Changing Matter Matter can be changed two ways: Physically Chemically Physical reaction Physical change Chemically Chemical reaction Chemical change

Physical Changes Do NOT CHANGE THE TYPE OF MATTER Nothing new or different is formed Could be a change in: Mass Volume Density Change in state Color Shape Size

Examples of Physical Changes Boiling, vaporization…. Any state change Dissolving Breaking Making a mixture 2 or more types of matter (substances) mixed together Not in specific amounts Can be separated physically

Chemical Changes Atoms have electrons arranged in energy levels or energy shells Electrons in the last (outermost) shell are called valence electrons Valence electrons let atoms bond with other atoms Ionic bonding Gaining or losing electrons Covalent bonding Sharing electrons

Chemical Changes Atoms that bond form molecules May be the same type (nonmetals) of atom or, Different types (metal + nonmetal) of atoms Different types  compounds

Chemical Changes Molecules can bond and “unbond” Atoms can re-arranged in different combinations For example: CaCO3 (1 atom Ca, 1 atom C, 3 atoms O) Add heat to re-arranged the atoms: CaO CO2

Chemical Changes Evidence of a chemical reaction Formation of gas Formation of precipitate Change in color Change in energy Endothermic Absorbs heat energy (gets cold) Exothermic Releases heat energy (gets hot)

Chemical Changes Chemical reactions can be represented by equations CaCO3  CaO + CO2 Reactants Products

Chemical Changes Atoms are re-arranged, NOT created or destroyed Law of Conservation of Matter Law of Conservation of Mass

Chemical Changes Matter is conserved  type of atoms does not change Nothing is created or destroyed Mass is conserved  amount of atoms cannot change

Chemical Changes To show conservation of mass  Balance equations Make sure there are the same number of each type of atom in the products and in the reactants

The equation for the burning of methane gas in oxygen is: Balancing Equations The equation for the burning of methane gas in oxygen is: CH4 + 2 O2 → CO2 + 2 H2O Subscript Shows # of atoms Coefficient Shows # of molecules

Balancing Equations 1 C  1 C 4 H  4 H 4 O  4 O No subscript or coefficient is understood to be 1 CH4 + 2 O2 → CO2 + 2 H2O = C1H4 + 2 O2 → C1O2 + 2 H2O1 1 C  1 C 4 H  4 H 4 O  4 O

Periodic Properties Post Lab Periodic Trends affect properties because of Zeff Ionization energy Energy required to remove an electron Electronegativity Ability to attract electrons Atomic Radius Distance of the valence electrons to the nucleus

Other Periodic Trends Groups have similar properties Valence electrons Members of the same representative family have the same number of valence electrons Reactivity Because they have the same number of electrons, they react similarly. CaCl2, MgCl2, SrCl2, etc. Density = mass/volume % error = I actual – theoretical I theoretical Sn = 7.265 Pb = 11.34 Si = 2.33 Ge = 5.323

Solubility Trends Common in Double Replacement Rxns

Double Replacement Rxns Two compounds react to form two new compounds Metal replaces metal Remember basic formula writing rules Ex: Ca(NO3)2 + Na2CO3  CaCO3 + NaNO3 2 ppt

Some Types of Chemical Reactions Type of Reaction Definition  Equation Synthesis Decomposition Single Replacement Double Replacement Two or more elements or compounds combine to make a more complex substance A + B → AB Compounds break down into simpler substances AB → A + B Occurs when one element replaces another one in a compound AB + C → AC + B Occurs when different atoms in two different compounds trade places AB + CD → AC + BD

Identifying Chemical Reactions S = Synthesis D = Decomposition SR = Single Replacement DR = Double Replacement   ____ P + O2 → P4O10 ____ Mg + O2 → MgO ____ HgO → Hg + O2 ____ Al2O3 → Al + O2 ____ Cl2 + NaBr → NaCl + Br2 ____ H2 + N2 → NH3

MOLAR MASS Molar Mass is shown below the element symbol on the Periodic Table. Units: grams mole Use molar mass to convert between: Number of Moles Mass (grams) Number of Molecules or Atoms

Molar Mass Examples NaHCO3 sodium bicarbonate sucrose NaHCO3 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol C12H22O11 12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mol

MOLES MASS in GRAMS MOLECULESATOMS Multiply moles by     Molar Mass to get   Divide the mass (in grams) by Molar Mass to get the number of moles. MASS in GRAMS MOLAR MASS grams/mol Multiply the number of moles by Avogadro’s Constant to get the number of atoms or molecules.   MOLES Multiply the moles by the Molar Mass to get the mass (in grams). AVOGADRO’S CONSTANT 6.022x1023 particles/mol MOLECULESATOMS Divide the number of molecules or atoms by Avogadro’s Constant to get the number of moles.

