX

  • One of the compounds formed is usually a

    precipitate, an insoluble gas that bubbles out of solution, or a molecular compound, usually water.

Double Replacement Rxn/ Metathesis

AX + BY  AY + BX

+  +

Write the sentence for these double replacement reactions

KOH (aq) + H2SO4 (aq) K2SO4 (aq) + H2O (l)

FeS (aq) + HCl (aq) FeCl2 (aq) + H2S (aq)

Combustion Reaction

A substance combines with oxygen, releasing a large amount of energy in the form of light and heat.

C3H8 (g)+5O2 (g)  3CO2 (g) + H2O (g)

Usually CO2 (carbon dioxide) / CO (carbon monoxide) and water are produced.

  • Reactive elements combine with oxygen

    P4(s) + 5O2(g)  P4O10 (s)

    (This is also a synthesis reaction)

  • The burning of natural gas, wood, gasoline

    C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)


  • Classifying types of Rxns worksheet

3.3 Formula Weights

  • Although we can’t “count atoms” in a molecule directly, we can count them indirectly if we know their masses.

Formula Weights/ Molecular Weight

  • Sum of atomic masses of each atom in a molecule.

    F.W of H2SO4 = 2(H) + S + 4(O)

    2(1) + 32 + 4(16) = 98amu

    98 g/mol

We can describe composition in two ways1. number of atoms (amu)

2. % (by mass) of its elements.

Percent Composition

Percent Composition

We can find % mass of an atom in a compound from formula mass, by comparing each element present in 1 mole of compound to the total mass of 1 mole of compound

Percent Composition Equation

% element = # of atoms element (atomic weight of the element ) * 100

Formula Weight


  • Calculate the percentage of nitrogen in Ca(NO3)2


% N = # N atoms (m.w N) X 100

m. w Ca(NO3)2


% N = 2(14.02 N amu) X 100

164.12 Ca(NO3)2amu

= 17%


  • Calculate the percent composition of each element in C12H22O11


Molar mass wks

Percent composition wks

3.4 The Mole!!!!!

  • The unit for dealing with, atoms, molecules, ions

  • Abbreviation = mol

    (oh the time you will save!)


  • Avogadro

  • Italian

  • 1776-1856

Molar Mass

  • A dozen eggs = 12

  • A dozen elephants = 12

  • But 12 eggs has a different weight than 12 elephants

  • Thus 1 mole of carbon is 6.02 x 1023 molecules but weighs 12 grams

  • 1 mole of sodium is 6.02 x 1023 molecules but weighs 23 g

  • If you had Avogadro's number of unpopped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles.

If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.

  • An Avogadro's number of standard soft drink cans would cover the surface of the earth to a depth of over 200 miles.

  • Molar mass of any substance is equal to its mass in atomic mass units (amu)






How many moles of water are in 5.380g of water?

  • Molar mass of water = 2(1) + 1(16) = 18g/mol

    5.380g H2O x 1 mol = 0.2989 moles H2O

    18g H2O

Using Moles in calculations

How many oxygen atoms are present in 4.20 grams NaHCO3?

4.20 g NaHCO3 x(1mole NaHCO3) x (6.02e 23molec)3 Oxygen atoms

84 g NaHCO3 1 mol 1 molec NaHCO3

= 9.03 x 10 22 atoms of Oxygen in 4.20 grams NaHCO3

Using Moles in calculations

  • How many nitrogen atoms are in 0.25 mol of Ca(NO3)2


0.25mol x 6.02 x 1023 molec Ca(NO3)2 x _2N atoms___

1 mol 1 molec Ca(NO3)2

= 3.0 x 10 23 moles


  • Molar mass worksheet

  • G-mol-molec atoms wks

3.5 Determining empirical formula from mass percent

  • Recall: Empirical formula: simplest whole # ratio of atoms in a compound.

  • Recall: We can find % mass from formula mass, by comparing each element present in 1 mole of compound to the total mass of 1 mole of compound


Example: Vitamin C is composed of 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula?

