Chapter #3 Chemical Composition. Chemical Reactions A chemical reactions is an abbreviated way to show a physical orchemical change A chemical change

Chapter #3Chemical Composition

Chemical Reactions• A chemical reactions is an abbreviated way to

show a physical orchemical change• A chemical change alters the physical and

chemical properties of a substance• Factors that indicate a chemical change

– Change in color– Temperature change– Change in odor– Change in taste (we do not taste chemicals)

• Reactions always contain an arrow that separates the reactants from the products

Reactants Products

Types of Chemical Reactions

• Combination reaction (synthesis)– Elements for reactants– Examples:

H2 + O2 H2O

N2 + H2 NH3

Al + O2 Al2O3

The Law of Conservation of matter, states matter cannot be created nor destroyed, that means equations must be balanced.


Balance the first equation

H2 + O2 H2O

Note two oxygen atoms on the reactant side and only one on the product side, therefore place a two in front of water

Combination reaction Continued



H2 + O2 2H2O


The two now doubles everything in water, thus 4 hydrogen and 2 oxygen. Now place a 2 in front of hydrogen.



2H2 + O2 2H2O


The two now doubles everything in water, thus 4 hydrogen and 2 oxygen. Now place a 2 in front of hydrogen.


Now balance the second equation

N2 + H2 NH3

Note two nigrogen atoms on the reactant side and only one on the product side.

Place a 2 in front of ammonia



N2 + H2 2NH3

Note two nitrogen atoms on the reactant side and only one on the product side.

Place a 2 in front of ammonia. This makes 2 nitrogen atoms and 6 hydrogen atoms. Now place a 3 in front of hydrogen to balance hydrogen atoms.



N2 + 3 H2 2NH3

Note two nitrogen atoms on the reactant side and only one on the product side.

Place a 2 in front of ammonia. This makes 2 nitrogen atoms and 6 hydrogen atoms. Now place a 3 in front of hydrogen to balance hydrogen atoms.


• Decomposition Reaction– Compounds form simpler compounds or

elements.– Examples

H2O H2 + O2


• Decomposition Reaction– Compounds form simpler compounds or

elements.– Examples

2H2O H2 + O2

Types of Chemical Reactions• Decomposition Reaction

– Compounds form simpler compounds or elements.

– Examples

2H2O 2H2 + O2

• Notice decomposition reactions are the opposite of combination reactions


Single Replacement reactions have an element and a compound for reactants.

Example:

Zn + HCl

How do we predict the products? Trade places with the metal or nonmetal with the metal or nonmetal in the compound



Example:

Zn + HCl

How do we predict the products? Trade places with the metal or nonmetal with the metal or nonmetal in the compound



Example:

Zn + HCl ZnCl + H

Now make the products stable. Slide with Clyde



Example:

Zn + HCl ZnCl2 + H2

Now make the products stable. Slide with Clyde



Example:

Zn + HCl ZnCl2 + H2

Now make the products stable.

Now Balance



Example:

Zn + 2HCl ZnCl2 + H2

Now make the products stable.

Now Balance



Another Example:

Cl2 + MgBr2

How do we predict the products? Trade places with the metal or nonmetal with the metal or nonmetal in the compound. In this case we are trading nonmetals



Another Example:

Cl2 + MgBr2 Br + MgCl




Another Example:

Cl2 + MgBr2 Br2 + MgCl2



Double Replacement reactions contain compounds as reactants.

HCl + Ca(OH)2 CaCl + HOH

Check formulas, and slide with Clyde when necessary



HCl + Ca(OH)2 CaCl2 + HOH




2HCl + Ca(OH)2 CaCl2 + 2HOH


Now Balance!


Combustion Reactions occur when an element or compound combine with oxygen to produce oxides of each element.

H2 + O2

CH4 + O2

What is the oxide of hydrogen?



H2 + O2

CH4 + O2

What is the oxide of hydrogen? Water



H2 + O2 H2O

CH4 + O2


And the oxide of carbon?



H2 + O2 H2O

CH4 + O2 CO2 + H2O


And the oxide of carbon? Carbon dioxide



2H2 + O2 2H2O

CH4 + O2 CO2 + H2O

Now balance



2H2 + O2 2H2O

CH4 + O2 CO2 + 2H2O

Now balance



2H2 + O2 2H2O

CH4 + 2O2 CO2 + 2H2O

Now balance

Ionic Solution Formation

KCN (S) K+ (aq) + CN- (aq) Ionic equation

Note: Not all ionic solutes are soluble in water.

