changing matter matter can be changed two ways: –physically physical reaction physical change...

Changing Matter• Matter can be changed two ways:

–Physically • 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– Nothing is created or destroyed

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

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

• 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 H4 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 = 7.265Pb = 11.34Si = 2.33Ge = 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 + NaNO32

ppt

Type of Reaction

Definition Equation

Synthesis

Decomposition

Single Replacement

Double Replacement

Some Types of Chemical Reactions

A + B → AB

AB → A + B

AB + C → AC + B

AB + CD → AC + BD

Two or more elements or compounds combine to make a more complex

substance

Compounds break down into simpler substances

Occurs when one element replaces another one in a

compound

Occurs when different atoms in two different

compounds trade places

Identifying Chemical Reactions

____ P + O2 → P4O10 ____ Mg + O2 → MgO

____ HgO → Hg + O2 ____ Al2O3 → Al + O2

____ Cl2 + NaBr → NaCl + Br2 ____ H2 + N2 → NH3

S = Synthesis D = Decomposition SR = Single Replacement DR = Double Replacement

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

• 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

MOLECULESATOMS

Multiply the number of moles

by Avogadro’s Constant to get the number of

atoms or molecules.

Divide the number of molecules or atoms by Avogadro’s Constant to get the number of

moles.

Divide the mass (in grams) by Molar Mass to get the

number of moles.

MOLARMASSgrams/

mol

MOLES

MASS inGRAMS

Multiply the moles

by the Molar Mass to get the

mass (in grams).

AVOGADRO’S CONSTANT6.022x1023

particles/mol

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?

2.50 mol 6.02 1023

molecules

1 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

= 290 g NaHCO3

84.01 g

1 mol

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

100mass total

element of massncompositio %

100 =

Percentage Composition

%Cu =127.10 g Cu

159.17 g Cu2S 100 =

%S =32.07 g S

159.17 g Cu2S

79.852% Cu

20.15% S

• Find the % composition of Cu2S.

%Fe =28 g

36 g 100 =78% Fe

%O =8.0 g

36 g 100 =22% O

• Find the percentage composition of a sample that is 28 g Fe and 8.0 g O.

Percentage Composition

• How many grams of copper are in a 38.0-gram sample of Cu2S?

(38.0 g Cu2S)(0.79852) = 30.3 g Cu

Cu2S is 79.852% Cu

Percentage Composition

100 =%H2O = 36.04 g

147.02 g 24.51%

H2O

• Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O?

Percentage Composition

Empirical Formula

C2H6

CH3

reduce subscripts

• Smallest whole number ratio of atoms in a compound

Empirical Formula1. 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

74.1 g 1 mol

16.00 g = 4.63 mol O

1.85 mol

1.85 mol

= 1 N

= 2.5 O

Empirical Formula

N1O2.5Need to make the subscripts whole

numbers multiply by 2

N2O5

Molecular Formula

• “True Formula” - the actual number of atoms in a compound

CH3

C2H6

empiricalformula

molecularformula

?

Molecular Formula1. 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.

nmass EF

mass MF nEF

Molecular Formula• The empirical formula for ethylene is CH2. Find the molecular

formula if the molecular mass is 28.1 g/mol?

28.1 g/mol

14.03 g/mol = 2.00

empirical mass = 14.03 g/mol

(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

• StoichiometryStoichiometry–mass relationships between

substances in a chemical reaction–based on the mole ratio

• Mole RatioMole Ratio–indicated by coefficients in a balanced

equation

2 Mg + O2 Mg + O22 2 MgO 2 MgO

43

2

23 m

1 mol

ol H

N

N2 + 3H2 2NH31 mol 2 mol3 mol

44

1 mol 2 mol3 molN2 + 3H2 2NH3

2

32 mo

3 mol H

l NH

Stoichiometry Steps1. 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

Core step in all stoichiometry problems!!

