Zumdahl • Zumdahl • DeCoste
World of
CHEMISTRY
Chapter 9
Chemical Quantities
Goals of Chapter 9
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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 Equations
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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 Equations
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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 9.1
Combustion of Propane
Propane reacts with oxygen to
produce heat and the products carbon
dioxide and water:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Interpretation of Equation
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1 molecule of C3H8 reacts with 5 molecules of
O2 to give 3 molecules of CO2 plus 4
molecules of H2O
1 mole of C3H8 reacts with 5 moles of O2 to
give 3 moles of CO2 plus 4 moles of H2O
Nuts & Bolts of Chemistry Activity
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2 Nuts (N) react with 1 bolt (B) to form a nutbolt molecule
• 2N + B → N2B
• Note difference between coefficient and subscript
• Construct nut-bolt molecules
• Which is limiting reactant? Why?
Average mass of bolt = 10.64 g & average
mass of nut = 4.35 g; If you are given about
1500 g of each, answer the following
questions:
1. How many bolts are in 1500 g? How
many nuts are in 1500 g?
2. Which is limiting reactant? Why?
3. What is largest possible mass of
product? How many products can you
make?
4. What is mass of leftover reactant?
Mole-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 Ratios: conversion factors based on
balanced 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 Calculations
What mass of oxygen
(O2) is required to react
with exactly 44.1 g of
propane (C3H8)?
Step 1: Write balanced equation
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Step 2: Convert grams of propane to
moles of propane
44.1 g C3H8 x 1 mol C3H8 = 1.00 mol C3H8
44.09 g C3H8
Step 3: Use coefficients in equation to
determine moles of oxygen required
1.00 mol C3H8 x 5 mol O2 = 5.00 mol O2
1 mol C3H8
Step 4: Use molar mass of O2 to calculate
the grams of oxygen
5.00 mol O2 x 32.0 g O2 = 160 g O2
1.00 mol O2
Can perform conversion in on long step:
44.1 g C3H8 x 1 mol C3H8
44.09 g C3H8
x 5 mol O2 x 32.0 g O2 = 160 g O2
1 mol C3H8
1.00 mol O2
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Mass Calculations Using Scientific
Notation
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
Stoichiometry
The process of using a
chemical equation to
calculate the relative masses
of reactants and products
involved in a reaction
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Mass Calculations: Comparing Two
Reactions
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
Limiting Reactants
Consider the reaction that forms ammonia:
N2(g) + 3H2(g) → 2NH3(g)
These gases are mixed in a closed
vessel and begin to react
Container (#1) of N2(g) and H2(g).
Before and after the reaction.
This reaction contained the exact number of
molecules to make ammonia molecules with
no unreacted molecules left over.
Before the reaction, there were 15 H2
molecules and 5 N2 molecules which gives
the exact ratio to make ammonia, 3:1.
This type of mixture is called a
stoichiometric mixture – contains the
relative amounts of reactants that matches
the numbers in balanced equation
What happens when the
ratio is not the same as in
the chemical equation?
Container (#2) of N2(g) and H2(g). (5:8 ratio)
Before and after the reaction.
(Some N2 molecules are left over)
Limiting Reactant
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The reactant that runs out first
and thus limits the amounts of
products that can form
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H2 is limiting reactant in
previous slide
Ammonia, which is used as a fertilizer, is
made by combining nitrogen from the air
with hydrogen. The hydrogen is produced
by reacting methane (CH4) with water. If
you have 249 grams of methane, how
much hydrogen will be produced and how
much water will you need to convert all of
the methane to hydrogen?
Step 1: Write balanced equation
CH4(g) + H2O(g) → 3H2(g) + CO(g)
Figure 9.1: A mixture of 5CH4 and 3H2O molecules.
Step 2: Convert mass of methane to moles
249 g CH4 x 1 mol CH4 = 15.5 mol CH4
16.04 g CH4
Step 3: Determine moles of H2O needed
15.5 mol CH4 x 1 mol H2O = 15.5 mol H2O
1 mol CH4
Step 4: Determine mass of water
15.5 mol H2O x 18.02 g H2O = 279 g H2O
1 mol H2O
Reacting 279 grams of water with 249 grams
of methane will cause both reactants to “run
out” at the same time.
If 300 grams of water is reacted with 249
grams of methane, the methane will “run out”
first = limiting reactant (it limits the reaction)
Figure 9.2: A map of the procedure used in
Example 9.7. (see for determining limiting reactant)
Steps for solving stoichiometric problems
involving limiting reactants:
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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 Yield
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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
Percent Yield
Actual Yield x 100% = percent
Theoretical Yield
yield