How can we tell if an ionic solute is soluble in water?

Ionic Solution Formation

KCN (S) K+ (aq) + CN- (aq) Ionic equation

Note: Not all ionic solutes are soluble in water.

How can we tell if an ionic solute is soluble in water?The solubility rules gives ionic solubility.

There are some more specific rules that allows us to better estimate the solubility of ionic compounds.

You will be given these if you need them.

Solubility Rules

Ionic EquationsUsing the solubility rules write the formula equation, the ionic equation and the net ionic equation when aqueous silver nitrate is combined with aqueous sodium chloride.

AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Formula Equation


AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Formula EquationAg+ (aq) + NO3

- (aq) + Na+ (aq) + Cl- (aq) AgCl (s) + Na+ (aq) + NO3- (aq)

Ionic Equation




Ionic Equation

Spectator ions are ions that are identical on the reactants and products side of the equation.




Ionic EquationSpectator ions are ions that are identical on the reactants and products side of the equation. Place a around the spectator ions.




Ionic EquationSpectator ions are ions that are identical on the reactants and products side of the equation. Place a around the spectator ions.




Ionic EquationSpectator ions are ions that are identical on the reactants and products side of the equation. Eliminating the spectator ions produces the netionic equation.

Ag+ (aq) + Cl- (aq) AgCl (s) Net ionic equation

Yet Another Ionic EquationWrite the formula, ionic and net ionic equationswhen aqueous sodium chloride combines withaqueous calcium bromide.


NaCl(aq) + CaBr2 CaCl2 + NaBrNow balance


NaCl(aq) + CaBr2 CaCl2 + NaBr(aq)Now balance

22


NaCl(aq) + CaBr2 CaCl2(aq) + NaBr(aq)Now balance

22



22

2Na+(aq) + 2Cl-(aq) + Ca2+(aq) + 2Br-(aq) Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) +2 Br-(aq)Ionic equation



22

2Na+(aq) + 2Cl-(aq) + Ca2+(aq) + 2Br-(aq) Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) +2 Br-(aq)Ionic equation

No net ionic equation No Reaction (NR)


REDOX reactions where the oxidation number changes from reactants to products.

Oxidation is when the oxidation number increases, by losing of electrons.

Reduction is when the oxidation number decreases by gaining electrons.

Consider the following equation:

H2 + O2 H2O

What are the oxidation numbers of hydrogen and oxygen?

0 0






H2 + O2 H2O


REDOX REACTIONS

H2 + O2 H2O

How about hydrogen and oxygen in water?

00 2(1+) 2- = 0

REDOX REACTIONS

H2 + O2 H2O

How about hydrogen and oxygen in water?Oxidation is caused by the oxygen molecule,

so it is referred to as the oxidizing agent (OA)

Reduction is caused by the hydrogen molecule, so it is referred to as the reducing agent (RA)

00 2(1+) 2- = 0

reducedoxidized

REDOX REACTIONSNote:

• All of the previously discussed reactions are REDOX except the double replacement reactions.

• The number of electrons lost is equal to the number of electrons gained in a reaction. Why?

• Most elements have variable oxidation numbers, except for hydrogen, oxygen, and the memorized polyatomic ions.

REDOX REACTIONSOxidation numbers for a compound must

add up to equal zero, while the oxidation numbers for a polyatomic ion must up to equal the charge of that ion.

Consider the following chlorine compounds

HClO4, HClO3, HClO2, HClO, Cl2, HCl

What is the oxidation number of chlorine in each of these compounds, assuming

H 1+ and oxygen is 2-

1+ 4(2-)=0





What is the oxidation number of chlorine in each of these compounds, assuming H is 1+ and oxygen is 2-

1+ 4(2-)=07+






1+ 4(2-)=07+ 5+






1+ 4(2-)=07+ 5+ 3+






1+ 4(2-)=07+ 5+ 3+ 1+






1+ 4(2-)=07+ 5+ 3+ 1+ 0






1+ 4(2-)=07+ 5+ 3+ 1+ 0 1-

REDOX REACTIONS

How about sulfur in SO3 2-

3(2-)=2-

REDOX REACTIONS


How about carbon in C6H12O6

3(2-)=2-4+

12(1+) +6(2-)=0

REDOX REACTIONS


How about carbon in C6H12O6

3(2-)=2-4+

12(1+) +6(2-)=00 +






H2 + O2 H2O







H2 + O2 H2O


0 0

REDOX REACTIONS

H2 + O2 H2O

How about hydrogen and oxygen in water?

