Chemical Bonding

Indicators of chemical reactions

Formation of a gas

Emission of light or heat

Formation of a precipitate

Color change

Emission of odor

All chemical reactions:

• have two parts• Reactants - the substances you start

with• Products- the substances you end up

with• The reactants turn into the products.• Reactants ® Products

Symbols used in equations

• (s) after the formula –solid Cu(s)

• (g) after the formula –gas H2 (g)

• (l) after the formula -liquid H2O(l)

• (aq) after the formula - dissolved in water, an aqueous solution. CaCl2 (aq)

Summary of Symbols

What is a catalyst?

• A substance that speeds up a reaction without being changed by the reaction.

• Enzymes are biological or protein catalysts.

All chemical reactions are accompanied by a change in energy. 

Exothermic - reactions that release energy to their surroundings (usually in the form of heat)

o ΔH (enthalpy) is negative – energy leaving system

Endothermic - reactions that need to absorb heat from their surroundings to proceed.

o ΔH (enthalpy) is positive – energy coming into the system

Reaction Energy

Diatomic elements• There are 8 elements that never want to

be alone.• They form diatomic molecules.• H2 , N2 , O2 , F2 , Cl2 , Br2 , I2 , and At2

• The –ogens and the –ines• 1 + 7 pattern on the periodic table

Convert this to an equation

Fe2S3 (s) + HCl(g) ® FeCl2 (s) + H2S(g)

HNO3 (aq) + Na2CO3 (s) ® NaNO3 (aq) + H2O(l)

Convert this to an equation

The other way

Fe(s) + O2(g) ® Fe2O3(s)

Solid iron reacts with oxygen gas to form solid iron oxide (rust).

A silver spoon tarnishes. The solid silver reacts with sulfur in the air to make solid silver

sulfide, the black material we call tarnish.

Ag (s) + H2S (g) + O2 (g) ® Ag2S (s) + H2O

Aluminum metal reacts with liquid bromine to form solid aluminum bromide

Translate Equation

___ Al(s) + ___ Br2(l) →___ AlBr3(s)2 3 2

Synthesis Reactions• Also called combination reactions• 2 elements, or compounds combine

to make one compound.• A + B ® AB• Na (s) + Cl2 (g) ® NaCl (s)

• Ca (s) +O2 (g) ® CaO (s) • We can predict the products if they

are two elements.• Mg (s) + N2 (g) ®

Mg3N2 (s)

A simulation of the reaction:

2H2 + O2   ®    2H2O  

Decomposition Reactions• decompose = fall apart• one compound (reactant) falls apart

into two or more elements or compounds.

• Usually requires energy• AB ® A + B

• NaCl Na + Cl2

• CaCO3 CaO + CO2

electricity

• Can predict the products if it is a binary compound

• Made up of only two elements• Falls apart into its elements

• H2O

• HgO

electricity

Decomposition Reactions

H2 (g) + O2

Single Replacement

• Also referred to as single displacement• One element replaces another• Reactants must be an element and a

compound.• Products will be a different element and

a different compound.• A + BC ® AC + B• 2Na + SrCl2 ® Sr + 2NaCl

Double Replacement

• Two things replace each other.• Reactants must be two ionic compounds or

acids.• Usually in aqueous solution

AB + CD ® AD + CB

AgNO3 + NaCl ® AgCl + NaNO3

ZnS + 2HCl ® ZnCl + H2S

Combustion

• A reaction in which a compound (often carbon) reacts with oxygen

• CH4 + O2 ® CO2 + H2O

• C3H8 + O2 ® CO2 + H2O

• C6H12O6 + O2 ® CO2 +

H2O

• The charcoal used in a grill is basically carbon. The carbon reacts with oxygen to yield carbon dioxide. The chemical equation for this reaction is C + O2 CO2

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STOICHIOMETRY

The measurement system in chemistry

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(1-2-3) General Approach For Problem Solving:

1. Clearly identify the Goal or Goals and the UNITS involved. (What are you

trying to do?) WHAT IS ASKED

2. Determine what is given and the UNITS. WHAT IS GIVEN!!!

3. Use conversion factors (which are really ratios) and their UNITS to CONVERT what is given into what is desired.

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Sample problem for general problem solving.Sam has entered into a 10 mile marathon. Use ALL of the following conversions (ratios) to determine how many inches there are in the race.5280 ft = 1 mile; 12 inches = 1 ft

1. What is the goal and what units are needed?

Goal = ______ inches

2. What is given and its units?

10 miles

3. Convert using factors (ratios).

10 miles = inches

mile 1

ft 5280ft 1

inches 12 633600

Units match

Goal

Menu

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Converting grams to moles.

Determine how many moles there are in 5.17 grams of Fe(C5H5)2.

Goal

= moles Fe(C5H5)2

Given

5.17 g Fe(C5H5)2

Use the molar mass to convert grams to moles. Fe(C5H5)2

2 x 5 x 1.001 = 10.012 x 5 x 12.011 = 120.11

1 x 55.85 = 55.85

mol

g 185.97

g 185.97

mol0.0278

units match

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Stoichiometry (more working with ratios)

Ratios are found within a chemical equation.

