Zumdahl • Zumdahl • DeCoste

Zumdahl • Zumdahl • DeCoste

CHEMISTRYWorld of

Chapter 9

Chemical Quantities

Goals of Chapter 9Goals of Chapter 9

• Understand molecular and molar mass given in balanced equation

• Use balanced equation to determine the relationships between moles of reactants and moles of products

• Relate masses of reactants and products in a chemical reaction

• Perform mass calculations that involve scientific notation• Understand the concept of limiting reactants• Recognize the limiting reactant in a reaction• Use the limiting reactant to do stoichiometric calculations

Information Given by Chemical EquationsInformation Given by Chemical Equations

• Reactions are described by equations • Give the identities of the reactants &

products • Show how much of each reactant and

product participates in the reaction. • Numbers or coefficients enable us to

determine how much product we get from a given quantity of reactants

Information Given by Chemical EquationsInformation Given by Chemical Equations

• Atoms are rearranged in a chemical reaction (not created or destroyed)

• Must have same number of each type of atom on both sides of equation.

• Coefficients give relative number of molecules, meaning we can multiply by any number and still have a balanced equation.

Table 9.1Table 9.1

Combustion of PropaneCombustion of Propane

Propane reacts with oxygen to produce heat and the products carbon

dioxide and water:

C3H

8(g) + 5O

2(g) → 3CO

2(g) + 4H

2O(g)

Interpretation of EquationInterpretation of Equation

• 1 molecule of C3H

8 reacts with 5 molecules of

O2 to give 3 molecules of CO

2 plus 4

molecules of H2O

• 1 mole of C3H

8 reacts with 5 moles of O

2 to

give 3 moles of CO2 plus 4 moles of H

2O

Nuts & Bolts of Chemistry ActivityNuts & Bolts of Chemistry Activity

• 2 Nuts (N) react with 1 bolt (B) to form a nut-bolt molecule• 2N + B → N

2B

• Note difference between coefficient and subscript• Construct nut-bolt molecules• Which is limiting reactant? Why?

Average mass of bolt = 10.64 g & average Average mass of bolt = 10.64 g & average mass of nut = 4.35 g; If you are given about mass of nut = 4.35 g; If you are given about 1500 g of each, answer the following 1500 g of each, answer the following questions: questions:

1. How many bolts are in 1500 g? How 1. How many bolts are in 1500 g? How many nuts are in 1500 g?many nuts are in 1500 g?

2. Which is limiting reactant? Why?2. Which is limiting reactant? Why?3. What is largest possible mass of 3. What is largest possible mass of

product? How many products can you product? How many products can you make?make?

4. What is mass of leftover reactant?4. What is mass of leftover reactant?

Mole-Mole RelationshipsMole-Mole Relationships

2H2O(l) → 2H2(g) + O2(g)

Equation tells us that 2 mol of H2O yields 2 mol of H2 and 1 mol of O2

If we decompose 4 mol of water, how many moles of

products do we get?

4H2O(l) → 4H2(g) + 2O2(g)

If we decompose 5.8 mol of water, how many moles of products do we get?

5.8H2O(l) → 5.8H2(g) + ?O2(g)

Mole RatiosMole Ratios: conversion factors based on : conversion factors based on balanced chemical equationsbalanced chemical equations

From initial equation:

2 mol H2O = 2 mol H2 = 1 mol O2

Can use equivalent statement & perform dimensional analysis

5.8 mol H2O x 1 mol O2_ = 2.9 mol O2

2 mol H2O

Mass CalculationsMass Calculations

What mass of oxygen (O2) is required to react with exactly 44.1 g of

propane (C3H8)?

Step 1: Write balanced equationStep 1: Write balanced equation

C3H

8(g) + 5O

2(g) → 3CO

2(g) + 4H

2O(g)

44.1 g C3H8 x 1 mol C3H8 = 1.00 mol C3H8

44.09 g C3H8

Step 2: Convert grams of propane to Step 2: Convert grams of propane to moles of propanemoles of propane

Step 3: Use coefficients in equation to Step 3: Use coefficients in equation to determine moles of oxygen requireddetermine moles of oxygen required

1.00 mol C3H8 x 5 mol O2 = 5.00 mol O

2

1 mol C3H8

Step 4: Use molar mass of OStep 4: Use molar mass of O22 to calculate to calculate the grams of oxygenthe grams of oxygen

5.00 mol O2 x 32.0 g O

2 = 160 g O2

1.00 mol O2

Can perform conversion in on long step:Can perform conversion in on long step:

44.1 g C3H8 x 1 mol C3H8 x 5 mol O2 x 32.0 g O

2 = 160 g O2

44.09 g C3H8 1 mol C3H8 1.00 mol O2

Mass Calculations Using Scientific Mass Calculations Using Scientific NotationNotation

• Step 1: Balance the equation for the reaction• Step 2: Convert the masses of reactants or

products to moles• Step 3: Use the balanced equation to set up

the appropriate mole ratio(s)• Step 4: Use the mole ratio(s) to calculate the

number of moles of the desired product or reactant.