00 2(1-) 2- = 0

REDOX REACTIONS

H2 + O2 H2O

How about hydrogen and oxygen in water?Oxidation is caused by the oxygen molecule,

so it is referred to as the oxidizing agent (OA)

Reduction is caused by the hydrogen molecule, so it is referred to as the reducing agent (RA)

00 2(1+) 2- = 0

reducedoxidized

REDOX REACTIONSNote:

• All of the previously discussed reactions are REDOX except the double replacement reactions.

• The number of electrons lost is equal to the number of electrons gained in a reaction. Why?

• Most elements have variable oxidation numbers, except for hydrogen, oxygen, and the memorized polyatomic ions.

Balancing Redox ReactionsI. Oxidation Number Method

a. Assign oxidation numbers to each elementb. Determine the elements oxidized and reducedc. Balance the atoms that are oxidized and reducedd. Balance the electrons lost or gained, to conform to the Law of

Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation.

e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both sides of the

equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.

HNO3 + Cu2O → Cu(NO3)2 + NO + H2O

Balancing REDOX ReactionsI. Oxidation Number Method



e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both

sides of the equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.

HNO3 + Cu2O → Cu(NO3)2 + NO + H2O1+ 3(2-)=0?




e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both sides of

the equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.

HNO3 + Cu2O → Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(?)+






HNO3 + Cu2O → Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+

Balancing Redox ReactionsI. Oxidation Number Method





HNO3 + Cu2O → Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0? +






HNO3 + Cu2O → Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=02+ +






HNO3 + Cu2O → Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 ? + 2- =0






HNO3 + Cu2O → Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 2 + 2- =0






HNO3 + Cu2O → Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 2 + 2- =02+

oxidized

reduced






HNO3 + Cu2O → 2 Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 2 + 2- =02+

oxidizedreduced






HNO3 + 3 Cu2O → 3 (2) Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 2 + 2- =02+

Oxidized3( -2e)Reduced 2(+3)e






2HNO3 + 3 Cu2O → 3 (2) Cu(NO3)2 + 2NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 2 + 2- =02+







2HNO3 + 3 Cu2O → 3 (2) Cu(NO3)2 + 2NO + H2O

1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 2 + 2- =02+







14HNO3 + 3 Cu2O → 3 (2) Cu(NO3)2 + 2NO + 7 H2O

1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 2 + 2- =02+







14 HNO3 + 3 Cu2O → 6 Cu(NO3)2 + 2 NO + 7 H2O

1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 2 + 2- =02+


OX # BALANCING EXAMPLE

MnO4 - + Cl- → Mn2+ + Cl2


MnO4 - + Cl- → Mn2+ + Cl2

7+


MnO4 - + Cl- → Mn2+ + Cl2

7+ 1-


MnO4 - + Cl- → Mn2+ + Cl2

7+ 1- 0


MnO4 - + Cl- → Mn2+ + Cl2

7+ 1- 0


MnO4 - + Cl- → Mn2+ + Cl2

7+ 1- 0

reducedoxidized


MnO4 - + Cl- → Mn2+ + Cl2

7+ 1- 0

reducedoxidized

Step C, balance atoms oxidized or reduced


MnO4 - + Cl- → Mn2+ + Cl2

7+ 1- 0

reducedoxidized

Step C, balance atoms oxidized or reduced

2


MnO4 - + Cl- → Mn2+ + Cl2

7+ 1- 0

reducedoxidized

Step d, balance electrons lost or gained. common denominator between 5 and

2 is 10. Therefore multiply Mn on both sides of the equation by two and Cl on both sides by 5.

2

+ 5 e-- 2 e-


MnO4 - + Cl- → Mn2+ + Cl2

7+ 1- 0

reducedoxidized

Step d, balance electrons lost or gained. The common denominator between 5 and

2 is 10. Therefore multiply Mn on both sides of the equation by 2 and Cl on both sides by 5.

5(2)

+ 5 e-- 2 e-

2 2 5


MnO4 - + Cl- → Mn2+ + Cl2

7+ 1- 0

reducedoxidized

Step e, balance remaining elements by inspection. There are 8 oxygen atoms on the left. Oxygen is balanced by adding water to the appropriate side. In this case since there are 8 oxygen atoms on the reactant side which requires adding 8 water molecules to the product side of the equation.

