Chapter 3 Stoichiometry

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Chapter 3 Stoichiometry

• Relative Mass
• Avogadros Number, Moles
• Mass Percent Composition
• Determination of Empirical and Molecular Formula
• Chemical Equation

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Chapter 3 Stoichiometry

• Stoichiometry Problems
• Mole lt-gt Mole
• Mole lt-gt Mass
• Mass lt-gt Mass
• Limiting Reagents
• Percent Yield

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Micro World atoms molecules
Macro World grams
Atomic mass is the mass of an atom in atomic mass
units (amu)
By definition 1 atom 12C weighs 12 amu
On this scale 1H 1.008 amu 16O 16.00 amu
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Molecular mass (or molecular weight) is the sum
of the atomic masses (in amu) in a molecule.
For any molecule molecular mass (amu) molar
mass (grams)
1 molecule SO2 64.07 amu 1 mole SO2 64.07 g
SO2
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Collection Terms

• 1 trio 3 singers
• 1 six-pack Cola 6 cans Cola drink
• 1 dozen donuts 12 donuts
• 1 gross of pencils 144 pencils

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A Moles of Particles

• Contains 6.02 x 1023 particles
• 1 mole C 6.02 x 1023 C atoms
• 1 mole H2O 6.02 x 1023 H2O molecules
• 1 mole NaCl 6.02 x 1023 Na ions and
• 6.02 x 1023 Cl ions

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Examples of Moles

• Moles of elements
• 1 mole Mg 6.02 x 1023 Mg atoms
• 1 mole Au 6.02 x 1023 Au atoms
• Moles of compounds
• 1 mole NH3 6.02 x 1023 NH3 molecules
• 1 mole C9H8O4 6.02 x 1023 aspirin molecules

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Avogadros Number

• 6.02 x 1023 particles
• 1 mole
• or
• 1 mole
• 6.02 x 1023 particles

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Learning Check

• Suppose we invented a new collection unit called
  a mep. One mep contains 8 objects.
• A. How many paper clips in 1 mep?
• 1) 1 2) 4 3) 8
• B. How many oranges in 2.0 meps?
• 1) 4 2) 8 3) 16
• C. How many meps contain 40 gummy bears?
• 1) 5 2) 10 3) 20

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Solution

• Suppose we invented a new collection unit called
  a mep. One mep contains 8 objects.
• A. How many paper clips in 1 mep?
• 3) 8
• B. How many oranges in 2.0 meps?
• 3) 16
• C. How many meps contain 40 gummy bears?
• 1) 5

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Avogadros Number
Chemistry An Environmental Perspective Buelll,
Phyllis and Girard, James Prentice Hall 1994,
p194
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Avogadros , NA6.022 X 10 23

• All of the following contain the same number of
  atoms (6.022 X 10 23)
• 51.996 g Cr -10.811 g B
• 88.905 g Y -79.909 g Br
• 39.102 g K -32.064 g S
• Take the mass atomic mass of one element and do
  the same for another element and theyll both
  have NA atoms

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Learning Check

• 1. Number of atoms in 0.500 mole of Al
• 1) 500 Al atoms
• 2) 6.02 x 1023 Al atoms
• 3) 3.01 x 1023 Al atoms
• 2.Number of moles of S in 1.8 x 1024 S atoms
• 1) 1.0 mole S atoms
• 2) 3.0 mole S atoms
• 3) 1.1 x 1048 mole S atoms

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Solution

• 1. Number of atoms in 0.500 mol of Al
• 3) 3.01 x 1023 Al atoms
• 0.500 mol Al x 6.02 x 1023 Al atoms
• 1 mol Al
• 2. Number of moles of S if a sample of S contains
  4.50 x 1024 S atoms
• 2) 3.0 mole S atoms
• 4.50 x 1024 S atoms x 1 mol S
  
• 6.02 x 1023 S atoms

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Na atoms
Pb atoms
Molar mass is the mass of 1 mole of
in grams
Kr atoms
Li atoms
1 mole C atoms 6.022 x 1023 atoms 12.011 g
1 mole lithium atoms 6.941 g of Li
For any element atomic mass (amu) molar mass
(grams)
Note Lab book calls molar mass the formula mass
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1 mol K 39.10 g K
1 mol K 6.022 x 1023 atoms K
0.551 g K
8.49 x 1021 atoms K
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Molar Mass

• Number of grams in 1 mole
• Equal to the numerical value of the atomic mass
• 1 mole of C atoms 12.0 g
• 1 mole of Mg atoms 24.3 g
• 1 mole of Cu atoms 63.5 g

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Learning Check

• Give the molar mass to 0.1 g
• A. 1 mole of Br atoms ________
• B. 1 mole of Sn atoms ________

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Solution

• Give the molar mass to 0.1 g
• A. 1 mole of Br atoms 79.9 g/mole
• B. 1 mole of Sn atoms 118.7 g/mole

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Molar Mass of Compounds

• Mass in grams of 1 mole equal numerically to the
  sum of the atomic masses
• 1 mole of CaCl2 111.1 g/mole
• 1 mole Ca x 40.1 g/mole
• 2 moles Cl x 35.5 g/mole
• 1 mole of N2O4 74.0 g/mole
• 2 moles N x 14.0 g/mole
• 4 moles O x 16.0 g/mole

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Learning Check

• A. 1 mole of K2O ______g
• B. 1 mole of antacid Al(OH)3 ______g

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Solution

• A. 1 mole of K2O 94.2 g
• 2 K x 39.1 g/mole 1 O x 16.0 g/mole
• B. 1 mole of antacid Al(OH)3 78.0 g
• 1 Al x 27.0 g/mole 3 O x 16.0 g/mole
• 3 x 1.0

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Learning Check

• Prozac, C17H18F3NO, is a widely used
  antidepressant that inhibits the uptake of
  serotonin by the brain. It has a molar mass of
• 1) 40.0 g/mole
• 2) 262 g/mole
• 3) 309 g/mole

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Solution

• Prozac, C17H18F3NO, is a widely used
  antidepressant that inhibits the uptake of
  serotonin by the brain. It has a molar mass of
• 3) 309 g/mole
• 17C (12.0) 18H (1.0) 3F (19.0) 1N (14.0)
  1 O (16.0)

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Moles, n
Avogadros Number
Molar Mass
Atoms
Mass, grams
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Mass
Moles
Molar Mass
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Converting Units

• Unit Factor Label Method
• Write Unit of Answer
• Write Starting Quantity from Problem
• Add appropriate unit factor(s) to cancel out
  starting quantity and put in unit of answer.

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Learning Check

• What is the percent carbon in C5H8NO4 (MSG
  monosodium glutamate), a compound used to flavor
  foods and tenderize meats?
• 1) 8.22 C
• 2) 24.3 C
• 3) 41.1 C

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Solution

• Molar mass 146.0 g/mole
• total g C x 100
• total g compound
• 60.0 g C x 100 41.1 C
• 146.0 g MSG

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Molar Mass Factors

• Methane CH4 known as natural gas is used in gas
  cook tops and gas heaters. Express the molar
  mass of methane in the form of conversion
  factors.
• Molar mass of CH4 16.0 g
• 16.0 g CH4 and 1 mole CH4
• 1 mole CH4 16.0 g CH4

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Learning Check

• Acetic acid CH3COOH is the acid in vinegar . It
  has a molar mass of 60.0 g/mole.
• 1 mole of acetic acid ____________
• 1 mole acetic acid or g
  acetic acid
• g acetic acid 1 mole acetic
  acid

