Stoichiometry 1 Formulas and the Mole

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Stoichiometry 1Formulas and the Mole

• L. Scheffler
• Lincoln High School

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The Mole

• Chemical reactions involve atoms and molecules.
• The ratios with which things combine depend on
  the number of atoms not on their mass.
• The masses of atoms or molecules vary depending
  on the substance.
• Atoms individual atoms and molecules are
  extremely small. Hence a larger unit is
  appropriate for measuring quantities of matter.
• A mole is equal to exactly the number of atoms in
  exactly 12.0000 grams of carbon 12. This number
  is known as Avogadros number. 1 mole is equal
  to 6.022 x 1023 particles.

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Definitions of the Mole

• 1 mole of a substance has a mass equal to the
  formula weight in grams.
• Examples
• 1 mole H2O is the number of molecules in 18.015 g
  H2O
• 1 mole H2 is the number of molecules in 2.016 g
  H2.
• 1 mole of atoms has a mass equal to the atomic
  weight in grams.
• 1 mole of particles 6.02214 x 1023 particles
  for any substance!
• The Molar mass is the mass of one mole of a
  substance
• Avogadro's number is the number of molecules in
  one mole for any substance

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The Formula Mass

• The formula mass is the sum of atomic masses in a
  formula.
• If the formula is a molecular formula, then the
  formula mass may also be called a molecular mass.

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Gram Formula Mass and Molar Mass

• If the formula mass is expressed in grams it is
  called a gram formula mass.
• The gram formula mass is also known as the Molar
  Mass.
• The molar mass is the number of grams necessary
  to make 1 mole of a substance.
• The units for Molar Mass are g mol-1.

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Formula Mass and the Mole

• The atomic mass of Carbon 12 is exactly 12.00000.
• 1 atomic mass unit 1/12 of the atomic mass of
  carbon 12.
• The periodic table gives the average atomic mass
  for an element relative to Carbon 12.
• 1 mole of a substance is 6.022 x 1023 particles.
  
• The mole of atomic mass units is equal to 1.000
  gram.

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Gram Formula Mass

• The formula mass is the sum of the atomic masses
  in a formula.
• A gram formula mass is the same number expressed
  in grams.
• It is also equal to Avogadros Number of
  particles
• Example H2O
• From the Periodic Table - Atomic Masses H
  1.00797, O 15.999
• The formula mass 2(1.00797)15.999 18.015
• Adding the unit grams to the formula mass
  transforms it into a gram formula mass or mole.

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The Mole

• The mole is connects the macro world that we can
  measure with the micro world of atoms and
  molecules.
• A Mole is also equal to
• 1 gram formula mass.
• 22.4 dm3 of any gas measured at 0o C and 1.0
  atmosphere of pressure.

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Example 1 Calculating the Molar Mass of a
Compound

• Calculate the gram formula mass or Molar Mass of
  Na3PO4.

Therefore the molar mass is 164.0 g mol-1
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Example 2

• Find the mass of 2.50 moles of Ca(OH)2
• Find the molar mass of Calcium hydroxide and
  multiply by 2.50 mol
• The molar mass of Ca(OH)2 is
• 1 Ca 1 x 40.08 40.08
• 2 O 2 x 16.00 32.00
• 2 H 2 x 1.01 2.02
• Molar Mass 74.10 g mol-1
• 2.50 mol x 74.10 g mol-1 185.25 g

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Calculating Moles

• The number of moles in a given mass of a
  substance can be determined by dividing the mass
  by the molar mass

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Example 3

• Find the number of moles in 44.46 grams of
  Ca(OH)2
• Find the molar mass of and divide it into the
  given mass
• From the previous example the molar mass of
  calcium hydroxide is 74.10 gmol-1.
• 44.46 g Ca(OH)2 . 0. 6000
  mol
• 74.10 g mol-1 Ca(OH)2

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Example 4 Calculating Moles

• Calculate the number of moles in 20.5 grams of
  Na3PO4

0.125 mol
Note Mol is the standard abbreviation for a mole
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Calculating Mass From Moles

• The mass of a quantity of a substance can be
  found by multiplying the number of moles by the
  molar mass

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Example 5 Calculating Mass from Moles

• Calculate the mass of 2.50 moles of Na3PO4

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Percentage Composition

• According to the law of definite proportions,
  compounds, contain definite proportions of each
  element by mass.
• The sum of all of the atomic masses of elements
  in a formula is called the formula mass.
• If it is expressed in grams, then it is called a
  gram formula mass or molar mass.
• If it represents the sum of all of the masses of
  all of the elements in a molecule then it is
  called a molecular mass.
• To find the percentage of each element in a
  compound it is necessary to compare the total
  mass of each element with the formula mass.

