Reactions, Equations, and Formulas (Chapter 11) Dr. Walker

What will you learn? SOL Ch 1 g, 3 b, c, e Classifying reaction types Balancing chemical reactions based on the law of conservation of mass Differentiating between molecular and empirical formulas Calculating percent composition of elements within compounds

Vocabulary Reactants – left side of a reaction; what you start with Products – right side of a reaction; what you finish with NaOH + HCl NaCl + H2O Reactants Products

Common chemical shorthand (l) – liquid (s) – solid (g) – gas (aq) – dissolved in water D – add heat DH – energy change (if necessary) Temperature – react at a specific temperature

Six Reaction Types Synthesis (or Combination) Decomposition Combustion Single Replacement Double Replacement Oxidation/Reduction

Synthesis Reactions Multiple Reactants One Product Examples H2O + CO2 H2CO3 4 Al + 3 O2 2 Al2O3 2 KCl + 3O2 2 KClO3

Decomposition Reactions One Reactant Multiple Products Examples 2 H2O2 2 H2O + O2 Ba(ClO3)2 BaCl2 + 3 O2

Combustion Reactions Reactants = hydrocarbon (or alcohol) + oxygen (CHO) Products = carbon dioxide + water Example 2 C8H18 + 25 O2 16 CO2 + 18 H2O C3H8 + 5 O2 3 CO2 + 4 H2O C6H12O6 + 6 O2 6 CO2 + 6 H2O

Single Replacement Reactions Reactants = Element + Compound Products = Different Element + Different Compound Examples 2 Na + MgCl2 2 NaCl + Mg Ca + 2 ZnBr CaBr2 + 2 Zn

Double Replacement Reactions Two Products Two Reactants Metals “switch” compounds Examples K2SO4 + 2 HCl 2 KCl + H2SO4 ZnCl + NaBr ZnBr + NaCl

Oxidation-Reduction Reactions (aka Redox Reactions) Oxidation – Loses electrons When atoms become cations, they are oxidized Zn+1 Zn+2 + 1e- Reduction – Gains electrons When atoms become anions, they are reduced Reduces the charge Cu+2 + 1e- Cu+1

So How Do We Know What Kind of Reaction Is Taking Place????? Ask a series of questions where reaction types can be eliminated.

First Question How many reactants? One Two You have a decomposition reaction Two Keep Going

Second Question We have multiple reactants. How many products? One This is a synthesis reaction Two or more Keep Going

Third Question Are the reactants oxygen gas and a hydrocarbon (or alcohol)? Yes This is a combustion reaction No Keep Going

Fourth Question Are the reactants and products two compounds? NO Yes This is a single replacement reaction. Yes This is a double replacement reaction.

Fourth Question Are the reactants and products two compounds? Yes This is a double replacement reaction.

Examples Zn + O2 ZnO2 Fe2O3 + Mg MgO + Fe

Examples Zn + O2 ZnO2 Fe2O3 + Mg MgO + Fe Synthesis Zn + O2 ZnO2 Single Replacement Fe2O3 + 3 Mg 3 MgO + 2 Fe

Examples Ca(ClO3)2 CaCl2 + O2 C5H12 + O2 CO2 + H2O

Examples Ca(ClO3)2 CaCl2 + O2 C5H12 + O2 CO2 + H2O Decomposition Combustion C5H12 + 8 O2 5 CO2 + 6 H2O

Examples KI + MgO K2O + MgI2 Ca + H2O H2 + CaO

Examples KI + MgO K2O + MgI2 Ca + H2O H2 + CaO Double Replacement Single Replacement Ca + H2O H2 + CaO

A little different Redox reactions Gaining or Losing Electrons Easy to distinguish from other reaction types

Redox Examples Fe+1 Fe3+ + 2e- Oxidation Sn+4 + 2e- Sn2+ Reduction

A little later… One more type… Acid-Base (Neutralization) Reactions We’ll talk about them in more detail during chapter 15 in a few weeks Will be on the SOL