5(2)

+ 5 e-- 2 e-

2 2 5

OX # BALANCING EXAMPLE MnO4

- + Cl- → Mn2+ + Cl27+ 1- 0

reducedoxidized

Step e, balance remaining elements by inspection. There are 8 oxygen atoms on the left. Oxygen is balanced by adding water to the appropriate side. In this case since there are 8 oxygen atoms on the reactant side which requires adding 8 water molecules to the product side of the equation. Now the hydrogen atoms need to be balanced by adding 16 H+ to the reactant side.

10

+ 5 e-- 2 e-

2 2 5 + 8H2O

OX # BALANCING EXAMPLE MnO4

-+ Cl- → Mn2++ Cl27+ 1- 0

reducedoxidized

Step e, balance remaining elements by inspection. There are 8 oxygen atoms on the left. Oxygen is balanced by adding water to the appropriate side. In this case since there are 8 oxygen atoms on the reactant side which requires adding 8 water molecules to the product side of the equation. Now the hydrogen atoms need to be balanced by adding 16 H+ to the reactant side.

10

+ 5 e-- 2 e-

2

2 5 + 8H2O16 H++

Balancing REDOX Equationsby

The Half Reaction Method

Half Reaction Steps1. Write separate equations (Half-reactions) for oxidized and reduced

substances.2. For each half-reaction balance all elements, except hydrogen and

oxygena. Balance oxygen using H2O

b. Balance hydrogen using H+

c. Balance charge in each half-reaction by adding electrons (reduction), or removing electrons (oxidation) to the appropriate half reaction.

3. Multiply each half-reaction by an integer so that the number of electrons lost equal the number of electrons gained

a. Add half-reactions, and simplifyb. For basic reactions add the same number of OH- ions to both

sides of the equation as there are H+ ions.c. Combine H+ and OH- ions to make waterd. Simplify again if necessary.

Half Reaction Example

MnO4- + Fe2+ → Mn2+ + Fe3+


MnO4- + Fe2+ → Mn2+ + Fe3+

Step 1, Write half reactions


MnO4- + Fe2+ → Mn2+ + Fe3+

Step 1, Write half reactions

MnO4- → Mn2+

Fe2+ → Fe3+


MnO4- + Fe2+ → Mn2+ + Fe3+

Step 2a, Balance Oxygen by adding water.

MnO4- → Mn2+

Fe2+ → Fe3+

+ 4 H2O


MnO4- + Fe2+ → Mn2+ + Fe3+

Step 2b, Balance hydrogen by adding H+.

MnO4- → Mn2+

Fe2+ → Fe3+

+ 4 H2O8 H+ +


MnO4- + Fe2+ → Mn2+ + Fe3+

Step 2c, Balance charge by adding/removing e’s

MnO4- → Mn2+

Fe2+ → Fe3+

+ 4 H2O8 H+ +

In the top half equation the reactants have 7+ and the products 2+, adding 5 e’s to the reactant side gives 2+ on both sides.In the bottom half equation the reactants have 2+ and the products have 2+, removing 1 e from the reactant side gives 2+ on both sides.


MnO4- + Fe2+ → Mn2+ + Fe3+


MnO4- → Mn2+

Fe2+ → Fe3+

+ 4 H2O8 H+ +


+ 5e-


MnO4- + Fe2+ → Mn2+ + Fe3+


MnO4- → Mn2+

Fe2+ → Fe3+

+ 4 H2O8 H+ +


+ 5e-

- e-


MnO4- + Fe2+ → Mn2+ + Fe3+


MnO4- → Mn2+

Fe2+ → Fe3+

+ 4 H2O8 H+ +


+ 5e-

- e-


MnO4- + Fe2+ → Mn2+ + Fe3+

Step 3, The common denominator between 5 and 1 is 5. Multiply the bottom half equation by 5

MnO4- → Mn2+

Fe2+ → Fe3+

+ 4 H2O8 H+ + + 5e-

- e-


MnO4- + Fe2+ → Mn2+ + Fe3+

Step 4, Add the two half equations together

MnO4- → Mn2+

Fe2+ → Fe3+

+ 4 H2O8 H+ + + 5e-

- e-5( )


MnO4- + Fe2+ → Mn2+ + Fe3+

Step 4, Add the two half equations together

MnO4- → Mn2+

Fe2+ → Fe3+

+ 4 H2O8 H+ + + 5e-

- e-5( )

8 H+ + MnO4- 5 Fe3++ 5 Fe2+ + Mn2+ + 4 H2O→

Other REDOX ExamplesHNO2 + Cr2O7

2- → Cr2+ + NO3- (acidic)

CN- + MnO4- → CNO- + MnO2 (basic)

Al(s) + OH- (aq) → Al(OH)4- (aq) + H2 (g) (acidic or basic)

Cl2 (g) → Cl- (aq) + ClO- (aq) (basic)

Ag (s) + CN- + O2 → Ag(CN)2 - (aq) (basic)

Real Life Examples of REDOX•REDOX reactions can be used to generate electricity.

