Understanding Chemical Equations: Mole, Mass Interpretation, and Balancing Redox, Study notes of Chemistry

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Chapter 5

Chemical Formulas

and

Chemical Reactions

Module 1

Physical Property

• (^) One that a sample of matter displays without changing its

composition or identity.

Describes physical attributes such as color, physical state,

hardness, and temperature.

Chemical Property

• (^) The ability of a sample of matter to undergo a change in

composition under stated conditions.

Describes the way matter acts during a chemical reaction.

Example: Ability of Zn to react with HCl to form hydrogen

and a salt (zinc chloride).

Physical and Chemical Properties

Sample Problem 1

Label each of the following as a physical or a chemical process:

(a) Rusting of iron.

(b) Condensation of water on a cold surface.

(c) Burning of paper.

(d) Pulverizing an aspirin.

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Sample Problem 2

Potassium is a soft, silver-colored metal that melts at 64

o C. It

reacts vigorously with water, with oxygen, and with chlorine.

Identify the physical and chemical properties stated in this problem.

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Physical and chemical Changes

Chemical Equations

Symbolic representation of a chemical reaction.

• (^) Chemical reaction: Oxygen plus hydrogen reacting to form water.
• (^) Chemical equation: 1 O

2

(g) + 2H 2

(g)  2 H 2

O (l)

Interpreting Chemical Equations

Chemical equation: 1 O 2

(g) + 2H 2

(g)  2 H 2

O (l)

Mole interpretation: 1mol O 2

• 2mol H 2

yield 2mol H 2

O

Molecular interpretation: 1molecule O 2

• 2molecules H 2

yield 2molecules H 2

O

Mass Interpretation: 32 grams O 2

• 4 grams H 2

yield 36 grams H 2

O

Chemical Equations

Reactants Products

State Coefficient Condition



Solution Continued

C is balanced and we now have 7 Os on the product side. So let’s

balance H.

C

3

H

8

+ O

2

 3 CO

2

+ 4 H

2

O

C = 3 C = 1 3

H = 8 H = 2 8

O = 2 O = 3 7 10

C and H are balanced and there are 10 Os on the product side. So

let’s balance O.

C

3

H

8

+ 5 O

2

 3 CO

2

+ 4 H

2

O (Balanced)

C = 3 C = 1 3

H = 8 H = 2 8

O = 2 10 O = 3 7 10

Sample Problem 3

Balance: Al + H 2

SO

4

Al 2

(SO

4

3

+ H

2

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Balancing Chemical Equations

Types of Chemical Reactions

Combination Reaction

• (^) One in which two or more reactants combine to form one product.
• Example: 2Mg(s) + O 2

(g)  2MgO(s)

Decomposition Reaction

• (^) One in which a single reactant breaks up to form two or more products.
• (^) Often occurs with the addition of heat.
• (^) Example: ZnCO 3

(s)  ZnO(s) + CO 2

(g)

Replacement Reaction

• (^) One in which an atom in a compound is replaced by another atom.
• Single Replacement Example: Cu + 2AgNO 3

Cu(NO 3

) 2

• 2Ag

• (^) Double Replacement Example: BaCl

2

• Na 2

SO 4

 BaSO 4

• 2NaCl

Combustion Reaction

• (^) Burning a substance in O 2

to produce heat and an oxide.

• Examples: 4Fe + 3O 2

 2Fe 2

O 3

and CH 4

• 2O 2

 CO 2

• 2H 2

O

Redox Reactions

Oxidation and Reduction

• (^) Occur simultaneously.

Total number of e

• loss must equal the total number of e - gained.

Oxidizing Agent

• (^) The substance that is reduced ( Fe

3+ ).

Reducing Agent

The substance that is oxidized (Al).

Example: Al(s) + Fe

3+  Al

3+

• Fe (s)

Oxidized Reduced

Reducing Oxidizing

Agent Agent

Rules

(1) The oxidation number of a free or uncombined element is zero.

Examples: Zn, H 2

, S

8

, and O 2

(2) The oxidation state of a monoatomic ion is equal to the charge on

the ion.

