Chapter 5 Chemical Quantities and Reactions 1
Avogadro's Number Small particles such as atoms, molecules, and ions are counted using the mole. 1 mole = 6.02 x 10 23 items Avogadro s number 602 000 000 000 000 000 000 000 = 6.02 x 10 23 1 mole of an element = 6.02 x 10 23 atoms of that element 1 mole of carbon = 6.02 x 10 23 atoms of carbon 1 mole of sodium = 6.02 x 10 23 atoms of sodium 2
Number of Particles in One-Mole Samples 3
Avogadro's Number Avogadro s number, 6.02 x 10 23, can be written as an equality and as two conversion factors. Equality: 1 mole = 6.02 x 10 23 particles Conversion Factors: 6.02 x 10 23 particles and 1 mole 1 mole 6.02 x 10 23 particles 4
Guide to Calculating Atoms or Molecules 5
Converting Moles to Particles Avogadro s number is used to convert moles of a substance to particles. How many CO 2 molecules are in 0.50 mole of CO 2? Step 1 State the needed and given quantities: Given: 0.50 mole of CO 2 Needed: molecules of CO 2 6
Converting Moles to Particles Step 2 Write a plan to convert moles to atoms or molecules: Avogadro s number moles of CO 2 molecules of CO 2 Step 3 Use Avogadro s number to write conversion factors. 1 mole of CO 2 = 6.02 x 10 23 molecules of CO 2 6.02 x 10 23 CO 2 molecules and 1 mole CO 2 1 mole CO 2 6.02 x 10 23 CO 2 molecules 7
Converting Moles to Particles Step 4 Set up the problem to calculate the number of particles. 0.50 mole CO 2 x 6.02 x 10 23 molecules CO 2 1 mole CO 2 = 3.01 x 10 23 molecules of CO 2 8
Learning Check The number of atoms in 2.0 mole of Al atoms is: A. 2.0 Al atoms B. 3.0 x 10 23 Al atoms C. 1.2 x 10 24 Al atoms 9
Solution The number of atoms in 2.0 moles of Al atoms is: Step 1 State the needed and given quantities: Given: 2.0 mole Al Needed: atoms of Al Step 2 Write a plan to convert moles to atoms or molecules: Avogadro s number Moles of Al Atoms of Al Step 3 Use Avogadro s number to write conversion factors. 1 mole of Al = 6.02 x 10 23 atoms of Al 6.02 x 10 23 atoms Al and 1 mole A 1 mole Al 6.02 x 10 23 atoms Al 10
Solution The number of atoms in 2.0 moles of Al atoms is: Step 4 Set up the problem to calculate the number of particles. C. 2.0 moles Al x 6.02 x 10 23 Al atoms 1 mole Al = 1.2 x 10 24 Al atoms 11
Molar Mass from Periodic Table is the mass of 1 mole of an element is the atomic mass expressed in grams Molar mass is the atomic mass expressed in grams. 1 mole of Ag 1 mole of C 1 mole of S = 107.9 g of Ag = 12.0 g of C = 32.1 g of S 12
Guide to Calculating Molar Mass 13
Molar Mass of the Compound C 2 H 6 O To calculate the molar mass of C 2 H 6 O: Step 1 Obtain the molar mass of each element. 1 mole of C = 12.0 g of C 1 mole of H = 1.01 g of H 1 mole of O = 16.0 g of O Step 2 Multiply each molar mass by the number of moles (subscripts) in the formula. 2 moles C x 12.0 g C = 24.0 g of C 1 mole C 6 moles H x 1.01 g H = 6.06 g of H 1 mole H 1 mole O x 16.0 g O = 16.0 g of O 1 mole O Step 3 Calculate the molar mass by adding the masses of the elements. 2 moles of C = 24.0 g of C 6 moles of H = 6.06 g of H 1 mole of O = 16.0 g of O Molar mass of C 2 H 6 O = 46.1 g of C 2 H 6 O 14
One-Mole Quantities 15
Learning Check What is the molar mass of each compound? A. K 2 O B. Al(OH) 3 16
Solution What is the molar mass of each compound? Step 1 Obtain the molar mass of each element. A. K 2 O K = 39.1 g/mole O = 16.0 g/mole B. Al(OH) 3 Al = 27.0 g/mole O = 16.0 g/mole H = 1.01 g/mole 17
Solution What is the molar mass of each compound? Step 2 Multiply each molar mass by the number of moles (subscripts) in the formula. A. K 2 O 2 moles K x 39.1 g K = 78.2 g of K 1 mole K 1 mole O x 16.0 g O = 16.0 g of O 1 mole O 18