•REDOX reactions can be used to protect metals from oxidation.

•REDOX reactions can be used to plate metals on to other metals or surfaces.

The Chemical Package

• The baker uses a package called the dozen. All dozen packages contain 12 objects.

• The stationary store uses a package called a ream, which contains 500 sheets of paper.

• So what is the chemistry package?

About Packages

The Chemical Package

• The baker uses a package called the dozen. All dozen packages contain 12 objects.

• The stationary store uses a package called a ream, which contains 500 sheets of paper.

• So what is the chemistry package? Well, it is called the mole (Latin for heap).

About Packages

Each of the above packages contain a number of objects that are convenient to work with, for that particular discipline.

The atomic weights listed on the periodic chart are the weights of a mole of atoms. For example a mole of hydrogen atoms weighs 1.00797 g and a mole of carbon atoms weighs 12.01 g which are weighted averages of the natural abundance of isotopes for that element.

The MoleA mole contains 6.022X1023 particles, which is the number of carbon-12 atoms that will give a mass of 12.00 grams, which is a convenient number of atoms to work with in the chemistry laboratory.

Moles of ObjectsSuppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows?

Moles of Objects

Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows?Would cover the entire 50 states 60 miles deep

Moles of Objects


How about a mole of computer paper instead of a ream of computer paper, how far would that stretch?

Moles of Objects


How about a mole of computer paper instead of a ream of computer paper, how far would that stretch? Way past the planet Pluto!

To calculate the molar mass of a compound we sum together the atomic weights of the atoms that make up the formula of the compound. This is called the formula weight (MW, M).

Formula weights are the sum of atomic weights of atoms making up the formula.

The following outlines how to find the formula weight of water

symbol weight numberHO

1.0116.0

21X

X== 2.02

16.018.0 g/mole

Formula Weight Calculation

Percent Composition

Find the formula weight and the percent composition of

glucose (C6H12O6)

symbol weight number

HO

C

16.01.01

12.0

6

12

6

x

x

x

=

=

=

72.0

12.1296.0

180.1 g/mole

%C =

%H =

%O =

72.0

12.12

96.0

180.1

180.1

180.1

X =

X =

X =

40.0 %C

6.73 %H

53.3 %O

A mole of glucose (C6H12O6) contains 6.022 X 1023 molecules of glucose. And 6 X 6.022 X 1023 atoms of C. Since a mole is 6.022 X 1023 particles then a mole of glucose must contain 6 moles of C atoms.

How many moles of hydrogen atoms are contained in a mole of glucose?

In 5 moles of H2SO4 how many moles of oxygen atoms is there?

Mole Concepts


How many moles of hydrogen atoms are contained in a mole of glucose? 12 Moles of hydrogen.

How many moles of oxygen and hydrogen are in one mole of H2O contains:

Mole Concepts



How many moles of oxygen and hydrogen are in one mole of H2O contains:One mole of oxygen atoms Two moles of hydrogen atoms

Mole Concepts





Mole Concepts





20 moles of O atoms.

Mole Concepts

In 50.0g of H2SO4 how many moles of sulfuric acid are there?

Mole Conversions

50.0g of H2SO4


Mole Conversions

50.0g of H2SO4 =98.0g of H2SO4

mole H2SO4


Mole Conversions

50.0g of H2SO4 = 0.510 mole H2SO498.0g of H2SO4

mole H2SO4

In 50.0g of H2SO4 how many moles of oxygen atoms are there?

Mole Conversions


Mole Conversions

50.0g of H2SO4 =


Mole Conversions

50.0g of H2SO4 =98.0g of H2SO4

mole H2SO4


Mole Conversions

50.0g of H2SO4 =mole H2SO4

4mole O98.0g of H2SO4

mole H2SO4


Mole Conversions

50.0g of H2SO4 =mole H2SO4

4mole O 2.04 mole O98.0g of H2SO4

mole H2SO4

In 5 moles of H2SO4 how many atoms of oxygen are present?