Chemical Reaction Vocabulary 4 P + 5 O2 2 P2O5 coefficients subscripts P 4 x 1 = 4 atoms O 5 x 2 = 10 atoms P 2 x 2 = 4 atoms O 2 x 5 = 10 atoms Number of atoms for each element = coefficient x subscript Lack of a coefficient or subscript denotes a value of 1

Chemical Reaction Vocabulary 4 P + 5 O2 2 P2O5 coefficients subscripts P 4 x 1 = 4 atoms O 5 x 2 = 10 atoms P 2 x 2 = 4 atoms O 2 x 5 = 10 atoms Law of conservation of mass – matter cannot be or destroyed in a chemical reaction Mass in = Mass out Atoms (reactants) = Atoms (products)

Balancing Chemical Reactions Balanced reactions are required to satisfy the law of conservation of mass You will balance reactions by adding coefficients to reactions Goal: Have the same number of atoms on each side of the reaction You may NOT change subscripts or you have a different reaction

Steps To Balance Reactions ___ Na + ___Cl2 ___ NaCl Step 1: Count all atoms on each side Remember, atoms = coefficient x subscript

Steps To Balance Reactions ___ Na + ___Cl2 ___ NaCl Na = 1 Na = 1 Cl = 2 Cl = 1 Step 1: Count all atoms on each side Remember, atoms = coefficient x subscript

Steps To Balance Reactions ___ Na + ___Cl2 _2_ NaCl Na = 1 Na = 2 Cl = 2 Cl = 2 Step 2: Add coefficients to balance atoms that are not equal - 2 is added in front of NaCl to balance Cl

Steps To Balance Reactions 2 Na + ___Cl2 _2_ NaCl Na = 2 Na = 2 Cl = 2 Cl = 2 Step 3: Adjust other coefficients for unequal elements 2 is added in front of Na to balance Na Verify that equation is balanced

Additional Examples ___ N2 + ___ H2 ___ NH3 Count atoms N N H H

Additional Examples ___ N2 + ___ H2 ___ NH3 Count atoms 2 N 1 N 2 H 3 H Common Factor method: 2 and 3 aren’t divisible. Make the total # of H’s a number that can be divided by 2 and 3

Additional Examples N2 + _3_ H2 _2_ NH3 Add coefficients 2 N 2 N 6 H 6 H Common Factor method: Notice that 2 and 3 are both divisible by 6. Balancing H works to balance N as well.

Additional Examples ___ H2O ___ H2 + ___ O2 Count atoms ___ H ____ H ___ O ____ O

Additional Examples ___ H2O ___ H2 + ___ O2 Count atoms _2_ H 2_ H _1_ O _2_ O

Additional Examples _2_ H2O _2_ H2 + ___ O2 Add coefficients _4_ H 4_ H _2_ O _2_ O Even-odd method Make all atoms even – multiply odd atoms by two

Additional Examples ___ CuNO3 + ___K2CO3 → ___ Cu2CO3 + ___ KNO3 Count atoms ___ Cu ___ Cu ___ NO3 ___ NO3 ___K ___ K ___CO3 ___ CO3 Keep intact polyatomics together – don’t split them apart!

Additional Examples ___ CuNO3 + ___K2CO3 → ___ Cu2CO3 + ___ KNO3 Count atoms _1_ Cu _2_ Cu _1_ NO3 _1_ NO3 _2_ K _1_ K _1_ CO3 _1_ CO3 Keep intact polyatomics together – don’t split them apart!

Additional Examples _2_CuNO3 + ___K2CO3 → ___ Cu2CO3 + _2_ KNO3 Add coefficients _2_ Cu _2_ Cu _2_ NO3 _2_ NO3 _2_ K _2_ K _1_ CO3 _1_ CO3 Keep intact polyatomics together – don’t split them apart!

Additional Examples ___ C3H8 + ___ O2 → ___ CO2 + ___ H2O Combustion reactions: Save oxygen for last!!

Additional Examples ___ C3H8 + ___ O2 → ___ CO2 + ___ H2O Count atoms 3 C 1 C 8 H 2 H 2 O 3 O Combustion reactions: Save oxygen for last!!

