• The study of the numerical relationship between chemical quantities in a chemical reaction is called stoichiometry.

Chapter 3

Chemical Reactions and Reaction Stoichiometry

許富銀 ( Hsu Fu-Yin)

1

Stoichiometry

• The study of the numerical relationship between chemical quantities in a chemical reaction is called stoichiometry.

EX:

2 C

H

(l) + 25 O

(g) → 16 CO

(g) + 18 H

O(g)

2 molecules of C₈H₁₈ react with 25 molecules of O₂ to form 16 molecules of CO₂ and 18 molecules of H₂O.

Chemical Equations

• Chemical equations are concise representations of chemical reactions

Reactants appear on the left side of the equation.

Products appear on the right side of the equation.

CH

(g) + 2O

(g) → CO

(g) + 2H

O(g)

The states of the reactants and products are written in parentheses to the right of each compound. (g) = gas; (l) = liquid; (s) = solid; (aq) = in aqueous solution

3

Chemical Equation

Coefficients are inserted to balance the equation

How to Write Balanced Chemical Equations

• If an element occurs in only one compound on each side of the equation, try balancing this element first.

• When one of the reactants or products exists as the free element, balance this element last.

• In some reactions, certain groups of atoms (for example, polyatomic ions) remain unchanged. In such cases,

balance these groups as a unit.

5

Question

• 1) When the following equation is balanced, the coefficients are ________.

C₈H₁₈ + O₂ → CO₂ + H₂O

Quiz

• 1) When the following equation is balanced, the coefficients are ________.

FeS₂ + O₂ → Fe₂O₃ + SO₂

7

Three Types of Reactions

• Combination reactions

• Decomposition reactions

• Combustion reactions

Combination reactions: two or more substances react to form one product.

• Examples:

 2 Mg(s) + O

(g) 2 MgO(s)

Decomposition Reactions

Decomposition reaction: one substance breaks down into two or more substances

9

• Examples:

Combustion Reactions

• Combustion reactions are generally rapid reactions that produce a flame.

• Combustion reactions most often involve oxygen in the air as a reactant.

 Examples:

 CH

(g) + 2 O

(g) CO

(g) + 2 H

O(g)

 C

H

(g) + 5 O

(g) 3 CO

(g) + 4 H

O(g)

Formula Weight

• A formula weight is the sum of the atomic weights for the atoms in a chemical formula.

• The formula weight of calcium chloride, CaCl

, would be

Ca: 1x(40.08 amu) + Cl: 2 x(35.453 amu)

110.99 amu

11

Molecular Weight

• A molecular weight is the sum of the

atomic weights of the atoms in a molecule.

• For the molecule ethane, C

H

, the molecular weight would be

•

C: 2 x (12.011 amu)

30.070 amu

+ H: 6 x (1.00794 amu)

Quiz

• The formula weight of potassium dichromate (K₂Cr₂O₇ ) is ________ amu

• The formula weight of calcium nitrate Ca(NO₃)₂ is ________

amu.

13

Ionic Compounds and Formulas

• I

onic compounds exist with a three-dimensional order of ions.

• Ionic compounds use empirical formulas and formula

weights (not molecular weights)

Percentage Composition from Chemical Formulas

• The mass percent composition or mass percent of an

element is that element’s percentage of the compound’s total mass.

15

Calculating Percentage Composition

EX: Calculate the percentage of carbon, hydrogen, and oxygen (by mass) in C

H

O

.

Sol:

%C =(12x(12.0 amu)/ 342.0 amu )* 100% = 42.1%

%H =(22x(1.0 amu)/342.0 amu) * 100% = 6.4%

%O =(11x(116.0 amu)/342.0 amu) * 100% = 51.5%

Question



What is the mass percent composition of halothane, C

HBrClF

?

17

Quiz

• The mass % of C in methane (CH₄) is ________?

• The mass % of H in methane (C₂H₆) is ________?

Quiz

• Calculate the percentage by mass of Pb, N, O in Pb(NO₃)₂

19

Avogadro’s Number and the Mole

• 6.02 × 10²³ atoms or molecules is ONE Mole.

• Avogadro’s number: 6.02 × 10²³

• One mole of 12C has a mass of 12.000 g.

Converting Moles to Number of Atoms

• Calculate the number of H atoms in 0.350 mol of C

H

O

.

Sol:

21

Quiz

• Calculate the number of H atoms in 0.50 mol of (NH₄)₂SO₄.

Molar Mass

• A molar mass is the mass of 1 mol of a substance (i.e., g/mol).

23

Mole Relationships

Calculating Molar Mass

• What is the molar mass of glucose, C

H

O

?

Sol:

25

C

H

O

Converting Grams to Moles

• Calculate the number of moles of glucose (C

H

O

) in 5.380 g of C

H

O

.