Solution What is the molar mass of each compound? Step 2 Multiply each molar mass by the number of moles (subscripts) in the formula. B. Al(OH) 3 1 mole Al x 27.0 g Al = 27.0 g of Al 1 mole Al 3 moles O x 16.0 g O = 48.0 g of O 1 mole O 3 moles H x 1.01 g H = 3.03 g of H 1 mole H 19
Solution What is the molar mass of each compound? Step 3 Calculate the molar mass by adding the masses of the elements. A. K 2 O 2 moles of K = 78.2 g of C 1 mole of O = 16.0 g of O Molar mass of K 2 O = 94.2 g of K 2 O B. Al(OH) 3 1 mole of Al = 27.0 g of Al 3 moles of O = 48.0 g of O 3 moles of H = 3.03 g of H Molar mass of Al(OH) 3 = 78.0 g of Al(OH) 3 20
Calculations Using Molar Mass Molar mass conversion factors are fractions (ratios) written from the molar mass. relate grams and moles of an element or compound. for methane, CH 4, used in gas stoves and gas heaters, is 1 mole of CH 4 = 16.0 g of CH 4 (molar mass equality) Conversion factors: 16.0 g CH 4 and 1 mole CH 4 1 mole CH 4 16.0 g CH 4 21
Guide to Calculating Moles from Mass 22
Converting Mass to Moles of Compound NaCl A box of table salt, NaCl, contains 737 g of NaCl. How many moles of NaCl are in the box? 23
Converting Mass to Moles of Compound NaCl A box of table salt, NaCl, contains 737 g of NaCl. How many moles of NaCl are in the box? Step 1 State the given and needed quantities. Given: 737 g of NaCl Need: moles of NaCl Step 2 Write a plan to convert grams to moles. molar mass Grams of NaCl Moles of NaCl 24
Converting Mass to Moles of Compound NaCl Step 3 Determine the molar mass and write conversion factors. Molar Mass 1 mole Na x 23.0 g Na = 23.0 g of Na 1 mole Na 1 mole Cl x 35.5 g Cl = 35.5 g of Cl 1 mole Cl 58.5 g of NaCl Conversion Factors 1 mole of NaCl = 58.5 g of NaCl 58.5 g NaCl and 1 mole NaCl 1 mole NaCl 58.5 g NaCl Step 4 Set up the problem to convert grams to moles. 737 g NaCl x 1 mole NaCl = 12.6 moles of NaCl 58.5 g NaCl 25
Map: Mass Moles Particles 26
Visible Evidence of a Chemical Reaction 27
Identifying a Balanced Equation In a balanced chemical equation, no atoms are lost or gained the number of reacting atoms is equal to the number of product atoms 28
Balancing a Chemical Equation 29
Balancing a Chemical Equation: Formation of Al 2 S 3 Step 1 Write an equation using the correct formulas of the reactants and products. Al(s) + S(s) Al 2 S 3 (s) Step 2 Count the atoms of each element in the reactants and products. Reactants Products 1 atom Al 2 atoms Al Not balanced 1 atoms S 3 atoms S Not balanced 30
Balancing a Chemical Equation: Formation of Al 2 S 3 Step 3 Use coefficients to balance each element. Starting with the most complex formula, change coefficients to balance equation. 2Al(s) + 3S(s) Al 2 S 3 (s) Step 4 Check the final equation to confirm it is balanced. Make sure coefficients are the lowest ratio. Reactants Products 2 atoms Al 2 atoms Al Balanced 3 atoms S 3 atoms S Balanced 31
Learning Check State the number of atoms of each element on the reactant side and the product side for each of the following balanced equations. A. P 4 (s) + 6Br 2 (l) 4PBr 3 (g) B. 2Al(s) + Fe 2 O 3 (s) 2Fe(s) + Al 2 O 3 (s) 32
Solution State the number of atoms of each element on the reactant side and the product side for each of the following balanced equations. A. P 4 (s) + 6Br 2 (l) 4PBr 3 (g) Reactants Products P atoms 4 4 Br atoms 12 12 B. 2Al(s) + Fe 2 O 3 (s) 2Fe(s) + Al 2 O 3 (s) Reactants Products Al atoms 2 2 Fe atoms 2 2 O atoms 3 3 33
Balancing Equations with Polyatomic Ions When balancing equations with polyatomic ions, balance each polyatomic ion as a unit. 2Na 3 PO 4 (aq) + 3MgCl 2 (aq) Mg 3 (PO 4 ) 2 (s) + 6NaCl(aq) Reactants Products PO 3 4 ions 2 2 Na + ions 6 6 Mg 2+ ions 3 3 Cl ions 6 6 34