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Percentage Composition

• The percent by mass of each element in a compound
  is equal to the percentage that its atomic mass
  is of the formula mass.
• Example Calculate the percentage of oxygen in
  potassium chlorate, KClO3
• Atomic masses K 39.09, Cl 35.45 and O
  16.00.
• Formula mass 39.09 35.45 3(16.00) 122.54
• Percent Oxygen (3(16.00)/122.54) (100)
• 39.17

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Example 2

• Calculate the percentage by mass of each element
  in potassium carbonate, K2CO3
• First calculate the formula mass for K2CO3
  . Find the atomic mass of each element from the
  periodic table. Multiply it by the number of
  times it appears in the formula and add up the
  total
• 2 Potassium atoms K 2 x 39.10
  78.20
• 1 carbon atom C 1 x 12.01
  12.01
• 3 Oxygen atoms O 3 x 16.00
  48.00
• 
  Total 138.21
• To find the percent of each element divide
  the part of the formula mass that pertains to
  that element with the total formula mass
• Percent of Potassium K 78.20 X 100
  56.58
• 
  138.21
• Percent of Carbon C 12.01
  X 100 8.69
  138.21
• Percent of Oxygen O 48.00 X
  100 34.73
• 
  138.21

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Empirical Formula Determination

• The empirical formula is the simplest ratio of
  the numbers of atoms of each element that make a
  compound.
• To find the empirical formula of a compound
• Divide the amount of each element (either in mass
  or percentage) by its atomic mass. This
  calculation gives you moles of atoms for each
  element that appears in the formula
• Convert the results to small whole number ratios.
  Often the ratios are obvious. If they are not
  divide all of the other quotients by the smallest
  quotient

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Example 1

• Analysis of a certain compound showed of 32.356
  grams of compound found that it contained 0.883
  grams of hydrogen, 10.497 grams of Carbon, and
  27.968 grams of Oxygen. Calculate the empirical
  formula of the compound.
• First divide the amount by the atomic mass to get
  the number of moles of each kind of atom in the
  formula
• Hydrogen H 0.883 g
  0.874 mol
• 1.01 g mol-1
• Carbon C 10.497 g 0.874 mol
• 12.01 g mol-1
• Oxygen O 27.968 g 1.748 mol
• 16.00 g mol-1
• Analysis of the ratio s shows that the first two
  are identical and that the third is twice the
  other two. Therefore the ratio of H to C to O is
  1 to 1 to 2. The empirical formula is HCO2

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Molecular Formula

• To calculate the molecular formula from the
  empirical formula it is necessary to know the
  molecular (molar) mass.
• Add up the atomic masses in the empirical formula
  to get the factor Divide this number into the
  molecular formula mass. If the number does not
  divide evenly you probably have a mistake in the
  empirical formula or its formula mass
• Multiply each subscript in the empirical formula
  by the factor to get the molecular formula

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Molecular Formula Example

• Example Suppose the molecular mass of the
  compound in the previous example, HCO2 is 90.0.
  Calculate the molecular formula.
• The empirical formula mass of is
  
• 1 H 1.0 x 1 1.0
• 1 C 12.0 x 1 12.0
• 2 O 16.0 x 2 32.0
• Total 45.0
• Note that 45 is exactly half of the molecular
  mass of 90. So the formula mass of HCO2 is
  exactly half of the molecular mass. Hence the
  molecular formula is double that of the empirical
  formula or H2C2O4

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Part 2 Stoichiometry Problems

• Mass-Mass Problems
• Mass-Volume

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Stoichiometry Problems

• Stoichiometry problems involve the calculation of
  amounts of materials in a chemical reaction from
  known quantities in the same reaction
• The substance whose amount is known is the given
  substance
• The substance whose amount is to be determined is
  the required substance

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Mass to Mass Problems

• Goal To calculate the mass of a substance that
  appears in a chemical reaction from the mass of a
  given substance in the same reaction.
• The given substance is the substance whose mass
  is known.
• The required substance is the substance whose
  mass is to be determined.

