Calculations and the Chemical Equation Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 4 Calculations and the Chemical Equation Denniston Topping Caret 6th Edition

4.1 The Mole Concept and Atoms Atoms are exceedingly small Unit of measurement for mass of an atom is atomic mass unit (amu) – unit of measure for the mass of atoms carbon-12 assigned the mass of exactly 12 amu 1 amu = 1.66 x 10-24 g Periodic table gives atomic weights in amu

Mass of Atoms What is the atomic weight of one atom of fluorine? Answer: 19.00 amu What would be the mass of this one atom in grams? Chemists usually work with much larger quantities It is more convenient to work with grams than amu when using larger quantities

The Mole and Avogadro’s Number A practical unit for defining a collection of atoms is the mole 1 mole of atoms = 6.022 x 1023 atoms This is called Avogadro’s number This has provided the basis for the concept of the mole

The Mole To make this connection we must define the mole as a counting unit The mole is abbreviated mol A mole is simply a unit that defines an amount of something Dozen defines 12 Gross defines 144

Atomic Mass The atomic mass of one atom of an element corresponds to: The average mass of a single atom in amu The mass of a mole of atoms in grams 1 atom of F is 19.00 amu 19.00 amu/atom F 1 mole of F is 19.00 g 19.00 g/mole F

Molar Mass Molar mass - The mass in grams of 1 mole of atoms What is the molar mass of carbon? 12.01 g/mol C This means counting out a mole of Carbon atoms (i.e., 6.022 x 1023) they would have a mass of 12.01 g One mole of any element contains the same number of atoms, 6.022 x 1023, Avogadro’s number

Calculating Atoms, Moles, and Mass Atomic mass unit converts amu to grams (other than give you an appreciation for the size of atoms, we won’t work with amu) Avogadro’s number converts moles to number of atoms Molar mass converts grams to moles

Strategy for Calculations Map out a pattern for the required conversion Example: given a number of grams of sulfur, find the number of atoms Two conversions are required Convert grams to moles 1 mol S/32.06 g S OR 32.06 g S/1 mol S Convert moles to atoms mol S x (6.022 x 1023 atoms S) / 1 mol S

Practice Calculations-assume 3 sig. figs for all. Calculate the number of atoms in 1.7 moles of boron. Find the mass in grams of 2.5 mol Na (sodium). Calculate the number of atoms in 5.0 g aluminum. Calculate the mass of 5,000,000 atoms of Au (gold)

4.2 The Chemical Formula, Formula Weight, and Molar Mass Chemical formula - a combination of symbols of the various elements that make up the compound Formula unit - the smallest collection of atoms that provide two important pieces of information The identity of the atoms The relative number of each type of atom

Chemical Formula Consider the following formulas: H2 – 2 atoms of hydrogen are chemically bonded forming diatomic hydrogen, subscript 2 H2O – 2 atoms of hydrogen and 1 atom of oxygen, lack of subscript means one atom NaCl – 1 atom each of sodium and chlorine Ca(OH)2 – 1 atom of calcium and 2 atoms each of oxygen and hydrogen, subscript outside parentheses applies to all atoms inside

Chemical Formula Consider the following formulas: (NH4)2SO4 – 2 ammonium ions and 1 sulfate ion Ammonium ion contains 1 nitrogen and 4 hydrogen Sulfate ion contains 1 sulfur and 4 oxygen Compound contains 2 N, 8 H, 1 S, and 4 O CuSO4.5H2O This is an example of a hydrate - compounds containing one or more water molecules as an integral part of their structure 5 units of water with 1 CuSO4

Comparison of Hydrated and Anhydrous Copper Sulfate Hydrated copper (II) sulfate Anhydrous copper (II) sulfate Marked color difference illustrates the fact that these are different compounds