• Mole ratio - moles moles

4. Check answer.

1 mol of a gas=22.4 Lat STP

Molar Volume at STP

Standard Temperature & Pressure0°C and 1 atm

Molar Volume at STP

Molar Mass(g/mol)

6.02 1023

particles/mol

MASSIN

GRAMSMOLES

NUMBEROF

PARTICLES

LITERSOF

SOLUTION

Molar Volume (22.4 L/mol)

LITERSOF GASAT STP

Molarity (mol/L)

48

Mole-Mole Mole-Mole CalculationsCalculations

49

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

50

Mole Ratio

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

Step 1 Moles starting substance: 10.0 mol H2SO4

Step 2 The conversion needed is moles H2SO4 moles H3PO4

1 mol 5 mol 3 mol 1 mol 5 mol

3 42 4

2 4

3 mol H PO10 mol H SO x =

5 mol H SO3 46 mol H PO

51

Step 2 The conversion needed is

moles Ca5(PO4)3F moles H2SO4

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

Mole Ratio

Step 1 The starting substance is 10.0 mol Ca5(PO4)3F

1 mol 5 mol 3 mol 1 mol 5 mol

2 45 4 3

5 4 3

5 mol H SO10 mol Ca (PO ) F x =

1 mol Ca (PO ) F 2 450 mol H SO

Stoichiometry Problems

• How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?

9 mol O2 2 mol KClO3

3 mol O2

= 6 mol KClO3

2KClO3 2KCl + 3O2 ? mol 9 mol

53

Mole-Mass CalculationsMole-Mass Calculations

54Mole Ratio

2 4

3 4

5 mol H SO =

3 mol H PO

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.

3 4

3 4

1 mol H PO =

98.0 g H PO

3 48.00 mol H PO

2 413.3 mol H SO

3 4784 g H PO

3 48.00 mol H PO

55

Mole Ratio

2 4

3 4

5 mol H SO =

3mol H PO

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

3 4784 g H PO 3 4

3 4

1 mol H PO

98.0 g H PO

2 413.3 mol H SO

• How many grams of KClO3 are req’d to

produce 9.00 L of O2 at STP?

9.00 LO2

1 molO2

22.4 L O2

= 32.8 g KClO3

2 molKClO3

3 molO2

122.55g KClO3

1 molKClO3

? g 9.00 L

Stoichiometry Problems

2KClO3 2KCl + 3O2

57

Mass-Mass CalculationsMass-Mass Calculations

58

Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.N2 + 3H2 2NH3Method 1 Step by Step

Step 1 The starting substance is 112 grams of H2. Convert 112 g of H2 to moles.

grams moles2

2

1 mol H

2.02 g H

2112 g H 255.4 moles H

Step 2 Calculate the moles of NH3 by the mole ratio method.

3

2

2 mol NH=

3 mol H

255.4 moles H 336.9 moles NH

59

Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.N2 + 3H2 2NH3Method 1 Step by Step

Step 3 Convert moles NH3 to grams NH3.

moles grams

336.9 moles NH 3

3

17.0 g NH=

1 mol NH

3629 g NH

60

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

grams H2 moles H2 moles NH3 grams NH3

2

2

1 mol H

2.02 g H

2112 g H 3

2

2 mol NH

3 mol H

Method 2 Continuous

3

3

17.0 g NH=

1 mol NH

3629 g NH

Stoichiometry Problems• How many grams of silver will be

formed from 12.0 g copper?

12.0g Cu

1 molCu

63.55g Cu

= 40.7 g Ag

Cu + 2AgNO3 2Ag + Cu(NO3)2

2 molAg

1 molCu

107.87g Ag

1 molAg

12.0 g ? g

63.55g Cu

1 molCu

Stoichiometry Problems

• How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3?

1.5L

.10 molAgNO3

1 L= 4.8 g

Cu

Cu + 2AgNO3 2Ag + Cu(NO3)2

1 molCu

2 molAgNO3

? g 1.5L0.10M

63

Limiting ReactantLimiting ReactantLimiting ReactantLimiting Reactant

64

• 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.