Additional Examples ___ C3H8 + ___ O2 → 3 CO2 + 4 H2O Balance, part 1 3 C 3 C 8 H 8 H 2 O 10 O Combustion reactions: Save oxygen for last!!

Additional Examples ___ C3H8 + 5 O2 → 3 CO2 + 4 H2O Balance, part 2 (oxygen) 3 C 3 C 8 H 8 H 10 O 10 O Combustion reactions: Save oxygen for last!!

Additional Examples ___ C8H18 + ___ O2 → ___ CO2 + ___ H2O

Additional Examples ___ C8H18 + ___ O2 → ___ CO2 + ___ H2O Count atoms 8 C 1 C 18 H 2 H 2 O 3 O

Additional Examples ___ C8H18 + ___ O2 → _8_ CO2 + 9 H2O Balance C and H 8 C 8 C 18 H 18 H 2 O 27 O Problem: Can’t balance both sides with a whole #

Additional Examples ___ C8H18 + ___ O2 → _8_ CO2 + 9 H2O Balance C and H 8 C 8 C 18 H 18 H 27 O 27 O Problem: Can’t balance both sides with a whole #

Additional Examples ___ C8H18 + 12.5 O2 → _8_ CO2 + 9 H2O Balance oxygen 8 C 8 C 18 H 18 H 25 O 25 O Problem: Temporarily use a decimal to balance

Additional Examples 2 C8H18 + 25 O2 → _16_ CO2 + 18 H2O Get rid of decimal coefficient 16 C 16 C 36 H 36 H 50 O 50 O Problem: Multiply ALL coefficients by 2 to get rid of decimal

Molar Mass The atomic mass of any element is the weighted average of all isotopes of an element These are shown in the periodic table This is also the molar mass of the element Examples H = 1.01 g/mole O = 16.00 g/mole C = 12.01 g/mole P = 30.97 g/mole Cl = 35.45 g/mole

Molar Mass The molar mass of a compound is the sum of all atomic weights of all atoms in the compound! Remember Compounds have more than one element When there are multiple atoms, you simply add the masses of each atom together to get the mass of a compound

Molar Mass Examples For compounds H2O H = 1.01 g/mole x 2 O = 16.00 g/mole x 1 =18.02 g/mole

Molar Mass Examples Another example CH4 C = 12.01 g/mole = 12.01 H = 1.01 g/mole x 4 = 4.04 16.05 g/mole

Molar Mass Examples Another Example C6H12O6 C = 12.01 g/mole x 6 = 72.06 H = 1.01 g/mole x 12 = 12.12 O = 16.00 g/mole x 6 = 96.00 180.18 g/mole

Molar Mass Examples Another Example Polyatomics Pb(SO4)2 Pb = 207.20 S = 32.07 g/mole x 2 = 64.14 O = 16.00 g/mole x 8 = 128.00 399.34 g/mole Notice you must multiply the subscript inside the parentheses by the number outside to get the right number of atoms

Practice Find the molar mass of the following compounds: LiOH SO2 Mg(NO3)2 Al2(CO3)3 C9H8O4 C8H10N4O2 C22H30N6O4S

Practice Find the molar mass of the following compounds: LiOH 6.94 + 16.00 + 1.01 = 23.95 g/mole SO2 32.06 + 2(16.00) = 64.06 g/mole Mg(NO3)2 24.31 + 2(14.01) + 6(16.00) = 148.33 g/mole Al2(CO3)3 2(26.98) + 3(12.01) + 9(16.00) = 233.99 g/mole C9H8O4 9(12.01) + 8(1.01) + 4(16.00) = 180.17 g/mole C8H10N4O2 8(12.01) + 10(1.01) + 4(14.01) + 2(16.00) = 194.22 g/mole C22H30N6O4S 22(12.01) + 30(1.01) + 6(14.01) + 4(16.01) + 1(32.06) = 474.68 g/mole

Empirical Formulas Molecular Formula – True number of atoms in a compound Empirical Formula – SIMPLEST ratio of atoms in a compound Benzene Molecular Formula = C6H6 Empirical Formula = CH Structural Formula Visual representation of a chemical formula showing bonds

Empirical and Molecular Formulas

Empirical and Molecular Formulas

Molecular Formulas from Empirical Formulas As previously demonstrated, multiple compounds can have the same empirical formula How do we figure out molecular formulas? We need another piece of information…the molar mass of the molecular formula

Molecular Formulas from Empirical Formulas Example What is the molecular formula of a compound with an empirical formula of CH2 and a molar mass of 70 g/mole?