Formula Weight and Molar Mass Formula weight - the sum of the atomic weights of all atoms in the compound as represented by its correct formula What is the formula weight of H2O? 16.00 g/mol + 2(1.008 g/mol) = 18.02 g/mol Molar mass – mass of a mole of compound in grams / mole What is the molar mass of H2O? 18.02 g/mol H2O

Formula Unit Formula unit – smallest collection of atoms from which the formula of a compound can be established When calculating the formula weight (or molar mass) of an ionic compound, the smallest unit of the crystal is used What is the molar mass of (NH4)3PO4? 3(N g/mol) + 12(H g/mol) + 1(P g/mol) + 4(O g/mol)= 3(14.01) + 12(1.008) + 30.97 + 4(16.00)= 149.10 g/mol (NH4)3PO4

4.3 The Chemical Equation and the Information It Conveys A Recipe For Chemical Change Chemical equation - shorthand notation of a chemical reaction Describes all of the substances that react and all the products that form, physical states, and experimental conditions Reactants – (starting materials) – the substances that undergo change in the reaction Products – substances produced by the reaction

Features of a Chemical Equation Identity of products and reactants must be specified using chemical symbols Reactants are written to the left of the reaction arrow and products are written to the right Physical states of reactants and products may be shown in parentheses Symbol  over the reaction arrow means that energy is necessary for the reaction to occur Equation must be balanced

Features of a Chemical Equation Products and reactants must be specified using chemical symbols Reactants – written on the left of arrow Products – written on the right  – energy is needed Physical states are shown in parentheses

The Experimental Basis of a Chemical Equation We know that a chemical equation represents a chemical change One or more substances changed into new substances Different chemical and physical properties

Evidence of a Reaction Occurring The following can be visual evidence of a reaction: Release of a gas CO2 is released when acid is placed in a solution containing CO32- ions Formation of a solid (precipitate) A solution containing Ag+ ions mixed with a solution containing Cl- ions Heat is produced or absorbed Acid and base are mixed together Color changes

Subtle Indications of a Reaction Heat or light is absorbed or emitted Changes in the way the substances behave in an electrical or magnetic field before and after a reaction Changes in electrical properties

Writing Chemical Reactions We will learn to identify the following patterns of chemical reactions: combination decomposition single-replacement double-replacement Recognizing the pattern will help you write and understand reactions

Combination Reactions The joining of two or more elements or compounds, producing a product of different composition A + B  AB Examples: 2Na(s) + Cl2(g)  2NaCl(s) MgO(s) + CO2(g)  MgCO3(s)

Types of Combination Reactions Combination of a metal and a nonmetal to form a salt Combination of hydrogen and chlorine molecules to produce hydrogen chloride Formation of water from hydrogen and oxygen molecules Reaction of magnesium oxide and carbon dioxide to produce magnesium carbonate

Decomposition Reactions Produce two or more products from a single reactant Reverse of a combination reaction AB  A + B Examples: 2HgO(s)  2Hg(l) + O2(g) CaCO3(s)  CaO(s) + CO2(g)

Types of Decomposition Reactions Heating calcium carbonate to produce calcium oxide and carbon dioxide Removal of water from a hydrated material

Replacement Reactions Single-replacement One atom replaces another in the compound producing a new compound Examples: Cu(s)+2AgNO3(aq)  2Ag(s)+Cu(NO3)2(aq) 2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g) A + BC  B + AC

Types of Replacement Reactions Replacement of copper by zinc in copper sulfate Replacement of aluminum by sodium in aluminum nitrate

Replacement Reactions Double-replacement Two compounds undergo a “change of partners” Two compounds react by exchanging atoms to produce two new compounds AB + CD  AD + CB

Example of Double-Replacement Formation of solid lead chloride from lead nitrate and sodium chloride Pb(NO3)2(aq) + 2NaCl(aq)  PbCl2(s) + 2NaNO3(aq) AB + CD  AD + CB

Types of Chemical Reactions Precipitation Reactions Chemical change in a solution that results in one or more insoluble products To predict if a precipitation reaction can occur it is helpful to know the solubilities of ionic compounds

Solubilities of Some Common Ionic Compounds

Predicting Whether Precipitation Will Occur Recombine the ionic compounds to have them exchange partners Examine the new compounds formed and determine if any are insoluble according to the rules in Table 4.1 Any insoluble salt will be the precipitate Pb(NO3)2(aq) + NaCl(aq)  PbCl2 (s) + NaNO3 ( aq)

Predict Whether These Reactions Form Precipitates Potassium chloride and silver nitrate Potassium acetate and silver nitrate Know the difference between overall and net ionic equations.