Learning Check Balance and list the coefficients from reactants to products. A. Fe 2 O 3 (s) + C(s) Fe(s) + CO 2 (g) 1) 2, 3, 2, 3 2) 2, 3, 4, 3 3) 1, 1, 2, 3 B. Al(s) + FeO(s) Fe(s) + Al 2 O 3 (s) 1) 2, 3, 3, 1 2) 2, 1, 1, 1 3) 3, 3, 3, 1 C. Al(s) + H 2 SO 4 (aq) Al 2 (SO 4 ) 3 (aq) + H 2 (g) 1) 3, 2, 1, 2 2) 2, 3, 1, 3 3) 2, 3, 2, 3 35
Solution A. 2) 2, 3, 4, 3 2Fe 2 O 3 (s) + 3C(s) B. 1) 2, 3, 3, 1 2Al(s) + 3FeO(s) 4Fe(s) + 3CO 2 (g) 3Fe(s) + 1Al 2 O 3 (s) C. 2) 2, 3, 1, 3 2Al(s) + 3H 2 SO 4 (aq) 1Al 2 (SO 4 ) 3 (aq) + 3H 2 (g) 36
Types of Reactions Chemical reactions can be classified as combination reactions decomposition reactions single replacement reactions double replacement reactions combustion reactions 37
Combination Reactions In a combination reaction, two or more elements form one product or simple compounds combine to form one product 2Mg(s) + O 2 (g) 2Na(s) + Cl 2 (g) SO 3 (g) + H 2 O(l) 2MgO(s) 2NaCl(s) H 2 SO 4 (aq) 38
Decomposition Reaction In a decomposition reaction, one substance splits into two or more simpler substances. 2HgO(s) 2KClO 3 (s) 2Hg(l) + O 2 (g) 2KCl(s) + 3O 2 (g) 39
Single Replacement Reaction In a single replacement reaction, one element takes the place of a different element in another reacting a compound. Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) Fe(s) + CuSO 4 (aq) FeSO 4 (aq) + Cu(s) 40
Double Replacement Reaction In a double replacement, two elements in the reactants exchange places. AgNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) ZnS(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 S(g) Boy 1/Girl 1 + Boy 2/Girl 2 Boy 1/Girl 2 + Boy 2/Girl 1 41
Combustion Reaction In a combustion reaction, a carbon-containing compound burns in oxygen gas to form carbon dioxide (CO 2 ) and water (H 2 O) energy is released as a product in the form of heat CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) + energy 42
Summary of Reaction Types 43
Learning Check Identify each reaction as combination, decomposition, combustion, single replacement, or double replacement. A. 3Ba(s) + N 2 (g) Ba 3 N 2 (s) B. 2Ag(s) + H 2 S(aq) Ag 2 S(s) + H 2 (g) C. 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(g) D. PbCl 2 (aq) + K 2 SO 4 (aq) 2KCl(aq) + PbSO 4 (s) E. K 2 CO 3 (s) K 2 O(aq) + CO 2 (g) 44
Solution Identify each reaction as combination, decomposition, combustion, single replacement, or double replacement. A. 3Ba(s) + N 2 (g) Ba 3 N 2 (s) Combination B. 2Ag(s) + H 2 S(aq) Ag 2 S(s) + H 2 (g) C. C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(g) Single Replacement Combustion 45
Solution Identify each reaction as combination, decomposition, combustion, single replacement, or double replacement. D. PbCl 2 (aq) + K 2 SO 4 (aq) 2KCl(aq) + PbSO 4 (s) Double Replacement E. K 2 CO 3 (s) K 2 O(aq) + CO 2 (g) Decomposition 46
Oxidation Reduction Reactions An oxidation reduction reaction provides us with energy from food provides electrical energy in batteries occurs when iron rusts: 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) 47
Oxidation Reduction In an oxidation reduction reaction, electrons are transferred from one substance to another. 48
Oxidation and Reduction 2Cu(s) + O 2 (g) 2CuO LEO says GER Loss of Electrons is Oxidation. 2Cu(s) 2Cu 2+ (s) + 4e Gain of Electrons is Reduction. O 2 (g) + 4e 2O 2 (s) The green patina on copper is due to oxidation 49
Oxidation Reduction in Biological Systems The oxidation of a typical biochemical molecule can involve the transfer of two hydrogen atoms (or 2H + and 2e ) to a proton acceptor such as the coenzyme FAD (flavin adenine dinucleotide). 50