Mole Conversions


Mole Conversions

5 moles H2SO4 =


Mole Conversions

5 moles H2SO4

mole H2SO4

4 mole O


Mole Conversions

5 moles H2SO4

mole H2SO4

4 mole O 6.02 x 1023 atoms O mole O

=

1.20 x 1025 atoms

Empirical FormulasEmpirical formula is the smallest whole number ratio between atoms and can be calculated from the percent composition.

Molecular formulas happen to be the exact number of atoms making up a molecule, and may or may no be the simplest whole number ratio. Molecular formulas are whole number multiples of the empirical formula.

Empirical Formula Steps1. Assume 100 g of compound.2. Convert percent to a mass number.3. Convert the mass to moles.4. Divide each mole number by the smallest mole

number.5. Rounding:

a. If the decimal is ≤ 0.1, then drop the decimalsb. If the decimal is ≥0.9, then round up.c. All other decimal need to be multiplied by a whole

number until roundable.

Empirical Formula ExampleA compound is composed of 75.0% C and 15.0% H. Find its empirical formula.

Step #1 Assume 100 g of compound

75.0 g C

25.0 g H


Step #2 Convert grams to moles.

75.0 g C

25.0 g H12.01 g

Mole C

1.008 g H

Mole H

= 6.225 mole C

= 24.802 mole


Step #3 Divide each mole number by the smallest.

75.0 g C

25.0 g H12.01 g

Mole C

1.008 g H

Mole H

= 6.225 mole C

= 24.802 mole

6.225 6.2256.22524.802

= 1.00 = 3.98


Step #4 Rounding; Decimal ≤ 0.1, drop decimals

75.0 g C

25.0 g H12.01 g

Mole C

1.008 g H

Mole H

= 6.225 mole C

= 24.802 mole

6.225 6.2256.22524.802

= 1.00 = 3.98


Step #4 Rounding; Decimal ≤ 0.1, drop decimals

75.0 g C

25.0 g H12.01 g

Mole C

1.008 g H

Mole H

= 6.225 mole C

= 24.802 mole

6.225 6.2256.22524.802

= 1 C = 3.98


Step #4 Rounding; Decimal ≥ 0.9, round up

75.0 g C

25.0 g H12.01 g

Mole C

1.008 g H

Mole H

= 6.225 mole C

= 24.802 mole

6.225 6.2256.22524.802

= 1 C = 3.98



75.0 g C

25.0 g H12.01 g

Mole C

1.008 g H

Mole H

= 6.225 mole C

= 24.802 mole

6.225 6.2256.22524.802

= 1 C = 3.98



75.0 g C

25.0 g H12.01 g

Mole C

1.008 g H

Mole H

= 6.225 mole C

= 24.802 mole

6.225 6.2256.22524.802

= 1 C = 4 H



75.0 g C

25.0 g H12.01 g

Mole C

1.008 g H

Mole H

= 6.225 mole C

= 24.802 mole

6.225 6.2256.22524.802

= 1 C = 4 H

Empirical Formula = CH4

Molecular Formulas

Empirical formula, is the smallest ratio between atoms in a molecular or formula unit.

Molecular formula, is the exact number of atoms in a molecule; a whole number multiple of an empirical formula