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Steps in a Mass to Mass Problem

• Find the gram formula masses for the given and
  the required substances
• Convert the given mass to moles by dividing it by
  the molar mass
• Multiple the given moles by the mole ratio to get
  the moles of the required substance
• Multiple the number of moles of the required
  substance by its molar mass to get the mass of
  the required substance

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Summary of Mass Relationships
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Example 1 Mass-Mass Problem

• Glucose burns in oxygen to form CO2 and H2O
  according to this equation
• C6H12O6 6 O2 ? 6 CO2 6 H2O
• How many grams of CO2 are produced from
  burning 45.0 g of glucose?

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Example 1 Mass-Mass Problem

• Glucose burns in oxygen to form CO2 and H2O
  according to this equation
• C6H12O6 6 O2 ? 6 CO2 6 H2O
• How many grams of CO2 are produced from
  burning 45.0 g of glucose?
• Make sure that the equations is balanced
• Divide the mass of the given by its molar mass

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Example 1 Mass-Mass Problem

• Glucose burns in oxygen to form CO2 and H2O
  according to this equation
• C6H12O6 6 O2 ? 6 CO2 6 H2O
• How many grams of CO2 are produced from
  burning 45.0 g of glucose?
• Make sure that the equations is balanced
• Divide the mass of the given by its molar
• 3. Multiply by the mole ratio

1.5 moles CO2
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Example 1 Mass-Mass Problem

• Glucose burns in oxygen to form CO2 and H2O
  according to this equation
• C6H12O6 6 O2 ? 6 CO2 6 H2O
• How many grams of CO2 are produced from
  burning 45 g of glucose?
• Make sure that the equations is balanced
• Divide the mass of the given by its molar
• 3. Multiply by the mole ratio
• 4. Multiply by the molar mass of the required

66.0 g of CO2
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Example 2 Mass-Mass Problem

• What mass of Barium chloride is required to
  react with 48.6 grams of sodium phosphate
  according to the following reaction
• 2 Na3PO4 3BaCl2 ? Ba3(PO4)2 6 NaCl

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Example 2

• What mass of Barium chloride is required to
  react with 48.6 grams of sodium phosphate
  according to the following reaction
• 2 Na3PO4 3BaCl2 ? Ba3(PO4)2 6 NaCl

Molar Masses Na3PO4 3(23.0)31.04(16.0) 164
g mol-1 BaCl2 137.3 2(35.5)
208.3 g mol-1
92.6 g of BaCl2
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Example 3

• What mass of carbon dioxide is produced from
  burning 100 grams of ethanol in oxygen according
  to the following reaction
• C2H5OH 3 O2 ? 2 CO2 3 H2O

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Example 3

• What mass of carbon dioxide is produced from
  burning 100 grams of ethanol in oxygen according
  to the following reaction
• C2H5OH 3 O2 ? 2 CO2 3 H2O

Molar Masses C2H5OH 2(12) 6(1) 16
46 CO2 12 2(16) 44.0
191.3 g CO2
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Mass to Volume Problems
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Mass to Volume Problems

• Goal To calculate the volume of a gas that
  appears in a chemical reaction from the mass of a
  given substance in the same reaction.
• The given substance is the substance whose mass
  is known.
• The required substance is the gas whose volume is
  to be determined.
• Remember 1 mole of any gas at STP is equal to
  22.4 dm3. STP is defined as 0 oC and 1
  atmosphere of pressure.

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Steps in a Mass to Volume Problem

• Find the gram formula masses for the given
  substance.
• Convert the given mass to moles by dividing it by
  the molar mass
• Multiple the given moles by the mole ratio to get
  the moles of the required substance
• Multiple the number of moles of the required
  substance by the molar volume, 22.4 dm3 mol-1, to
  get the volume of the required substance.
• This procedure is only valid if the required
  substance is a gas. It does not work for solids,
  liquids, or solutions.