Sol:

• Calculate the mass, in grams, of 0.433 mol of calcium nitrate.

Sol:

Quiz

• How many moles of carbon dioxide (CO₂) are there in 52.06 g of carbon dioxide?

27

Interconverting Masses and Numbers of Particles

EX: How many glucose molecules are in 5.23 g of C

H

O

?

How many oxygen atoms are in this sample?

Question



An analytical balance can detect a mass of 0.1 mg. How many ions are present in this minimally detectable quantity of MgCl

?

29

Empirical Formulas from Analyses

• One can determine the empirical formula from the percent composition by following these three steps.

Calculating an Empirical Formula

• Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula of ascorbic acid?

Sol:

In 100.00 g of ascorbic acid we have 40.92 g C, 4.58 g H, and 54.50 g O.

31

Quiz

• What is the empirical formula of a compound that contains 27.0% S, 13.4% O, and 59.6% Cl by mass?

Molecular Formulas from Empirical Formulas

• The number of atoms in a molecular formula is a multiple of the number of atoms in an empirical formula.

EX: The empirical formula of a compound was found to be CH. It has a molar mass of 78 g/mol. What is its molecular formula?

Solution:

• Whole-number multiple = 78/13 = 6

The molecular formula is C₆H₆.

33

Whole-number multiple

=𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡/𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑤𝑒𝑖𝑔ℎ𝑡

Question

• Dibutyl succinate is an insect repellent used against household ants and roaches. Its composition is 62.58% C, 9.63% H, and

27.79% O. Its experimentally determined molecular mass is 230 u.

What are the empirical and molecular formulas of dibutyl succinate?

35

Quiz

• A compound contains 40.0% C, 6.71% H, and 53.29% O by mass. The molecular weight of the compound is 60.05 amu.

The molecular formula of this compound is ________.

Combustion Analysis

• One technique for determining empirical formulas in the laboratory is combustion analysis, commonly used for compounds containing principally carbon and hydrogen

37

Determining an Empirical

Formula by Combustion Analysis

• Isopropyl alcohol, sold as rubbing alcohol, is composed of C, H, and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g of CO₂ and 0.306 g of H₂O. Determine the empirical formula of isopropyl alcohol.

Sol:

• all of the carbon in the sample is converted to CO₂

Determining an Empirical

Formula by Combustion Analysis

• all of the hydrogen in the sample is converted to H₂O

The mass of the sample, 0.255 g, is the sum of the masses of C, H, and O.

Mass of O = mass of sample – (mass of C + mass of H)

= 0.255 g – (0.153 g + 0.0343 g) = 0.068 g O

39 Moles C = (0.153 g )(1 mol C/12.0 g C) = 0.0128 mol C

Moles H = (0.0343 g)(1 mol H/1.01 g H)=0.0340 mol H Moles O = (0.068 g )(1 mol O/16.0 g O)= 0.0043 mol O

The first two numbers are very close to the whole numbers 3 and 8, giving the empirical Formula C₃H₈O.

Question

• Vitamin C is essential for the prevention of scurvy. Combustion of a 0.2000 g sample of this carbon– hydrogen–oxygen

compound yields 0.2998 g CO2 and 0.0819 g H2O. What are the percent composition and the empirical formula of vitamin C?

Sol

41

43

Quiz

• A compound is composed of only C, H, and O. The combustion of a 0.519-g sample of the compound yields 1.24 g of CO₂ and 0.255 g of H₂O. What is the empirical formula of the

compound?

Quantitative Information from Balanced Equations

45

The coefficients in a balanced chemical equation indicate both the relative numbers of molecules (or formula units) in the

reaction and the relative numbers of moles.

The coefficients are called stoichiometrically equivalent

quantities.

Balanced chemical equation

quantitatively.

Balanced chemical equation quantitatively.

47

Ex: Determine how many grams of water are produced in the oxidation of 1.00 g of glucose, C

H

O

Moles C₆H₁₂O₆ = (1.00 g C₆H₁₂O₆)(1 mol C₆H₁₂O₆/ 180.0 g C₆H₁₂O₆)

Quiz

The combustion of ammonia in the presence of excess oxygen yields NO

and H

O:

4 NH

(g) + 7 O

(g) → 4 NO

(g) + 6 H

O (g) The combustion of 43.9 g of ammonia produces ________ g of NO

.

49

Limiting Reactants

• The reactant that is completely consumed in a reaction is called the limiting reactant. The other reactants are

sometimes called excess reactants

Sample Exercise 3.18

51 Sol:

Theoretical and Percent Yields

• The quantity of product calculated to form when all of a limiting reactant is consumed is called the theoretical yield.

• The amount of product actually obtained, called the actual yield

• The percent yield of a reaction relates actual and theoretical yields

Percent yield = (actual yield/theoretical yield) * 100%

Calculating Theoretical Yield and Percent Yield

53 Sol: