Chapter Three

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Chapter Three

• StoichiometryChemical Calculations

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Molecular Masses AndFormula Masses

• Molecular Mass
• The sum of the masses of the atoms represented in
  a molecular formula.
• Specifically for molecules.
• Formula Mass
• The sum of the masses of the atoms or ions
  present in a formula unit.
• Used for ionic compounds (dont exist as
  molecules)
• What are the units of molecular/formula mass?

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Molecular Masses AndFormula Masses

• Carbon dioxide, molecular mass 44.0 u
• Oxygen gas, molecular mass 32.0 u
• Bromine trichloride, molecular mass 186.3 u
• Magnesium oxide, formula mass 40.3 u
• Calcium carbonate, formula mass 100.1 u
• Copper(II) sulfate pentahydrate
• formula mass 249.7 u

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The Mole Avogadros Number

• A mole (mol) is an amount of substance that
  contains as many elementary entities as there are
  atoms in exactly 12 g of the carbon-12 isotope.
• Atoms are small, so this is a BIIIIG number
• Avogadros Constant (NA)
• NA 6.02214199 x 1023 mol-1
• 1 mol 6.02214199 x 1023 elementary entities.
• Warning! Mole is NOT abbreviated as either M or
  m.

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The Mole and Formation of Carbon Dioxide
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Molar Mass

• The mass of one mole of that substance.
• The molar mass is numerically equal to the atomic
  mass, molecular mass, or formula mass.
• The units of molar mass are grams/mole (g/mol).
• Examples
• 1 mol Na 22.99 g/mol
• 1 mol CO2 44.01g/mol

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Converting mass ?? moles
Use molar mass as a conversion factor!
Moles of A
Conversion factor from
molar mass of A
Mass of A
Note preliminary and followup calculations may
be needed
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Converting mass ?? moles
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Mass Percent CompositionFrom Chemical Formulas

• The mass percent composition of a compound refers
  to the proportion of the constituent elements
  expressed as the number of grams of each element
  per 100 grams of the compound.
• Mass fraction mass of element / mass of
  compound
• Mass percent mass fraction x 100

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Mass Percent CompositionFrom Chemical Formulas
SCHEMATIC REPRESENTATION OF
Percentage of carbon in butane, C4H10
Percentage of hydrogen in butane, C4H10
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Determining Empirical Formulas

• Reverse of finding percentage composition.
• Determine moles of each element (from percentage
  or mass).
• Find the smallest whole-number ratio of moles by
  dividing each number of moles by the smallest
  number.

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Determining Empirical Formulas
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Determining Empirical Formulas (contd)
Caution dont round off these values
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Relating Molecular FormulasTo Empirical Formulas

• Molecular formulas are simple integer multiples
  of the empirical formulas.
• That is, an empirical formula of CH2 means that
  the molecular formula is CH2 or C2H4 or C3H6 or
  C4H8 etc.
• Find the molecular formula by
• molecular formula mass/empirical formula mass
  number of empirical formula units in molecular
  formula
• Example A compound has an empirical formula of
  CH and a molecular mass of about 77 u. What is
  the molecular formula?
• 77 u / 13 u 5.92 or 6 empirical formula units
  C6H6

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Elemental Analysis

• A method of determining empirical formula in the
  laboratory.
• Used primarily for organic compounds (contain
  carbon, hydrogen, oxygen).
• The products of combustion are weighed, and the
  amount of each element is determined.
• Question When an organic compound is burned,
  exactly what are the products of combustion?

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Elemental Analysis
Oxygen gas enters the combustion apparatus and
streams over the sample under analysis in a
high-temperature furnace. The absorbers collect
the water vapor and carbon dioxide that are
produced in the combustion.
How do we get mass of oxygen?
(Gives mass of hydrogen)
(Gives mass of carbon)
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Writing Chemical Equations

• A chemical equation is a shorthand description of
  a chemical reaction, using symbols and formulas
  to represent the elements and compounds involved.
• Starting substances are called reactants.
• The substances formed from the reaction are
  called products.
• A plus sign () is used between each reactant and
  between each product.
• An arrow (?) points from the reactants to the
  products.