Examples: K

and Br

(3) The oxidation number of hydrogen is +1 when bonded to a

nonmetal and -1 when bonded to a metal.

Examples: HCl, NaH

Oxidation Number (State)

The charge assigned to an atom according to a prescribed set of rules.

Oxidation Numbers

Oxidation Number (State)

Illustrative Example

• (^) Determine the oxidation number of each element in C

2

H

4

O

2

and

SO

4

2- .

Solution

• (^) Consider C

2

H

4

O

2

. Remember that the oxidation numbers for a

neutral compound must add to zero.

• (^) There are 4 H’s. Hence the charge contribution of H is (+1)4= +4.

There are 2 O’s. Hence the charge contribution of O is 2(-2) = -4.

• (^) The oxidation number of C is therefore zero: C

2

H

4

O

2

Final check: 2C + 4H + 2O should give 2(0) + 4(+1) + 2(-2) = 0.

Oxidation Number (State)

Solution Continued

• (^) Consider SO 4

2- .

The net charge for this ion is -2. Thus, the sum of the oxidation

numbers for all atoms must equal -2, the charge of the ion.

Oxygen has a common oxidation number of -2. Thus, the total

charge contribution of oxygen is 4(-2) or -8.

• (^) Assigning a + 6 to sulfur will give a net charge of -2:

• (^) Thus, the oxidation number is -2 for O, and + 6 for S.

Sample Problem 4

Determine the oxidation number of each element in KClO 3

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Chemical Formulas

Chemical Formula

• (^) Symbolic representation of the relative number of atoms in a

molecule.

• (^) Example: H 2

O

2

Types of Chemical Formulas

(a) Molecular formula

• (^) Shows the actual number of different atoms in the molecule.

Example: C

6

H

12

O

6

(b) Empirical formula

• (^) Gives the simplest whole number ratio of atoms in the molecule.

Example: The empirical formula for C

6

H

12

O

6

is CH

2

O.

(c) Structural formula

Shows how atoms in the molecule are bonded.

• (^) Example:

Molecular and Formula Weights

Formula Weight (Mass)

• (^) Mass of a formula unit in atomic mass units ( amu ).

Molecular Weight (Mass)

• (^) Mass of a molecule in atomic mass units.

Illustrative Example

Calculate the molecular weight of glucose, C 6

H

12

O

6

Solution

6 C = 6 x 12 = 72 amu

12 H = 12 x 1 = 12 amu

6 O = 6 x 16 = 96 amu

180 amu

Sample Problem 6

• (^) Calculate the formula weight of calcium nitrate, Ca(NO

3

2

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Converting between grams and moles

If we are given the # of grams of a compound

we can determine the # of moles, & vise-versa

In order to convert from one to the other you

must first calculate molar mass

g = mol x g/mol

mol = g  g/mol

• This can be represented in an “equation triangle”

g

mo

l

g/

mol

HCl 0.25 g= g/mol x mol

H 53.

2

SO

4

NaCl 3.

Cu 1.

Formula g/mol g mol (n) Equation

98.08 0.5419 mol= g  g/mol

58.44 207 g= g/mol x mol

63.55 0.0200 mol= g  g/mol

Molar Mass

• (^) Mass in grams of 1 mole of a substance.
• (^) Numerically equal to the formula weight (in amu) of the

substance.

Examples

• (^) The atomic weight of a carbon-12 atom = 12 amu.
• (^) The mass of 1 mole of carbon-12 atoms = 12 g.
• (^) The atomic weight of a carbon-13 atom = 13 amu.
• (^) The mass of 1 mole of carbon-13 atoms = 13 g.

The formula weight of water, H 2

O = 18 amu.

• (^) The mass of 1 mole of water molecules = 18 g.

1 molecule C 6

H

12

O

6

1 mol C 6

H

12

O

6

6 atoms 12 atoms 6 atoms 6 mol 12 mol 6 mol

FW = 180 amu Molar mass = 180 g

Mole Calculations