Reactions with Oxygen Reactions with oxygen generally release energy Combustion of natural gas Organic compounds CO2 and H2O are usually the products CH4+2O2CO2+2H2O Rusting or corrosion of iron 4Fe + 3O2  2Fe2O3

Acid-Base Reactions These reactions involve the transfer of a hydrogen ion (H+) from one reactant (acid) to another (base) HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) The H+ on HCl was transferred to the oxygen in OH-, giving H2O

Oxidation-Reduction Reactions Reaction involves the transfer of one or more electrons from one reactant to another Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq) Two electrons are transferred from Zn to Cu2+

Writing Chemical Reactions Consider the following reaction: hydrogen reacts with oxygen to produce water Write the above reaction as a chemical equation H2 + O2  H2O Don’t forget the diatomic elements

4.4 Balancing Chemical Equations A chemical equation shows the molar quantity of reactants needed to produce a particular molar quantity of products The relative number of moles of each product and reactant is indicated by placing a whole-number coefficient before the formula of each substance in the chemical equation

Law of Conservation of Mass Law of conservation of mass - matter cannot be either gained or lost in the process of a chemical reaction The total mass of the products must equal the total mass of the reactants

Coefficient - how many of that substance are in the reaction Balancing Coefficient - how many of that substance are in the reaction The equation must be balanced All the atoms of every reactant must also appear in the products Number of Hg on left? 2 on right 2 Number of O on left? 2

Examine the Equation H2 + O2  H2O Is the law of conservation of mass obeyed as written? NO Balancing chemical equations uses coefficients to ensure that the law of conservation of mass is obeyed You may never change subscripts! WRONG: H2 + O2  H2O2

Steps in Equation Balancing The steps to balancing: H2 + O2  H2O Step 1. Count the number of moles of atoms of each element on both product and reactant sides Reactants Products 2 mol H 2 mol H 2 mol O 1 mol O

Steps in Equation Balancing H2 + O2  H2O Step 2. Determine which elements are not balanced – do not have same number on both sides of the equation Oxygen is not balanced Step 3. Balance one element at a time by changing the coefficients H2 + O2  2H2O This balances oxygen, but is hydrogen still balanced?

Steps in Equation Balancing H2 + O2  2H2O How will we balance hydrogen? 2H2 + O2  2H2O Step 4. Check! Make sure the law of conservation of mass is obeyed Reactants Products 4 mol H 4 mol H 2 mol O 2 mol O

Practice Equation Balancing Balance the following equations: C2H2 + O2  CO2 + H2O AgNO3 + FeCl3  Fe(NO3)3 + AgCl C2H6 + O2  CO2 + H2O N2 + H2  NH3

4.5 Calculations Using the Chemical Equation Calculation quantities of reactants and products in a chemical reaction has many applications Need a balanced chemical equation for the reaction of interest The coefficients represent the number of moles of each substance in the equation

General Principles Chemical formulas of all reactants and products must be known Equation must be balanced to obey the law of conservation of mass Calculations of an unbalanced equation are meaningless Calculations are performed in terms of moles Coefficients in the balanced equation represent the relative number of moles of products and reactants

Using the Chemical Equation Examine the reaction: 2H2 + O2  2H2O Coefficients tell us? 2 mol H2 reacts with 1 mol O2 to produce 2 mol H2O What if 4 moles of H2 reacts with 2 moles of O2? It yields 4 moles of H2O