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Balancing Equations Illustrated
Equations may be balanced only by changing the
coefficients.
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Chemical Equation Notations

• Extra notation sometimes added to each substance
• (g) gas
• (l) liquid
• (s) solid
• (aq) aqueous (water) solution
• Extra notation added above the arrow
• ? heat
• Sometimes the actual temperature is used instead

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Balancing Chemical Equations

• Stoichiometric coefficients are numbers placed in
  front of formulas in a chemical equation to
  balance the equation.
• The coefficients indicate the combining ratios of
  the reactants and the ratios in which products
  are formed.
• For each element, the same number of atoms must
  appear on each side of the equation.

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Guidelines For BalancingChemical Equations

• If an element is present in just one compound on
  each side of the equation, try balancing that
  element first.
• Balance any reactants or products that exist as
  the free element last.
• In some reactions, certain groupings of atoms
  (such as in polyatomic ions) remain unchanged. In
  such cases, balance these groupings as a unit.
• At times, an equation can be balanced most
  readily by first using a fractional
  coefficient(s) and then clearing the fractions by
  multiplying all coefficients by a common
  multiplier.

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Stoichiometric EquivalenceAnd Reaction
Stoichiometry

• Stoichiometric factors, or mole ratios, are
  conversion factors formed from the stoichiometric
  coefficients in a chemical equation.
• In the equation CO(g) 2H2(g) ? CH3OH(l)
• 1 mol CO is chemically equivalent to 2 mol H2
• 1 mol CO is chemically equivalent to 1 mol CH3OH
• 2 mol H2 is chemically equivalent to 1 mol CH3OH
• These conversion factors are commonly used in
  problem-solving.

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Concept of Stoichiometric Equivalence
One car may be equivalent to either 25 feet or 10
feet, depending on the equation (method of
parking).
One mole of CO may be chemically equivalent to
one mole of CH3OH or to one mole of CO2 it
depends on the equation.
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Outline Of Simple Reaction Stoichiometry
Moles of A
Conversion factor from
balanced equation
Moles of B
Note preliminary and followup calculations may
be needed
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Simple Reaction Stoichiometry
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Outline of stoichiometry involving mass
Mass of A
Conversion factor from
molar mass of A
Moles of A
Conversion factor from
balanced equation
Moles of B
Conversion factor from
molar mass of B
Mass of B
Note preliminary and followup calculations may
be needed
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Stoichiometry involving mass
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Stoichiometry involving mass
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Limiting Reactants

• Many reactions are carried out with a limited
  amount of one reactant and a plentiful amount of
  the other(s).
• The reactant that is completely consumed in the
  reaction limits the amounts of products and is
  called the limiting reactant, or limiting
  reagent.
• Keep in mind that the limiting reactant is not
  necessarily the one present in smallest amount.

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Limiting Reactant Analogy
Limiting reactant determines
the amount of product
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Recognizing and Solving Limiting Reactant Problems

• Recognize by the fact that amounts of two (or
  more) reactants are given.
• Solve the same as a normal stoichiometry problem
  except
• preliminary calculations are needed to
  determine which of the two reactants is limiting
  (which reactant should be used to do the
  stoichiometry calculation).

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Recognizing and Solving Limiting Reactant
Problems (contd)

• Select one reactant.
• Determine the amount (moles, grams, mL, etc.) of
  the other reactant is needed to consume the first
  reactant.
• Compare the amount needed to the amount you
  have, and ask
• IS THERE ENOUGH?
• The reactant thats not enough is consumed
  completely it is the limiting reactant.
• Now do the stoichiometry based on the limiting
  reactant.