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Example 1 Mass-Volume Problem

• Sucrose burns in oxygen to form CO2 and H2O
  according to this equation
• C12H22O11 12 O2 ? 12 CO2 11 H2O
• What volume of CO2 measured at STP is produced
  from burning 100 g of sucrose?

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Example 1 Mass-Volume Problem

• Sucrose burns in oxygen to form CO2 and H2O
  according to this equation
• C12H22O11 12 O2 ? 12 CO2 11 H2O
• What volume of CO2 measured at STP is
  produced from burning 100 g of sucrose?
• 1. Find the molar mass of the given substance

Molar mass of C12H22O11 12 (12.0) 22 (1.0)

11 (16.0) 342.0 g mol-1
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Example 1 Mass-Volume Problem

• Sucrose burns in oxygen to form CO2 and
  H2O according to this equation
• C12H22O11 12 O2 ? 12 CO2 11 H2O
• What volume of CO2 measured at STP is
  produced from burning 100 g of sucrose?
• 2. Find moles of the given

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Example 1 Mass-Volume Problem

• Sucrose burns in oxygen to form CO2 and H2O
  according to this equation
• C12H22O11 12 O2 ? 12 CO2 11 H2O
• What volume of CO2 measured at STP is
  produced from burning 100 g of sucrose?
• 3. Multiply by the mole ratio

3.51 moles CO2
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Example 1 Mass-Volume Problem

• Sucrose burns in oxygen to form CO2 and H2O
  according to this equation
• C12H22O11 12 O2 ? 12 CO2 11 H2O
• What volume of CO2 measured at STP is
  produced from burning 100 g of sucrose?
• 4. Multiply by the molar volume, 22.4 dm3 mol-1.

78.6 dm3
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Example 2 Mass-Volume Problem

• What volume of carbon dioxide gas would be
  produced by reacting 25.0 g of sodium carbonate
  with hydrochloric acid according to the following
  reaction
• Na2CO3 2 HCl ? 2 NaCl CO2 H2O

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Example 2 Mass-Volume Problem

• What volume of carbon dioxide gas would be
  produced by reacting 25.0 g of Sodium carbonate
  with hydrochloric acid according to the following
  reaction
• Na2CO3 2 HCl ? 2 NaCl CO2 H2O
  

Molar Mass Na2CO3 2(23.0) 12.0 3(16.0)
106.0
5.28 dm3 of CO2
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Summary of Stoichiometric Relationships
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Solutions and Stoichiometry

• Many times the reactants and/or products of
  chemical reactions are water solutions.
• In these cases the concentration of the solution
  must be determined in order to determine amounts
  of reactants or products
• The concentration of a solution is a measure of
  the amount of solute that is dissolved in a given
  amount of solution

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Molarity

• The most common concentration unit is Molarity

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Molarity Calculations

• How many grams of NaOH are required to prepare
  250 cm3 of 0.500 M solution?
• Molar Mass of NaOH 23161 40.0 g/mol
• 250 cm3 0.250 dm3

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Molarity Calculations

• Calculate the concentration of a NaCl solution
  that contains 24.5 g of NaCl in 250 cm3 of
  solution.
• Molar mass of NaCl 23.0 35.5 58.5

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Stoichiometry Calculations Involving Solutions

• 20.0 cm3 of a 0.500 M AgNO3 solution is
  required to precipitate the sodium chloride in
  10.0 cm3 of a salt solution. What is the
  concentration of the solution?
• AgNO3 (aq) NaCl (aq) ?AgCl (s) KNO3
  (aq)

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Stoichiometry Calculations Involving Solutions

• 20.0 cm3 of a 0.500 M AgNO3 solution is
  required to precipitate the sodium chloride in 10
  cm3 of a salt solution. What is the
  concentration of the solution?
• AgNO3 (aq) NaCl (aq) ?AgCl (s) KNO3
  (aq)
• -- Molar Mass NaCl 23.0 35.5 58.5 g/mol

0.585 g of NaCl