Using the Chemical Equation 2H2 + O2  2H2O The coefficients of the balanced equation are used to convert between moles of substances How many moles of O2 are needed to react with 4.26 moles of H2? Use the factor-label method to perform this calculation

Use of Conversion Factors 2H2 + O2  2H2O 1 2.13 mol O2 2 Digits in the conversion factor come from the balanced equation

Conversion Between Moles and Grams Requires only the formula weight Convert 1.00 mol O2 to grams Plan the path Find the molar mass of oxygen 32.0 g O2 = 1 mol O2 Set up the equation Cancel units 1.00 mol O2 x 32.0 g O2 1 mol O2 Solve equation 1.00 x 32.0 g O2 = 32.0 g O2 moles of Oxygen grams of Oxygen

Conversion of Mole Reactants to Mole Products Use a balanced equation C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 1 mol C3H8 results in: 5 mol O2 consumed 1 mol C3H8 /5 mol O2 3 mol CO2 formed 1 mol C3H8 /3 mol CO2 4 mol H2O formed 1 mol C3H8 /4 mol H2O This can be rewritten as conversion factors

Calculating Reacting Quantities Calculate grams O2 reacting with 1.00 mol C3H8 Use 2 conversion factors Moles C3H8 to moles O2 Moles of O2 to grams O2 Set up the equation and cancel units 1.00 mol C3H8 x 5 mol O2 x 32.0 g O2 = 1 mol C3H8 1 mol O2 1.00 x 5 x 32.0 g O2 = 1.60 x 102 g O2 moles C3H8 moles Oxygen grams Oxygen

Calculating Grams of Product from Moles of Reactant Calculate grams CO2 from combustion of 1.00 mol C3H8 Use 2 conversion factors Moles C3H8 to moles CO2 Moles of CO2 to grams CO2 Set up the equation and cancel units 1.00 mol C3H8 x 3 mol CO2 x 44.0 g CO2 = 1 mol C3H8 1 mol CO2 1.00 x 3 x 44.0 g CO2 = 1.32 x 102 g CO2 moles C3H8 moles CO2 grams CO2

Relating Masses of Reactants and Products Calculate grams C3H8 required to produce 36.0 grams of H2O Use 3 conversion factors Grams H2O to moles H2O Moles H2O to moles C3H8 Moles of C3H8 to grams C3H8 Set up the equation and cancel units = 22.0 g C3H8 grams H2O moles H2O moles C3H8 grams C3H8

Calculating a Quantity of Reactant Ca(OH)2 neutralizes HCl Calculate grams HCl neutralized by 0.500 mol Ca(OH)2 Write chemical equation and balance Ca(OH)2(s) + 2HCl(aq) CaCl2(s) + 2H2O(l) Plan the path Set up the equation and cancel units 0.500 mol Ca(OH)2 x 2 mol HCl x 36.5 g HCl 1 mol Ca(OH)2 1 mol HCl = 36.5 g HCl moles Ca(OH)2 moles HCl grams HCl

Sample Calculation Na + Cl2  NaCl Balance the equation Calculate the moles Cl2 reacting with 5.00 mol Na Calculate the grams NaCl produced when 5.00 mol Na reacts with an excess of Cl2 Calculate the grams Na reacting with 5.00 g Cl2 2Na + Cl2  2NaCl

Theoretical and Percent Yield Theoretical yield - the maximum amount of product that can be produced Pencil and paper yield Actual yield - the amount produced when the reaction is performed Laboratory yield Percent yield: = 125 g CO2 actual x 100% = 97.4% 132 g CO2 theoretical

Sample Calculation If the theoretical yield of iron was 30.0 g and actual yield was 25.0 g, calculate the percent yield: 2 Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(l) [25.0 g / 30.0 g] x 100% = 83.3% Calculate the % yield if 26.8 grams iron was collected in the same reaction