• The limiting reactant is one of the reactants in a chemical reaction.

65

How many bicycles can be assembled from the parts shown?

From eight wheels four bikes can be constructed.

From four frames four bikes can be constructed.

From three pedal assemblies three bikes can be constructed.

The limiting part is the number of pedal assemblies.

9.2

66

H2 + Cl2 2HCl

+

7 molecules H2 can form 14 molecules HCl

4 molecules Cl2 can form 8 molecules HCl 3 molecules of H2 remain

H2 is in excess Cl2 is the limiting reactant

9.3

67

Steps Used to Determine Steps Used to Determine the Limiting Reactantthe Limiting Reactant

Steps Used to Determine Steps Used to Determine the Limiting Reactantthe Limiting Reactant

68

1. Calculate the amount of product (moles or grams, as needed) formed from each reactant.

2. Determine which reactant is limiting. (The reactant that gives the least amount of product is the limiting reactant; the other reactant is in excess.

3. 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.

69

ExamplesExamplesExamplesExamples

70

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?

Step 1 Calculate the moles of HCl that can form from each reactant.

24.0 mol H2

2 mol HCl

1 mol H

8.0 mol HCl

23.5 mol Cl2

2 mol HCl

1 mol Cl

7.0 mol HCl

H2 + Cl2 → 2HCl

Step 2 Determine the limiting reactant.

The limiting reactant is Cl2 because it produces less HCl than H2.

71

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?

Step 1 Calculate the grams of AgBr that can form from each reactant.

MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)

The conversion needed isg reactant → mol reactant → mol AgBr → g AgBr

250.0 g MgBr 102 g AgBr2

2

1 mol MgBr

184.1 g MgBr

2

2 mol AgBr

1 mol MgBr

187.8 g AgBr

1 mol AgBr

3100.0 g AgNO 110.5 g AgBr3

3

1 mol AgNO

169.9 g AgNO

3

2 mol AgBr

2 mol AgNO

187.8 g AgBr

1 mol AgBr

72

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?

Step 2 Determine the limiting reactant.

MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)

250.0 g MgBr 102 g AgBr2

2

1 mol MgBr

184.1 g MgBr

2

2 mol AgBr

1 mol MgBr

187.8 g AgBr

1 mol AgBr

3100.0 g AgNO 110.5 g AgBr3

3

1 mol AgNO

169.9 g AgNO

3

2 mol AgBr

2 mol AgNO

187.8 g AgBr

1 mol AgBr

The limiting reactant is MgBr2 because itforms less Ag Br.

73

How many grams of the excess reactant (AgNO3) remain unreacted?

Step 3 Calculate the grams of unreacted AgNO3. First calculate the number of grams of AgNO3 that will react with 50 g of MgBr2.

MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)

250.0 g MgBr 392.3 g AgNO2

2

1 mol MgBr

184.1 g MgBr

3

2

2 mol AgNO

1 mol MgBr

3

3

169.9 g AgNO

1 mol AgNO

The conversion needed isg 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

74

Reaction YieldReaction YieldReaction YieldReaction Yield

75

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

76

Many reactions fail to give a 100% yield of product.

This occurs because of side reactions and the fact that many reactions are reversible.

77

• 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.

78

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

actual yield x 100 = percent yield

theoretical yield

79

187.8 g AgBr

1 mol AgBr

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 isg MgBr2 → mol MgBr2 → mol AgBr → g AgBr

2200.0 g MgBr 408.0 g AgBr2

2

1 mol MgBr

184.1 g MgBr

2

2 mol AgBr

1 mol MgBr

80

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.

actual yieldpercent yield = x 100

theoretical yield

percent yield = 375.0 g AgBr

x 100 =408.0 g AgBr

91.9%

must have same units

must have same units