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Solving Limiting Reactant Problems
Question What mass of nitrogen will react with
35.00 g of Mg?
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Solving Limiting Reactant Problems (contd)
Question What mass of nitrogen will react with
35.00 g of Mg?
Answer 1 mol Mg 1 mol N2 28.0 g
N2 35.00 g Mg x x
x 13.4 g N2 24.3 g Mg 3 mol Mg
1 mol N2
We need 13.4 g N2. We have 15.00 g N2. Clearly,
there is more N2 than is needed. Therefore Mg is
the limiting reactant.
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Solving Limiting Reactant Problems (contd)
Now solve the stoichiometry problem How many
grams of Mg3N2 can be made from 35.00 g Mg? 1
mol Mg 1 mol Mg3N2 100.9 g Mg3N2 35.00 g Mg
x x x
24.3 g Mg 3 mol Mg 1 mol Mg3N2
48.4 g Mg3N2 The preliminary limiting-reactant
calculation was needed to find out which reactant
to use in this calculation the Mg.
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Yields Of Chemical Reactions

• Theoretical yield of a chemical reaction the
  calculated quantity of product in the reaction.
• Actual yield the amount you actually get when
  you carry out the reaction.
• Actual yield will always be less than the
  theoretical yield, for many reasons (can you name
  some?)
• Percent yield

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Solutions AndSolution Stoichiometry

• Solute the substance being dissolved
• Solvent the substance doing the dissolving
• The concentration of a solution refers to the
  quantity of a solute in a given quantity of
  solvent.
• A concentrated solution contains a relatively
  large amount of solute as compared to the
  solvent.
• A dilute solution contains a relatively small
  concentration of solute as compared to the
  solvent.

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Molar Concentration

• Molarity (M), or molar concentration, is the
  amount of solute, in moles, per liter of
  solution.
• Keep in mind that molarity signifies moles of
  solute per liter of solution, not liters of
  solvent.

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Molar Concentration
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Dilution Of Solutions

• Dilution is the process of preparing a more
  dilute solution by adding solvent to a more
  concentrated one.
• Addition of solvent does not change the amount of
  solute in a solution but does change the solution
  concentration.

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Visualizing The Dilution Of A Solution
Same amount of solute in a larger volume of
solution
means lower concentration.
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Dilution of Copper (II) sulfate
Notice that the solute particles are ions
copper(II) and sulfate
Same amount of solute in a larger volume of
solution.
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Dilution calculations
Moles of solute does not change on dilution, so
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Dilution calculations (contd)
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Solutions In Chemical Reactions

• Molarity provides an additional tool in
  stoichiometric calculations based on chemical
  equations.
• Molarity provides conversion factors for
  converting between moles of solute and liters of
  solution.

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Mass of A
Outline of Stoichiometry Including Solutions
Conversion factor from molar mass of A
Mmol/L
Moles of A
V and M of solution A
Conversion factor from
balanced equation
V or M of solution B
Moles of B
Mmol/L
Conversion factor from molar mass of B
Mass of B
Note preliminary and followup calculations may
be needed
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Example of Solution Stoichiometry
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Example of Solution Stoichiometry (contd)
1. Volume and concentration of A
2. Conversion factor from balanced equation
3. We stop at mol of B because thats the unit
we want.
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Example of Solution Stoichiometry (contd)
Start with mol of A
Molarity
From balanced equation
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Summary

• Molecular and formula masses relate to the masses
  of molecules and formula units.
• A mole is an amount of a substance containing
  Avogadros number of elementary entities.
• Avogadros constant (NA) 6.022 x 1023.
• The molar mass of a substance is the mass in
  grams of one mole of that substance.
• Formulas and molar masses can be used to
  calculate mass percent composition.

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Summary (continued)

• A chemical equation uses symbols and formulas for
  the elements and/or compounds involved in a
  reaction.
• Calculations involving reactions use
  stoichiometric factors based on stoichiometric
  coefficients in the balanced equation.
• The limiting reactant determines the amounts of
  products in a reaction.
• The molarity of a solution is the number of moles
  of a solute per liter of solution.