How to attack the problem

1. Convert mass % into grams (assume 100g ie: 40.92% = 40.92 g)

2. Convert grams to moles using molar mass.

3. Divide moles of each element by the smallest number of moles present. You may round to nearest whole # . (This establishes a ratio of comparison)


40.92 g C 1mol C= 3.4 moles C /3.4 = 1 C

12 g C

54.40 g O 1mol O = 3.4 moles O / 3.4 = 1 O

16g O

4.58 g H 1mol H = 4.58 moles H/ 3.4 = 1.3 = 1 H

1 g H

  • CHO

Empirical formula

Mass % elements



Calculate mole ratio

Assume 100g of sample

Grams of each element

Moles of each element

Use atomic weights


Compound X is composed of 55.3% K, 14.6% P , and 30.1% O.

What is the empirical formula of compound X?

55.3 g K 1mol K= 1.4 moles K /.47 = 3 K

39 g K

14.6 g P 1mol P = 0.47 moles P/ .47 = 1 P

31 g P

30.1 g O 1mol O = 1.9 moles O / .47 = 4 O

16g O

  • K3PO4

Determine Molecular formula from Empirical Formula


Molecular formula: the exact formula of a molecules, giving types of atoms and the number of each type.

1. Using mass % and molar mass, determine mass of each element in 1 mole of compound (same)

2. Determine number of moles of each element in 1 mole of compound. (same)

3. The integers from the previous step represent the subscripts in the molecular formula ( you just don’t divide by smallest mole ratio)

Let’s look back at our work

40.92 g C1mol C= 3.4 moles C /3.4 = 1 C

12 g C

54.40 g O1mol O = 3.4 moles O / 3.4 = 1 O

16g O

4.48 g H1mol H = 4.48 moles H/ 3.4 = 1 H

1 g H

C3H4O3= molecular formula


n = Molecular Formula Weight

Empirical Formula Weight

(where n = # of atoms)

Try it …..

The molecular weight of butyric acid is

88 amu. If the empirical formula is C2H4O. What is the molecular formula?

  • The empirical formula was given, us it to find the empirical formula weight.

  • C2H4O = 12 + 12+ 1+1+1+1+16 = 44 amu

  • 2. The molecular formula weight was given (88amu) plug it into the short cut formula.

  • n = 88 amu = 2

  • 44

  • 3. Apply the new number of atoms.

  • Molecular formula = (empirical) n

  • (C2H4O)2 = Molecular Formula = C4H8O2


  • Chan pg 108 43, 44, 45, 49, 50, 52, 53

  • BLPg 106-107: 37a, 39, 42, 44, 45, 46

3.6 Qualitative information from balanced equations.

  • Stoichiometry: mixing exactly enough chemical so that all is used

    Mass-Mass problems a new highway!!!!!!

    g given  mol given  mol required  g required

    Think: (grams to moles to moles to grams)

Time out for mole ratios

2H2 (g) + O2 (g)  2H2O (l)

Coefficients tell us that 2 molecules of H2 react with each molecule of O2 to form 2 molecules of H2O.

(recall: 6.02 x 1023 molecule = 1 mol)

These coefficients can be used to convert between quantities of reactants and products.

Example of mole ratios

  • Calculate the number of moles of H2O produced from 1.57 mol of O2?

    2H2 (g) + O2 (g)  2H2O (l)

2H2 (g) + O2 (g)  2H2O (l)

1.57 mol O2 ( 2 mol H2O) =

1 mol O2

= 3.14 mole H2O

Silicon carbide is made by heating silicon dioxide to high temperatures.

SiO2 (s) + 3C (s)  SiC(s) + 2CO (g)

How many grams of CO are formed by complete rxn of 5.00 g SiO2?

HINT: always make sure your equation is balanced first or mole ratios will be wrong.