2HCl + Ba(OH)2 2H2O + BaCl2 1 1

2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O and 1 mole of BaCl2

coefficients give MOLAR RATIOS

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When N2O5 is heated, it decomposes:

2N2O5(g) 4NO2(g) + O2(g)

a. How many moles of NO2 can be produced from 4.3 moles of N2O5?

= moles NO2

4.3 mol N2O5

52

2

ON mol2

NO mol48.6

b. How many moles of O2 can be produced from 4.3 moles of N2O5?

= mole O2

4.3 mol N2O5

52

2

ON 2mol

O mol12.2

2N2O5(g) 4NO2(g) + O2(g)4.3 mol ? mol

2N2O5(g) 4NO2(g) + O2(g)4.3 mol ? mol

Mole – Mole Conversions

Units match

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When N2O5 is heated, it decomposes:2N2O5(g) 4NO2(g) + O2(g)

a. How many moles of N2O5 were used if 210g of NO2 were produced?

= moles N2O5

210 g NO2

2

52

NO mol4

ON mol22.28

b. How many grams of N2O5 are needed to produce 75.0 grams of O2?

= grams N2O5

75.0 g O2

2

52

O 1mol

ON mol2506

2

2

NO g0.46

NO mol

2

2

O g 32.0

O mol

52

52

ON mol

ON g108

gram ↔ mole and gram ↔ gram conversions

2N2O5(g) 4NO2(g) + O2(g)210g? moles

2N2O5(g) 4NO2(g) + O2(g)75.0 g? grams

Units match

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Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?

First write a balanced equation.

Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 3

Gram to Gram Conversions

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Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?

Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 3

Now let’s get organized. Write the information below the substances.

3.45 g ? grams

Gram to Gram Conversions

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Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?

Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 33.45 g ? grams

Let’s work the problem.

= g AlCl3

3.45 g Al

Alg 27.0

Almol

We must always convert to moles.Now use the molar ratio.

Almol 2

AlClmol 2 3

Now use the molar mass to convert to grams.

3

3

AlClmol

AlClg 133.317.0

Units match

gram to gram conversions

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Molarity

Molarity is a term used to express concentration. The units of molarity are moles per liter (It is abbreviated as a capital M)

When working problems, it is a good idea to change M into its units.

mL 1000

moles

Liter

moles M

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37

A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution.

What type of problem(s) is this?

Molarity followed by dilution.

Solutions

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A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution.

1st:= mol L

3.73 g

g 133.4

mol

200.0 x 10-3 L0.140

2nd: M1V1 = M2V2

(0.140 M)(10.0 mL) = (? M)(100.0 mL)0.0140 M = M2

molar mass of AlCl3

dilution formula

final concentration

Solutions

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40

50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used?

H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g)

Solution Stoichiometry

41

50.0 mL6.0 M

L

mol 6.0

? g

Look! A conversion factor!

50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used?

H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g)

Solution Stoichiometry

=

Our Goal

42

50.0 mL6.0 M

L

mol 6.0

? g

50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used?

H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g)

Solution Stoichiometry

=

Our Goal

= g NaHCO3

H2SO4

50.0 mL

1000mL

SOH mol 6.0

42SOH

42

1 molH2SO4

NaHCO3

2 molNaHCO3

84.0 gmolNaHCO3

50.4

43

44

Solution Stoichiometry:

Determine how many mL of 0.102 M NaOH solution are needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.

First write a balancedEquation.

____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 1

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Solution Stoichiometry:

Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.

Now, let’s get organized. Place numerical Information and accompanying UNITS below each compound.

____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 1

0.102 ML

mol

? mL

35.0 mL

mL 1000

mol 0.125

L

mol 0.125

Since 1 L = 1000 mL, we can use this to save on the number of conversions

Our Goal

46

Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.

Now let’s get to work converting.

____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 1

0.102 ML

mol

? mL

35.0 mL

mL1000

mol 0.125

L

mol 0.125

= mL NaOH

H2SO4

35.0 mL H2SO4

0.125 mol 1000 mL H2SO4

NaOH2 mol1 mol H2SO4

1000 mL NaOH0.102 mol NaOH

85.8

Units Match

Solution Stoichiometry:

shortcut

47

48

What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2?

1st write out a balanced chemicalequation

Solution Stoichiometry

49

What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2?

2HCl(aq) + Ba(OH)2(aq) 2H2O(l) + BaCl2

0.40 M 47.1 mL0.75 M? mL

= mL HCl

Ba(OH)2

47.1 mL

2

2

Ba(OH)

Ba(OH)

mL 1000

0.75mol

1 mol Ba(OH)2

HCl2 mol

0.40 mol HCl

HCl1000 mL 176

Units match

Solution Stoichiometry

50

51

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Solution Stochiometry Problem:

A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?

First write a balanced chemical reaction.

____HCl(aq) + ____Ba(OH)2(aq) ____H2O(l) + ____BaCl2(aq)2 1 2 1

23.28 mL

0.135 mol L

25.00 mL

? mol L

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Solution Stochiometry Problem:

A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?

____HCl(aq) + ____Ba(OH)2(aq) ____H2O(l) + ____BaCl2(aq)2 1 2 1

23.28 mL

0.135 mol L

25.00 mL

? mol L

= mol Ba(OH)2

L Ba(OH)225.00 x 10-3 L Ba(OH)2

Units Already Match on Bottom!

HClmL 23.28

HCl

HCl

mL 1000

mol 0.135

HCl

Ba(OH)

mol 2

mol l2 0.0629

Units match on top!

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55

48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution.

We must first write a balanced equation.

Solution Stochiometry Problem:

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48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution.

Ca(OH)2(aq) + HNO3(aq) H2O(l) + Ca(NO3)2(aq)2 248.0 mL 19.2 mL

0.385 ML

mol 0.385

= mol(Ca(OH)2)

L (Ca(OH)2)

19.2 mLHNO3

3

3

HNO

HNO

mL 1000

mol0.385

3

2

HNO 2mol

Ca(OH) 1mol

48.0 x 10-3L

? M

units match!

0.0770

Solution Stochiometry Problem:

57

58

Limiting/Excess/ Reactant and Theoretical Yield Problems :

Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?

b. Determine the limiting reactant.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

First copy down the the BALANCED equation!

Now place numerical the information below the compounds.

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Limiting/Excess/ Reactant and Theoretical Yield Problems :

Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?

b. Determine the limiting reactant.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 0.15 mol 0.10 mol ? moles

Two starting amounts? Where do we start?

Hide

one

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Limiting/Excess/ Reactant and Theoretical Yield Problems :

Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?b. Determine the limiting reactant.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 0.15 mol 0.10 mol ? molesHide

Based on:KO2 = mol O2

0.15 mol KO2

2

2

KO 4mol

O mol30.1125

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Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?b. Determine the limiting reactant.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 0.15 mol 0.10 mol ? moles

Based on:KO2 = mol O2

0.15 mol KO2

2

2

KO 4mol

O mol30.1125

Hide

Based on: H2O

= mol O20.10 mol H2O

OH 2mol

O mol3

2

2 0.150

Limiting/Excess/ Reactant and Theoretical Yield Problems :

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Limiting/Excess/ Reactant and Theoretical Yield Problems :

Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?

Determine the limiting reactant.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 0.15 mol 0.10 mol ? moles

Based on:KO2 = mol O2

0.15 mol KO2

2

2

KO 4mol

O mol30.1125

Based on: H2O

= mol O20.10 mol H2O

OH 2mol

O mol3

2

2 0.150

What is the theoretical yield? Hint: Which is the smallest amount? The is based upon the limiting reactant?

It was limited by theamount of KO2.

H2O = excess (XS) reactant!

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Theoretical yield vs. Actual yield

Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 12.3 grams of product were recovered. Determine the % yield.

Theoretical yield = 19.5 g based on limiting reactantActual yield = 12.3 g experimentally recovered

100x yield ltheoretica

yield actual yield %

yield 63.1% 100x 19.5

12.3 yield %

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4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 120.0 g 47.0 g ? gHide one

Based on:KO2

= g O2 120.0 g KO2

g1.71

mol2

2

KO 4mol

O mol3

2

2

O mol

O g0.32 40.51

Limiting/Excess Reactant Problem with % Yield

65

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 120.0 g 47.0 g ? g

Based on:KO2

= g O2 120.0 g KO2

g1.71

mol2

2

KO 4mol

O mol3

2

2

O mol

O g0.32 40.51

Based on:H2O

= g O2

Question if only 35.2 g of O2 were recovered, what was the percent yield?

yield 86.9% 100x 51.40

2.35 100x

ltheoretica

actual

Hide

47.0 g H2O

OH g 02.18

OH mol

2

2

OH mol 2

O mol 3

2

2

2

2

O mol

O g0.32 125.3

Limiting/Excess Reactant Problem with % Yield

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If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 120.0 g 47.0 g ? g

Based on:KO2

= g O2 120.0 g KO2

g1.71

mol2

2

KO 4mol

O mol3

2

2

O mol

O g0.32 40.51

Based on:H2O

= g O247.0 g H2O

OH g 02.18

OH mol

2

2

OH mol 2

O mol 3

2

2

2

2

O mol

O g0.32 125.3

Determine how many grams of Water were left over.The Difference between the above amounts is directly RELATED to the XS H2O.

125.3 - 40.51 = 84.79 g of O2 that could have been formed from the XS water.

= g XS H2O84.79 g O2

2

2

O g 32.0

O mol

2

2

O mol 3

OH mol 2

OH mol 1

OH g 02.18

2

2 31.83

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68

Calculate the molarity of a solution prepared by dissolving 25.6 grams of Al(NO3)3 in 455 mL of solution.

L

mol 0.264

L 10x 455g 213

moleg 25.63-

After you have worked the problem, click here to see setup answer

Try this problem (then check your answer):

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