Molecular Formulas from Empirical Formulas Example What is the molecular formula of a compound with an empirical formula of CH2 and a molar mass of 70 g/mole? Step 1 – Figure out molar mass of empirical formula. Molar mass of CH2 = 12 + (2 x 1) = 14 g/mole Mass from periodic table for carbon (12) and H (1)

Molecular Formulas from Empirical Formulas Example What is the molecular formula of a compound with an empirical formula of CH2 and a molar mass of 70 g/mole? Step 2 – Divide given mass by the molar mass of the empirical formula you found in the previous step (70 g/mole)/(14 g/mole) = 5

Molecular Formulas from Empirical Formulas Example What is the molecular formula of a compound with an empirical formula of CH2 and a molar mass of 70 g/mole? Step 3 – Multiply subscripts of empirical formula by number found in step 2 CH2 x 5 = C5H10 You can double check this formula – if it doesn’t have the molar mass given in the problem, it’s wrong

Another Example A compound is found to have an empirical formula of CH2O and a molar mass of 120.12 g/mole. What is the molecular formula?

Another Example A compound is found to have an empirical formula of CH2O and a molar mass of 120.12 g/mole. What is the molecular formula? Step 1 – CH2O – molar mass = 30.03 g/mole Step 2 – (120.12 g/mole)/(30.03 g/mole) = 4 Step 3 – 4 x (CH2O) = C4H8O4

Percentage Composition The percentage by mass of each element in a compound “Mass of whole” = the molar mass of the whole compound “Mass of part” = the molar mass of a single element

How do we find percentage composition? Example Give the percentage composition of each element of H2O.

How do we find percentage composition? Example Give the percentage composition of each element of H2O. Step 1 = Find molar mass of compound H = 1.01 g/mole x 2 = 2.02 g/mole O = 16.00 g/mole x 1 = 16.00 g/mole 18.02 g/mole

How do we find percentage composition? Example Give the percentage composition of each element of H2O. Step 2: Divide the mass of each element by the total mass and multiply by 100. H = [ (2.02 g/mole)/(18.02 g/mole)] x 100 = 11.2% O = [ (16.00 g/mole)/(18.02 g/mole)] x 100 = 88.8%

Percentage Compositon Example: Find the percentage composition of (NH4)2SO4

Percentage Compositon Example – (NH4)2SO4 Molar mass N = 14.01 g/mole x 2 = 28.02 g/mole H = 1.01 g/mole x 8 = 8.08 g/mole S = 32.06 g/mole x 1 = 32.06 g/mole O = 16.00 g/mole x 4 = 64.00 g/mole Total molar mass = 132.16 g/mole

Percentage Compositon Example – (NH4)2SO4 Percentage Composition N = [(28.02 g/mole)/(132.16 g/mole) x 100 = 21.2% H = [(8.08 g/mole)/(132.16 g/mole) x 100 = 6.1 % S = [(32.06 g/mole)/(132.16 g/mole) x 100 = 24.3 % O = [(64.00 g/mole)/(132.16 g/mole) x 100 = 48.4 % Total molar mass = 132.16 g/mole Double check – what is our total percentage? It should be 100%

Terms To Know Reactants Products Synthesis (or Combination) Decomposition Combustion Single Displacement (Replacement) Double Displacement (Replacement) Oxidation/Reduction Law of conservation of mass Empirical formula Structural formula Percentage composition

Skill To Master Determine reaction type from given chemical equation Balance chemical reactions by changing coefficients Determine empirical and molecular formulas from the other Determine percentage composition of each element from a chemical formula