Grams substance A

Grams of substance B



Use molar mass of B

Use molar mass of A

Moles of substance A

Moles of substance B

Use coefficients of A and B from balanced equation

Every line of dimensional analysis should have a unit AND a chemical formula!!!!!

Given: 5.00 g SiO2

Find : CO g

SiO2 (s) + 3C (s)  SiC(s) + 2CO (g)

think: grams to moles to moles to grams

5.00 g SiO21mol SiO22 mol CO28 g CO = 4.67 g CO

60 g SiO21 mol SiO21 mol CO

Mole ratio

How many moles of sulfuric acid would be needed to produce 4.80 moles of molecular iodine (I2) according to the following balanced equation.

10HI + 2KMnO4 + 3H2SO4 5I2 + 2MnSO4 + K2SO4 + 8H2O

4.80 mol I23 mol H2SO4= 2.88 mol H2SO4

5 mol I2


  • Chan pg: pg 109 46, 66,67,68,69

  • BLPg: 107 #’s 51, 53, 55, 57, 59, 60

    Note: you must be able to pass a pop quiz with 80% accuracy to be able to move on to the next section. If not additional problems and help will be mandatory.

3.7 Limiting Reactant

The number of products that can form is limited by the amount of reactant present.

The limiting reactant is the one that gives the least amount of product.

Reactants  Products

  • Lets say you want to make a sandwich using 2 slices of bread (Bd), 1 slice of ham (Hm).

    2Bd + 1Hm  Bd2Hm

If you have 10 slices of bread and 7 slices of ham, how many sandwiches will you be able to make according to the previous reaction?

2Bd + 1Hm  Bd2Hm

What is your limiting reagent?

Extra Tasty Ham Slabs

What is the limiting reactant, Blue or Red?

Attacking limiting reactant problems

  • Using the grams of each reactant and their mole ratios calculate how many grams of product will be formed.

  • Compare the amount of product formed by each reactant.

  • The reactant that give the lesser amount of product is the limiting reactant.

Try One….

When a mixture of silver and sulfur is heated, silver sulfide is formed:

16 Ag (s) + S8 (s)  8 Ag2S (s)

What mass of Ag2S is produced from a mixture of 2.0 g of Ag and 2.0 g of S8?

2 g Ag 1 mol Ag 8 mol Ag2S 248 g Ag2S

108 g Ag 16 mol Ag 1 mol Ag2S

= 2.3g Ag2S can be formed from 2 g Ag

2 g S81 mol S88 mol Ag2S 248 g Ag2S

256 g S8 1 mol S8 1 mol Ag2S

= 15.5 g Ag2S can be formed from 2.0 g S8

= 15.5 g Ag2S can be formed from 2.0 g S8

Most amount of product can be formed (aka: ham slabs!!!)

= 2.3g Ag2S can be formed from 2 g Ag

Least amount of product can be formed Thus Ag is the limiting reactant!!!! (aka: bread)

Try one again…. (man these are fun!!!)

How many grams of urea can be produced from 10.0 g of NH3 and 10.0 g of CO2?

2NH3 + CO2 CH4N2O + H2O



17.6 g of Urea from 10g NH3

13.6 g of urea from 10 g CO2 LIMITNG REACANT

Theoretical Yield

  • The amount of product calculated based on the limiting reactant.

    % yield = collected yield X 100

    Predicted yield

    Actual/collected = what you really get

    Theoretical/Predicted = What you might get or predict you will get.

Try one!

In a chemical reaction the theoretical yield is 145 g. If the percent yield was 92.0%, what was the actual yield?


92.0 = actual x 100


= 133 g

Calculating Theoretical Yield

When you are doing a limiting reactant equation,

(g – mol-mol-g) you are calculating the theoretical yield. It is not until you actually run the experiment in the lab that you will get your actual yield.

Some times not all the reactants react, or they may react in a way different than you desired. Chemistry is not perfect ( unlike you guys)

Homework 3.7

Chang: pg 107-108 # 63, 67, 69, 70, 77, 78

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