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Solution

• Acetic acid CH3COOH is the acid in vinegar . It
  has a molar mass of 60.0 g/mole.
• 1 mole of acetic acid 60.0 g acetic acid
• 1 mole acetic acid or 60.0 g acetic
  acid
• 60.0 g acetic acid 1 mole acetic acid

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Calculations with Molar Mass

• molar mass
• Grams Moles

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Moles and Grams

• Aluminum is often used for the structure of
  light-weight bicycle frames. How many grams of
  Al are in 3.00 moles of Al?
• 3.00 moles Al ? g Al

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• 1. Molar mass of Al 1 mole Al 27.0 g Al
• 2. Conversion factors for Al
• 27.0g Al or 1 mol Al
• 1 mol Al 27.0 g Al
• 3. Setup 3.00 moles Al x 27.0 g Al
• 1 mole Al
• Answer 81.0 g Al

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Learning Check

• The artificial sweetener aspartame
  (Nutri-Sweet) formula C14H18N2O5 is used to
  sweeten diet foods, coffee and soft drinks. How
  many moles of aspartame are present in 225 g of
  aspartame?

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Solution

• 1. Molar mass of Aspartame C14H18N2O5
• (14 x 12.0) (18 x 1.01) (2 x 14.0) (5 x
  16.0) 294 g/mole
• 2. Setup
• 225 g aspartame x 1 mole aspartame
• 294 g aspartame
• 0.765 mole aspartame

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Mass Percent composition of an element in a
compound
n is the number of moles of the element in 1 mole
of the compound
52.14 13.13 34.73 100.0
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Weights of Atoms and Weights of Moles of Atoms
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Relation Between Weight, Kind, and Number of Atoms
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Weights of Molecules and Weights of Moles of
Molecules
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Calculation of the Number of Moles of Molecules
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Calculation of Moles of Silver and Moles of
Copper in Silver Dollars

• Weight of Silver Dollars 108.0 grams
• 90 of this weight is silver, or 0.90 x 108.0
  97.2 g Ag.
• 108.0 g 97.2 g 10.8 g Cu.
• Moles silver 97.2 g/ 108.0 g/mole 0.900 moles
  Ag
• Moles copper 10.8 g/63.5 g/ mole 0.170 moles
  Cu

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Determination of the Simple Formula of a Metallic
Sulfide
Calculation of Simple Formula 1.107 g silver
combined with 0.164 g sulfur A mole of silver
atoms weighs 108.0 g, so 108.0 g silver combined
with 1/1.107 x 0.164 x 108.0 16.0 g
sulfur 1 mole of silver atoms combined with
16.0/32.0 0.500 moles sulfur atoms 2 moles
silver atoms combined with 1 mole sulfur
atoms. Therefore, 2 silver atoms combined with 1
sulfur atom and the simple formula is Ag2S.
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Determination of Empirical Formula from Mass

• Assume 100 grams of material
• Determine moles of each element in compound
• Divide all subscripts in empirical formula by
  lowest number because the number of atoms must be
  an integer

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Determination of Molecular Formula from Empirical
Formula

Only Empirical Formula can be Determined by Mass
Percent
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Chemical Equations
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Some Values Determined by an Equation
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Balancing Chemical Equations

• Write the correct formula(s) for the reactants on
  the left side and the correct formula(s) for the
  product(s) on the right side of the equation.

Ethane reacts with oxygen to form carbon dioxide
and water

• Change the numbers in front of the formulas
  (coefficients) to make the number of atoms of
  each element the same on both sides of the
  equation. Do not change the subscripts.

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Balancing Chemical Equations

• Start by balancing those elements that appear in
  only one reactant and one product.

start with C or H but not O
multiply CO2 by 2
multiply H2O by 3
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Balancing Chemical Equations

• Balance those elements that appear in two or more
  reactants or products.

7 oxygen on right
remove fraction multiply both sides by 2
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Balancing Chemical Equations

• Check to make sure that you have the same number
  of each type of atom on both sides of the
  equation.

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