Possible Molecular Formulas

Assume an empirical formula of C3H5O

Empirical formula Integer Molecular Formula

C3H5O 1

2

3

4

5

C3H5O

C3H5O

C3H5O

C3H5O

C3H5O




C3H5O 1

2

3

4

5

C3H5O

C3H5O

C3H5O

C3H5O

C3H5O

C6H10O




C3H5O 1

2

3

4

5

C3H5O

C3H5O

C3H5O

C3H5O

C3H5O

C6H10O2

C9H15O3




C3H5O 1

2

3

4

5

C3H5O

C3H5O

C3H5O

C3H5O

C3H5O

C6H10O2

C9H15O3

C12H20O4

C15H25O5

Sample ProblemCalculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole


Step #1 Assume 100g of compound

83.6 g C

16.3 g H


Step #2 Convert grams to moles

83.6 g C

16.3 g H12.01 g Cmole

mole1.008 g H


Step #2 Convert grams to moles

83.6 g C

16.3 g H12.01 g Cmole

mole1.008 g H

= 6.961 mole

= 16.17 mole



83.6 g C

16.3 g H12.01 g Cmole

mole1.008 g H

= 6.961 mole

= 16.17 mole



83.6 g C

16.3 g H12.01 g Cmole

mole1.008 g H

= 6.961 mole

= 16.17 mole

6.961 6.961

= 1.00

= 2.32


Step #4 Round if---Not Roundable

83.6 g C

16.3 g H12.01 g Cmole

mole1.008 g H

= 6.961 mole

= 16.17 mole

6.961 6.961

= 1.00

= 2.32

Step #4, Multiply by an integer until roundable

1.00 X 3 = 3

2.32 X 3 = 7Empirical formula C3H7

Molecular Formula IntegerDivide empirical weight into molecular weight

3x12 + 7x1 =43

43 862

Now multiply the empirical formula by 2

Molecular Formula IntegerDivide empirical weight into molecular weight

3x12 + 7x1 =43

43 862

Now multiply the empirical formula by 2

Molecular Formula is C6 H14

Stoichiometry

Sotichiometery is the process of converting quantities of reactants or products to other participants of a chemical or physical change using the coefficients of a balanced equation.

STOICHIOMETRY

Stoichiometry is the use of balanced chemical equations in the conversion process.

Example

Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required.

STOICHIOMETRY


Example


2 H2 + O2 2 H2O

STOICHIOMETRY


Examples


2 H2 + O2 2H2O6.33 g H2

STOICHIOMETRY


Example


2 H2 + O2 2 H2O6.33 g H2

2.016 g H2O

Mole H2

STOICHIOMETRY


Example


2 H2 + O2 2 H2O6.33 g H2

2.016 g H2O

Mole H2 2 Mole H2O

2 Mole H2

STOICHIOMETRY


Example


2 H2 + O2 2 H2O6.33 g H2

2.016 g H2O

Mole H2 2 Mole H2O

2 Mole H2 Mole H2O

18.02 g H2O

STOICHIOMETRY


Examples


2 H2 + O2 2 H2O6.33 g H2

2.016 g H2O

Mole H2 2 Mole H2O

2 Mole H2 Mole H2O

18.02 g H2O= 56.6 g H2O

Excess and Limiting

Reactants are substances that can be changed into something else. For example, nails and boards are reactants for carpenters, while thread and fabric are reactants for the seamstress. And for a chemist hydrogen and oxygen are reactants for making water.

Building HousesSuppose, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?

Building HousesOk, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?

Yes, only one house!


What reactant is in excess? And how many more houses could we use if we had enough boards?


What reactant is in excess? And how many more houses could we use if we have enough boards?


What reactant is in excess? And how many more houses could we use if we have enough boards?

Yes, nails are in excess!

Building HousesOk, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?What reactant is in excess? And how many more houses could we use if we have enough boards?Yes, nails are in excess! Nine more houses if we have an adequate amount of boards.

Making WaterIf we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?


Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.



2 H2 + O2 2 H2O

10.0 g O2



2 H2 + O2 2 H2O

10.0 g O2

32.0 g O2

mole O2



2 H2 + O2 2 H2O

10.0 g O2

32.0 g O2

mole O2

mole O2

2 mole H2



2 H2 + O2 2 H2O

10.0 g O2

32.0 g O2

mole O2

mole O2

2 mole H2

mole H2

2.02 g H2



2 H2 + O2 2 H2O

10.0 g O2

32.0 g O2

mole O2

mole O2

2 mole H2

mole H2

2.02 g H2 = 1.26 g H2

Making WaterOnly 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen 10.0 -1.26 = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.


10.0 g O2mole O2

32.0 g O2


10.0 g O2mole O2

32.0 g O2 mole O2

2 mole H2O


10.0 g O2mole O2

32.0 g O2 mole O2

2 mole H2O 18.0 g H2Omole H2O


10.0 g O2mole O2

32.0 g O2 mole O2

2 mole H2O 18.0 g H2Omole H2O

= 11.3 g H2O

Percentage YieldThe percent yield is a comparison of the laboratory answer to the correct answer which is determined by the conversion process. Suppose a student combined 10.0 g of oxygen and 10.0 g of hydrogen in the lab and recovered 8.66 g of water. What would be the percent yield?

Percentage YieldThe percent yield is a comparison of the laboratory answer to the correct answer which is determined by the conversion process. Suppose a student combined 10.0 g of oxygen and 10.0 g of hydrogen in the lab and recovered 8.66 g of water. What would be the percent yield?

percent yield = Yield (the lab amount)Theoretical Yield (by conversions)

X 100

percent yield = 8.6611.3

X 100 = 76.6%

Combustion AnalysisEmpirical formulas of hydrocarbons can be determined by combustion analysis. The complete combustion of a hydrocarbon produces carbon dioxide and water. Measuring the mass of the carbon dioxide and water produced can give the mass of hydrogen and carbon present in the compound. Subtracting the mass of carbon and hydrogen from the weight of the starting hydrocarbon gives the mass of the oxygen. Ascarite™ a commercial name for sodium or potassium hydroxide absorbs between 0-1 ppm of the carbon dioxide.

