Unit V:Chemical Quantities

Information in Chemical EquationsAs we have seen in the last unit, chemistry is

about reactionsReactions are described by equations that give the

identities of the reactants and productsThe amount of each is shown by the coefficients

Example: Making a sandwich – Ratio 2:3:1

Using the equation in a reaction permits us to determine the amounts of reactants needed give a particular amount of product

The equation would look like this…

2 + 3 + 1 → 1The #’s are all coefficients that would be

placed in front of the elements and or compounds

Mol-Mol RelationshipsWe can use an equation to predict the mols of products that a given number of mols of reactants will yieldExample: 2 H2O (l) → 2 H2 (g) + O2 (g)If we decompose 4 mol of water, how many

mol of product do we get?

4 mol H2O

mol H2O

mol H2

2

2= 4 mol H2

Using the same equation, suppose we were to decompose 5.8 mol of waterEquation: 2 H2O (l) → 2 H2 (g) + O2 (g)The answer is to use mol ratios

Examples: 2 mol H2O = 2 mol H2 (1:1 ratio)

2 mol H2O = 1 mol O2 (2:1 ratio)

Set up a mol ratio: 1 mol O2 = 2 mol H2O

5.8 mol H2O • (1 mol O2 = 2 mol H2O)

= 2.9 mol O2

Final Answer: 5.8 H2O (l) → 5.8 H2 (g) + 2.9 O2 (g)

mol – mol Conversions

Sample Problems:1A. Iron (II) oxide reacts with oxygen to form

iron (III) oxideEquation: 4 FeO (s) + O2 (g) → 2 Fe2O3 (s)

2.4 mol FeO → ? mol O2

2.4mol FeO

mol FeO

mol O2

4

1=

6.0 x 10-1 mol O2

Equation: 4 FeO (s) + O2 (g) → 2 Fe2O3 (s)

1B. What number of moles of iron (III) oxide will be produced by reacting 9.2 mol of iron (II) oxide with excess oxygen?

9.2 mol FeO → ? Fe2O3

9.2mol FeO

mol FeO

mol Fe203

4

2= 4.6 x 100

mol Fe2O3

Mass CalculationsWe know that moles represent the number of moleculesWe can not count molecules directly

We count by weighingGrams to grams stoichiometry

Consider the following problem: C3H8 (g)+ 5 O2 (g)→ 3 CO2 (g) + 4 H20 (g)

2A. What mass of oxygen will be required to react exactly with 54.1 g of propane?

54.1 g C3H8 → ? g O2 (grams → grams)

(54.1 • 1 • 5 • 32) / (44.11 • 1 • 1) = 196.2366Round to 3 SF’s = 196 → SSN = 1.96 x 102

Add units of measure =

54.1g C3H8

g C3H8

mol C3H8

mol C3H8

mol O2

mol O2

g O2

44.11

1

1

5

1

32

1.96 x 102 g O2

Consider the following problem: C3H8 (g)+ 5 O2 (g)→ 3 CO2 (g) + 4 H20 (g)

2B. What mass of water will be required to react exactly with 24.2 g of oxygen gas?

24.2 g O2 → ? g H2O (grams → grams)

(24.2 • 1 • 4 • 18.02) / (32 • 5 • 1) = 10.9021Round to 3 SF’s = 10.9 → SSN = 1.09 x 101

Add units of measure =

24.2g O2

g O2

mol O2

mol O2

mol H2O

mol H2O

g H2O

32

1

5

4

1

18.02

1.09 x 101 g H2O

Consider the following problem: C3H8 (g)+ 5 O2 (g)→ 3 CO2 (g) + 4 H20 (g)

2C. How many L of CO2 will be required to react exactly with 7.5 x 1024 atoms of O2 gas?

7.5 x 1024 atoms O2 → ? L CO2 (4 steps)

(7.5 x 1024•3•22.4) / (6.022 x 1023•5) = 167.38625Round to 2 SF’s = 170 → SSN = 1.7 x 102

Add units of measure =

7.5 x 1024

atoms O2

atoms O2

mol O2

mol O2

mol CO2

mol CO2

L CO2

6.022 x 1023

1

5

3

1

22.4

1.7 x 102 L CO2

Mass Calculations:Comparing Two Reactions

Question: Which antacid can consume the most stomach acid, 1.00 g of NaHCO3 or 1.00 g of Mg(OH)2?Remember – moles are used to compare!Reactions:

A. NaHCO3(s)+HCl (aq)→ NaCl(aq) + H2O(l) + CO2(g)

B. Mg(OH)2 (s) + 2 HCl (aq)→ 2 H2O (l) + MgCl2 (aq)

Problems: (3 steps: g → mol)A1. 1.00 g NaHCO3 → ? mol HCl

A2. 1.00 g Mg(OH)2 → ? mol HCl

The Concept of Limiting Reactants

Earlier in the unit, we discussed making sandwiches…2 pieces of bread + 3 slices of meat + 1 slice

of cheese → 1 sandwichSuppose you came to work and found the

following: 20 slices of bread

24 slices of meat

12 slices of cheese

Example:

Available Tools

Sets of Tools

Extra Tools

Calculations Involving a Limiting Reactant

3. What mass of water is required exactly to react exactly with 249 g of methane?249 g CH4 → ? g H2O (4 steps)Balanced Equation:

CH4 (g) + H20 (g) → 3 H2 (g) + CO (g)

249

g CH4

g CH4

mol CH4

mol CH4

mol H2O

mol H2O

g H2O

16.05

1

1

1

1

18.02

Problem: 249 g CH4 → ? g H2O

Equation: CH4 (g) + H20 (g) → 3 H2 (g) + CO (g)

Calculation:(249 •18.02) / (16.05) = 279.56261

Round to 3 SF’s = 280. g H2O

Normally we would place the answer in SSN

Question: If 249 g of CH4 were mixed with 300 g of water…

Remember: 249 g CH4 → 280 g H2O

The CH4 will be consumed before the H2O runs out

The CH4 is the limiting reactant

H2O will be in excess

Percent YieldIn the previous sections, we learned many aspects about chemical reactionsEssentially, products stop running out when

one reactant runs outThe amount calculated in this way is called

the theoretical yieldIt is the amount of product

predicted from the amounts of reactant used

The actual yield of the product (which is actually obtained) is called the percent yield

AY

TY%

Actual Yield (g)

Theoretical Yield (g)

What actually occurs during the experiment

Based on Stoichiometry

Percent Yield (g)Expressed as a decimal

Sample Problem 4: if 124 g of Iron is produced from 222 g of Iron (III) oxide, what is the A) Theoretical Yield, B) Percent Yield?

Equation: Iron (III) oxide decomposes…

2 Fe2O3 (s) → 4 Fe (s) + 3 O2 (g)Actual Yield = 124 g Fe

Have to convert 222 g Fe2O3 to g FeGrams to grams problem (4 steps)This will be the Theoretical Yield

(222 • 4 • 55.85) / (159.7 • 2) = 155.2748904Round to 3 SF’s = 155 → SSN = 1.55 x 102

Add units of measure =

% Yield = (AY / TY) • 100

= (124 g Fe / 155 g Fe) • 100

= 80% Yield

222

g Fe2O3

g Fe2O3

mol Fe2O3

mol Fe2O3

mol Fe

mol Fe

g Fe

159.7

1

2

4

1

55.85

1.55 x 102 g Fe