Characteristics of Oxidation and Reduction 51
Characteristics of Oxidation and Reduction Methyl alcohol hydrogen gas) formaldehyde Oxidation (loss of H as CH 3 OH H 2 CO + 2H 2 Formaldehyde formic acid 2H 2 CO + O 2 2H 2 CO 2 Oxidation (addition of O) 52
Law of Conservation of Mass The law of conservation of mass indicates that in an ordinary chemical reaction, matter cannot be created or destroyed no change in total mass occurs the mass of products is equal to mass of reactants 53
Conservation of Mass 54
Information from a Balanced Equation 55
Reading Equations in Moles Consider the following equation: 2Fe(s) + 3S(s) Fe 2 S 3 (s) This equation can be read in moles by placing the word moles of between each coefficient and formula. 2 moles of Fe + 3 moles of S 1 mole of Fe 2 S 3 56
Learning Check Consider the following equation: 3H 2 (g) + N 2 (g) 2NH 3 (g) A. A mole mole factor for H 2 and N 2 is: 1) 3 moles N 2 2) 1 mole N 2 3) 1 mole N 2 1 mole H 2 3 moles H 2 2 moles H 2 B. A mole mole factor for NH 3 and H 2 is: 1) 1 mole H 2 2) 2 moles NH 3 3) 3 moles N 2 2 moles NH 3 3 moles H 2 2 moles NH 3 57
Solution 3H 2 (g) + N 2 (g) 2NH 3 (g) A. A mole mole factor for H 2 and N 2 is: 2) 1 mole N 2 3 moles H 2 B. A mole mole factor for NH 3 and H 2 is: 2) 2 moles NH 3 3 moles H 2 58
Calculations with Mole Factors How many moles of Fe 2 O 3 can form from 6.0 moles of O 2? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Relationship: 3 moles of O 2 = 2 moles of Fe 2 O 3 Use a mole mole factor to determine the moles of Fe 2 O 3. 6.0 moles O 2 x 2 moles Fe 2 O 3 = 4.0 moles of Fe 2 O 3 3 moles O 2 59
Guide to Using Mole Mole Factors 60
Learning Check How many moles of Fe are needed for the reaction of 12.0 moles of O 2? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) A. 3.00 moles of Fe B. 9.00 moles of Fe C. 16.0 moles of Fe 61
Solution In a problem, identify the compounds given and needed. How many moles of Fe are needed for the reaction of 12.0 moles of O 2? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Step 1 State the given and needed quantities. Given: 12.0 moles of O 2 The possible mole factors for the solution are: 4 moles Fe and 3 moles O 2 3 moles O 2 4 moles Fe Needed:? moles of Fe 62
Solution Step 2 Write a plan to convert the given to the needed moles. Mole mole factor Moles of O 2 Moles of Fe Step 3 Use coefficients to write relationships and mole mole factors. 4 moles of Fe = 3 moles of O 2 4 moles Fe and 3 moles O 2 3 moles O 2 4 moles Fe 63
Solution Step 4 Set up the problem using the mole mole factor that cancels given moles. 12.0 moles O 2 x 4 moles Fe = 16.0 moles of Fe 3 moles O 2 The answer is C, 16.0 moles of Fe. 64
Mass Calculations in Equations 65
Moles to Grams Suppose we want to determine the mass (g) of NH 3 that can be produced from 32 grams of N 2. N 2 (g) + 3H 2 (g) 2NH 3 (g) Step 1 Use molar mass to convert grams of given to moles. 1 mole of N 2 = 28.0 g of N 2 1 mole N 2 and 28.0 g N 2 28.0 g N 2 1 mole N 2 32 g N 2 x 1 mole N 2 = 1.1 mole of N 2 28.0 g N 2 66
Moles to Grams Step 2 Write a mole mole factor from the coefficients in the equation. 1 mole of N 2 = 2 moles of NH 3 1 mole N 2 and 2 moles NH 3 2 moles NH 3 1 mole N 2 Step 3 Convert moles of given to moles of needed using the mole mole factor. 1.1 mole N 2 x 2 moles NH 3 = 2.2 moles of NH 3 1 mole N 2 67
Moles to Grams Step 3 Convert moles of given to moles of needed using the mole mole factor. 1.1 mole N 2 x 2 moles NH 3 = 2.2 moles of NH 3 1 mole N 2 68
Moles to Grams Step 4 Convert moles of needed substance to grams using molar mass. 1 mole of NH 3 = 17.0 g of NH 3 1 mole NH 3 and 17.0 g NH 3 17.0 g NH 3 1 mole NH 3 2.2 moles NH 3 x 17.0 g NH 3 = 37 g of NH 3 1 mole NH 3 69
Learning Check How many grams of O 2 are needed to produce 45.8 grams of Fe 2 O 3 in the following reaction? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) A. 38.4 g of O 2 B. 13.8 g of O 2 C. 1.38 g of O 2 70
Solution How many grams of O 2 are needed to produce 45.8 grams of Fe 2 O 3 in the following reaction? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Step 1 Use molar mass to convert grams of given to moles. 1 mole of Fe 2 O 3 = 159.7 g of Fe 2 O 3 1 mole Fe 2 O 3 and 159.7 g Fe 2 O 3 159.7 g Fe 2 O 3 1 mole Fe 2 O 3 45.8 g Fe 2 O 3 x 1 mole Fe 2 O 3 = 0.287 mole of Fe 2 O 3 159.7 g Fe 2 O 3 71
Solution Step 2 Write a mole mole factor from the coefficients in the equation. 3 moles of O 2 = 2 moles of Fe 2 O 3 3 moles O 2 and 2 moles Fe 2 O 3 2 moles Fe 2 O 3 3 moles O 2 Step 3 Convert moles of given to moles of needed using the mole mole factor. 0.287 mole Fe 2 O 3 x 3 moles O 2 = 0.430 mole of O 2 2 moles Fe 2 O 3 72
Solution Step 4 Convert moles of needed substance to grams using molar mass. 1 mole of O 2 = 32.0 g of O 2 1 mole O 2 and 32.0 g O 2 32.0 go 2 1 mole O 2 0.430 mole O 2 x 32.0 g O 2 = 13.8 g of O 2 1 mole O 2 The answer is B, 13.8 g of O 2. 73
Molecules Must Collide for Reaction Three conditions for a reaction to occur are: 1. collision: The reactants must collide. 2. orientation: The reactants must align properly to break and form bonds. 3. energy: The collision must provide the energy of activation. 74
Exothermic Reaction In an exothermic reaction, heat is released the energy of the products is less than the energy of the reactants heat is a product C(s) + 2H 2 (g) CH 4 (g) + 18 kcal 75
Endothermic Reactions In an endothermic reaction, heat is absorbed the energy of the products is greater than the energy of the reactants heat is a reactant (added) N 2 (g) + O 2 (g) + 43.3 kcal 2NO(g) 76
Reaction Rate The reaction rate is the speed at which reactant is used up is the speed at which product forms increases when temperature rises because reacting molecules move faster, providing more colliding molecules with energy of activation increases with increase in concentration of reactants 77
Catalyst A catalyst increases the rate of a reaction lowers the energy of activation is not used up during the reaction
Factors That Increase Reaction Rate 79
Learning Check State the effect of each on the rate of reaction as increases, decreases, or has no effect: A. increasing the temperature B. removing some of the reactants C. adding a catalyst D. placing the reaction flask in ice E. increasing the concentration of one of the reactants 80
Solution State the effect of each on the rate of reaction as increases, decreases, or has no effect: A. increasing the temperature increases B. removing some of the reactants decreases C. adding a catalyst increases D. placing the reaction flask in ice decreases E. increasing the concentration of increases one of the reactant 81
Learning Check Indicate the effect of each factor listed on the rate of the following reaction as increases, decreases, or no change. 2CO(g) + O 2 (g) 2CO 2 (g) A. raising the temperature B. adding O 2 C. adding a catalyst D. lowering the temperature 82
Solution Indicate the effect of each factor listed on the rate of the following reaction as increases, decreases, or no change. 2CO(g) + O 2 (g) 2CO 2 (g) A. raising the temperature increases B. adding O 2 increases C. adding a catalyst increases D. lowering the temperature decreases 83
Chemical Quantities and Reactions 84