Molar Conversion Examples How many moles of carbon are in 26 g of carbon? 26 g C 1 mol C 12.01 g C = 2.2 mol C

Molar Conversion Examples How many molecules are in 2.50 moles of C12H22O11? 6.02  1023 molecules 1 mol 2.50 mol = 1.51  1024 molecules C12H22O11

Molar Conversion Examples Find the mass of 2.1  1024 molecules of NaHCO3. 2.1  1024 molecules 1 mol 6.02  1023 molecules 84.01 g 1 mol = 290 g NaHCO3

Sample Problems – Moles and Atoms Determine the number of atoms present in 2.50 moles of strontium. Convert 5.01 x 1024 atoms of strontium to moles of strontium.

Sample Problems – Moles and Mass Determine the mass in grams of 2.50 moles of strontium. Determine the number of moles represented by 943.5 grams of strontium.

Sample Problems – Moles, Atoms, Mass Determine the mass in grams of (exactly) 5 atoms of strontium. Determine the number of atoms represented by 43.5 grams of strontium.

Percentage Composition the percentage by mass of each element in a compound

Percentage Composition Find the % composition of Cu2S. 127.10 g Cu 159.17 g Cu2S 79.852% Cu %Cu =  100 = 32.07 g S 159.17 g Cu2S 20.15% S %S =  100 =

Percentage Composition Find the percentage composition of a sample that is 28 g Fe and 8.0 g O. 28 g 36 g %Fe =  100 = 78% Fe 8.0 g 36 g %O =  100 = 22% O

Percentage Composition How many grams of copper are in a 38.0-gram sample of Cu2S? Cu2S is 79.852% Cu (38.0 g Cu2S)(0.79852) = 30.3 g Cu

Percentage Composition Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O? 36.04 g 147.02 g %H2O =  100 = 24.51% H2O

C2H6 CH3 Empirical Formula Smallest whole number ratio of atoms in a compound C2H6 reduce subscripts CH3

Empirical Formula 1. Find mass (or %) of each element. 2. Find moles of each element. 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

Empirical Formula Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g 1 mol 14.01 g = 1.85 mol N = 1 N 1.85 mol 74.1 g 1 mol 16.00 g = 4.63 mol O = 2.5 O

N2O5 N1O2.5 Empirical Formula Need to make the subscripts whole numbers  multiply by 2 N2O5

CH3 C2H6 Molecular Formula “True Formula” - the actual number of atoms in a compound CH3 empirical formula ? C2H6 molecular formula

Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3.

(CH2)2  C2H4 Molecular Formula The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is 28.1 g/mol? empirical mass = 14.03 g/mol 28.1 g/mol 14.03 g/mol = 2.00 (CH2)2  C2H4

Using the Basics… The empirical formula of a compound is found to be P2O5. The molar mass of the compound is 284 grams/mole. What is the molecular formula for the compound?

Proportional Relationships Stoichiometry mass relationships between substances in a chemical reaction based on the mole ratio Mole Ratio indicated by coefficients in a balanced equation 2 Mg + O2  2 MgO

N2 + 3H2  2NH3 1 mol 2 mol 3 mol

N2 + 3H2  2NH3 1 mol 2 mol 3 mol

Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. Mole ratio - moles  moles Molar mass - moles  grams Molarity - moles  liters soln Molar volume - moles  liters gas Mole ratio - moles  moles Core step in all stoichiometry problems!! 4. Check answer.

Standard Temperature & Pressure Molar Volume at STP 1 mol of a gas=22.4 L at STP Standard Temperature & Pressure 0°C and 1 atm

Molar Volume at STP (g/mol) particles/mol LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES Molar Mass (g/mol) 6.02  1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION

Mole-Mole Calculations

Phosphoric Acid Phosphoric acid (H3PO4) is one of the most widely produced industrial chemicals in the world. Most of the world’s phosphoric acid is produced by the wet process which involves the reaction of phosphate rock, Ca5(PO4)3F, with sulfuric acid (H2SO4). Ca5(PO4)3F(s) + 5H2SO4  3H3PO4 + HF + 5CaSO4

Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4 Calculate the number of moles of phosphoric acid (H3PO4) formed by the reaction of 10 moles of sulfuric acid (H2SO4). Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4 1 mol 5 mol 3 mol 1 mol 5 mol Step 1 Moles starting substance: 10.0 mol H2SO4 Step 2 The conversion needed is moles H2SO4  moles H3PO4 Mole Ratio

Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4 Calculate the number of moles of sulfuric acid (H2SO4) that react when 10 moles of Ca5(PO4)3 react. Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4 1 mol 5 mol 3 mol 1 mol 5 mol Step 1 The starting substance is 10.0 mol Ca5(PO4)3F Step 2 The conversion needed is moles Ca5(PO4)3F  moles H2SO4 Mole Ratio

Stoichiometry Problems How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas? 2KClO3  2KCl + 3O2 ? mol 9 mol 9 mol O2 2 mol KClO3 3 mol O2 = 6 mol KClO3

Mole-Mass Calculations

Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4 Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4. Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4 Method 1 Step by Step Step 1 The starting substance is 784 grams of H3PO4. Step 2 Convert grams of H3PO4 to moles of H3PO4. Step 3 Convert moles of H3PO4 to moles of H2SO4 by the mole-ratio method. Mole Ratio

Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4 Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4 Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4 Method 2 Continuous grams H3PO4  moles H3PO4  moles H2SO4 The conversion needed is Mole Ratio

Stoichiometry Problems How many grams of KClO3 are req’d to produce 9.00 L of O2 at STP? 2KClO3  2KCl + 3O2 ? g 9.00 L 9.00 L O2 1 mol O2 22.4 L 2 mol KClO3 3 mol O2 122.55 g KClO3 1 mol KClO3 = 32.8 g KClO3

Mass-Mass Calculations

Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2. N2 + 3H2  2NH3 Method 1 Step by Step Step 1 The starting substance is 112 grams of H2. Convert 112 g of H2 to moles. grams  moles Step 2 Calculate the moles of NH3 by the mole ratio method.

Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2. N2 + 3H2  2NH3 Method 1 Step by Step Step 3 Convert moles NH3 to grams NH3. moles  grams

Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2. N2 + 3H2  2NH3 Method 2 Continuous grams H2  moles H2  moles NH3  grams NH3

Stoichiometry Problems How many grams of silver will be formed from 12.0 g copper? Cu + 2AgNO3  2Ag + Cu(NO3)2 12.0 g ? g 12.0 g Cu 1 mol Cu 63.55 g Cu 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag = 40.7 g Ag

Stoichiometry Problems How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3? Cu + 2AgNO3  2Ag + Cu(NO3)2 ? g 1.5L 0.10M 1.5 L .10 mol AgNO3 1 L 1 mol Cu 2 mol AgNO3 63.55 g Cu 1 mol Cu = 4.8 g Cu

Limiting Reactant

The limiting reactant limits the amount of product that can be formed. The limiting reactant is one of the reactants in a chemical reaction. It is called the limiting reactant because the amount of it present is insufficient to react with the amounts of other reactants that are present. The limiting reactant limits the amount of product that can be formed.

How many bicycles can be assembled from the parts shown? From eight wheels four bikes can be constructed. From three pedal assemblies three bikes can be constructed. From four frames four bikes can be constructed. The limiting part is the number of pedal assemblies. 9.2

H2 + Cl2  2HCl  + Cl2 is the limiting reactant 4 molecules Cl2 can form 8 molecules HCl Cl2 is the limiting reactant 3 molecules of H2 remain H2 is in excess 7 molecules H2 can form 14 molecules HCl 9.3

Steps Used to Determine the Limiting Reactant

Calculate the amount of product (moles or grams, as needed) formed from each reactant. Determine which reactant is limiting. (The reactant that gives the least amount of product is the limiting reactant; the other reactant is in excess. Calculate the amount of the other reactant required to react with the limiting reactant, then subtract this amount from the starting quantity of the reactant. This gives the amount of the substance that remains unreacted.

Examples

How many moles of HCl can be produced by reacting 4. 0 mol H2 and 3 How many moles of HCl can be produced by reacting 4.0 mol H2 and 3.5 mol Cl2? Which compound is the limiting reactant? H2 + Cl2 → 2HCl Step 1 Calculate the moles of HCl that can form from each reactant. Step 2 Determine the limiting reactant. The limiting reactant is Cl2 because it produces less HCl than H2.

Step 1 Calculate the grams of AgBr that can form from each reactant. How many moles of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr2 and 100.0 g of AgNO3 are mixed together? How many grams of the excess reactant remain unreacted? MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq) Step 1 Calculate the grams of AgBr that can form from each reactant. The conversion needed is g reactant → mol reactant → mol AgBr → g AgBr

Step 2 Determine the limiting reactant. How many moles of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr2 and 100.0 g of AgNO3 are mixed together? How many grams of the excess reactant remain unreacted? MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq) Step 2 Determine the limiting reactant. The limiting reactant is MgBr2 because it forms less Ag Br.

How many grams of the excess reactant (AgNO3) remain unreacted? MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq) Step 3 Calculate the grams of unreacted AgNO3. First calculate the number of grams of AgNO3 that will react with 50 g of MgBr2. The conversion needed is g MgBr2 → mol MgBr2 → mol AgNO3 → g AgNO3 The amount of MgBr2 that remains is 100.0 g AgNO3 - 92.3 g AgNO3 = 7.7 g AgNO3

Reaction Yield

The quantities of products calculated from equations represent the maximum yield (100%) of product according to the reaction represented by the equation.

Many reactions fail to give a 100% yield of product. This occurs because of side reactions and the fact that many reactions are reversible.

The theoretical yield of a reaction is the calculated amount of product that can be obtained from a given amount of reactant. The actual yield is the amount of product finally obtained from a given amount of reactant.

The percent yield of a reaction is the ratio of the actual yield to the theoretical yield multiplied by 100.

Silver bromide was prepared by reacting 200 Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if 375.0 g of silver bromide was obtained from the reaction: MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq) Step 1 Determine the theoretical yield by calculating the grams of AgBr that can be formed. The conversion needed is g MgBr2 → mol MgBr2 → mol AgBr → g AgBr

MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq) Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if 375.0 g of silver bromide was obtained from the reaction: MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq) Step 2 Calculate the percent yield. must have same units