• Step 5: Convert from moles back to mass

StoichiometryStoichiometry

The process of using a chemical equation to

calculate the relative masses of reactants and products

involved in a reaction

Mass Calculations: Comparing Two Mass Calculations: Comparing Two ReactionsReactions

• Antacids are used to neutralize excess hydrochloric acid secreted by the stomach. Which antacid is more effective: baking soda, NaHCO3, or milk of magnesia, Mg(OH)2?

• Determine how many moles of stomach acid (HCl) will react with 1.00 g of each acid.

The Concept of The Concept of Limiting ReactantsLimiting Reactants

Consider the reaction that forms ammonia:Consider the reaction that forms ammonia:

N2(g) + 3H

2(g) → 2NH

3(g)

These gases are mixed in a closed vessel and begin to react

Container (#1) of NContainer (#1) of N22((gg) and H) and H22((gg).).

Before and after the reaction. Before and after the reaction.

This reaction contained the exact number of This reaction contained the exact number of molecules to make ammonia molecules with molecules to make ammonia molecules with

no unreacted molecules left over.no unreacted molecules left over.

Before the reaction, there were 15 HBefore the reaction, there were 15 H22

molecules and 5 Nmolecules and 5 N22 molecules which gives molecules which gives

the exact ratio to make ammonia, 3:1.the exact ratio to make ammonia, 3:1.

This type of mixture is called a This type of mixture is called a stoichiometric mixturestoichiometric mixture – contains the – contains the

relative amounts of reactants that matches relative amounts of reactants that matches the numbers in balanced equationthe numbers in balanced equation

What happens when the What happens when the ratio is not the same as in ratio is not the same as in

the chemical equation?the chemical equation?

Container (#2) of NContainer (#2) of N22((gg) and H) and H22((gg). (5:8 ratio)). (5:8 ratio)

Before and after the reaction.Before and after the reaction. (Some N (Some N

22 molecules are left over) molecules are left over)

Limiting ReactantLimiting Reactant

• The reactant that runs out first and thus limits the amounts of products that can form

• H2 is limiting reactant in

previous slide

Ammonia, which is used as a fertilizer, is Ammonia, which is used as a fertilizer, is made by combining nitrogen from the air made by combining nitrogen from the air with hydrogen. The hydrogen is produced with hydrogen. The hydrogen is produced

by reacting methane (CHby reacting methane (CH44) with water. If ) with water. If you have 249 grams of methane, how you have 249 grams of methane, how

much hydrogen will be produced and how much hydrogen will be produced and how much water will you need to convert all of much water will you need to convert all of

the methane to hydrogen?the methane to hydrogen?

Step 1: Write balanced equationStep 1: Write balanced equation

CH4(g) + H

2O(g) → 3H

2(g) + CO(g)

Figure 9.1:Figure 9.1: A mixture of 5CH A mixture of 5CH44 and 3H and 3H22O molecules.O molecules.

Step 2: Convert mass of methane to molesStep 2: Convert mass of methane to moles

249 g CH4 x 1 mol CH

4 = 15.5 mol CH

4

16.04 g CH4

Step 3: Determine moles of HStep 3: Determine moles of H22O neededO needed

15.5 mol CH4 x 1 mol H

2O = 15.5 mol H

2O

1 mol CH4

Step 4: Determine mass of waterStep 4: Determine mass of water

15.5 mol H2O x 18.02 g H

2O = 279 g H

2O

1 mol H

2O

Reacting 279 grams of water with 249 grams Reacting 279 grams of water with 249 grams of methane will cause both reactants to “run of methane will cause both reactants to “run

out” at the same time.out” at the same time.

If 300 grams of water is reacted with 249 If 300 grams of water is reacted with 249 grams of methane, the methane will “run out” grams of methane, the methane will “run out” first = first = limiting reactantlimiting reactant (it (it limits limits the reaction) the reaction)

Figure 9.2:Figure 9.2: A map of the procedure used in A map of the procedure used in Example 9.7.Example 9.7. (see for determining limiting reactant) (see for determining limiting reactant)

Steps for solving stoichiometric problems Steps for solving stoichiometric problems involving limiting reactants:involving limiting reactants:

• Step 1: Write and balance chemical equation• Step 2: Convert known masses to moles• Step 3: Using number of moles of reactants

and mole ratios, determine limiting reactant• Step 4: Use amount of limiting reactant and

mole ratios to calculate number of moles of product

• Step 5: Convert from moles to mass

Theoretical YieldTheoretical Yield

• Calculated yield from chemical reaction (amount of product form mole ratio calculations)

• Maximum amount that can be produced• Amount predicted is seldom obtained• Side reactions occur• Actual yield: amount of product actually

obtained

Actual Yield x 100% = percent

Theoretical Yield yield

Percent YieldPercent Yield