Combustion AnalysisAscarite the weight increase of Ascarite gives the mass of the carbon dioxide according to the following equation.

CO2 + 2 KOH K2CO3 + 2 H2O

Vitamin C is essential for the prevention of scurvy. Combustion of a 0.2000 g sample of this hydrocarbon, which may or may not contain oxygen, gave 0.2998 g of carbon dioxide and 0.0819 g of water. What is the empirical formula of vitamin C?

Combustion AnalysisVitamin C is essential for the prevention of scurvy. Combustion of a 0.2000 g sample of this hydrocarbon, which may or may not contain oxygen, gave 0.2998 g of carbon dioxide and 0.0819 g of water. What is the empirical formula of vitamin C?

First covert the mass of carbon dioxide and water into grams of carbon and of carbon and hydrogen. Subtract these masses from the sample weight, if the difference is zero then vitamin C does not contain any oxgen.

Combustion Analysis

0.2298 g CO2

Combustion Analysis

0.2298 g CO2

44.010 g CO2

Mole CO2

Combustion Analysis

0.2298 g CO2

44.010 g CO2

Mole CO2

Mole CO2

Mole C

Combustion Analysis

0.2298 g CO2

44.010 g CO2

Mole CO2

Mole CO2

Mole C

Mole C

12.011 g C= 0.08182 g C

Combustion Analysis

0.2298 g CO2

44.010 g CO2

Mole CO2

Mole CO2

Mole C

Mole C

12.011 g C= 0.08182 g C

0.0819 g H2O

18.02 g H2O

mole H2O

mole H2O

2 mole H

mole H

1.008 g H= 0.00916 g H

Combustion Analysis

0.2298 g CO2

44.010 g CO2

Mole CO2

Mole CO2

Mole C

Mole C

12.011 g C= 0.08182 g C

0.0819 g H2O

18.02 g H2O

mole H2O

mole H2O

2 mole H

mole H

1.008 g H= 0.00916 g H

0.2000 - 0.08182 - 0.00916 = 0.1090 g O

Combustion Analysis

0.2298 g CO2

44.010 g CO2

Mole CO2

Mole CO2

Mole C

Mole C

12.011 g C= 0.08182 g C

0.0819 g H2O

18.02 g H2O

mole H2O

mole H2O

2 mole H

mole H

1.008 g H= 0.00916 g H

0.2000 - 0.08182 - 0.00916 = 0.1090 g O

0.08182 g C

12.011 g C

Mole C= 0.006812 mole

0.00916 g H

1.008 g H

mole H= 0.00909 mole

0.1090 g O

16.00 g O

Mole O= 0.006813 mole

Combustion Analysis

0.2298 g CO2

44.010 g CO2

Mole CO2

Mole CO2

Mole C

Mole C

12.011 g C= 0.08182 g C

0.0819 g H2O

18.02 g H2O

mole H2O

mole H2O

2 mole H

mole H

1.008 g H= 0.00916 g H

0.2000 - 0.08182 - 0.00916 = 0.1090 g O

0.08182 g C

12.011 g C


0.00916 g H

1.008 g H


0.1090 g O

16.00 g O


0.006812 0.006812

0.006812 0.006813

0.006812 0.00909

= 1.000 X 3 = 3

= 1.000 X 3 = 3

= 1.333 X 3 = 4

Combustion Analysis

0.2298 g CO2

44.010 g CO2

Mole CO2

Mole CO2

Mole C

Mole C

12.011 g C= 0.08182 g C

0.0819 g H2O

18.02 g H2O

mole H2O

mole H2O

2 mole H

mole H

1.008 g H= 0.00916 g H

0.2000 - 0.08182 - 0.00916 = 0.1090 g O

0.08182 g C

12.011 g C


0.00916 g H

1.008 g H


0.1090 g O

16.00 g O


0.006812 0.006812

0.006812 0.006813

0.006812 0.00909

= 1.000 X 3 = 3

= 1.000 X 3 = 3

= 1.333 X 3 = 4

C3H4O3

The End

Documents

Chapter #3 Chemical Composition. Chemical Reactions A chemical reactions is an abbreviated way to show a physical orchemical change A chemical change