Stoichiometry

1
Stoichiometry

• Atomic Mass
• The Mole concept
• Molar Mass
• Percent Composition of Compounds
• Determination of Formula of Compounds
• Writing and Balancing Chemical Equations
• Interpreting balance equations, and
• Stoichiometric Calculations

2
Atomic Masses

• Relative atomic masses are determined using mass
  spectrometer
• Carbon-12 is used as atomic mass reference and is
  assigned an atomic mass of 12 amu exactly
• Other atoms are assigned masses relative to
  carbon-12 that is, how many times they are
  heavier (or lighter) than carbon-12
• Recorded atomic mass of an element is the
  weighted average mass of its naturally occurring
  isotopes

3
Calculation of Relative Atomic Masses

• Example-1
• The ratio of atomic mass of an oxygen atom to
  carbon-12 is 1.3329. If the atomic mass of
  carbon-12 is 12 amu exactly, what is the atomic
  mass of oxygen?
• Atomic mass of oxygen 1.3329 x 12 amu
• 
  15.995 amu

4
Calculation of Average Atomic Masses

• Example-2
• Copper is composed of two naturally occurring
  isotopes copper-63 (69.09 62.93 amu) and
  copper-65 (30.91 64.93 amu). What is the
  average atomic mass of copper?
• Atomic mass of copper
• (0.6909 x 62.93 amu) (0.3091 x
  64.93 amu)
• 63,55 amu (as given in the periodic
  table)

5
Molar Quantity

• The Mole
• A quantity that contains the Avogadros number of
  items
• Avogadros number 6.022 x 1023
• 12 g of carbon-12 isotope contains the Avogadros
  number of carbon atoms.
• then, 12 g of carbon-12 1 mole

6
Gram-Atomic Mass

• Mass of a carbon-12 atom 12 amu (exactly)
• Mass of 1 mole of carbon-12 12 g
• Mass of an oxygen atom 16.00 amu
• Mass of 1 mole of oxygen 16.00 g
• Gram-atomic mass mass (in grams) of 1 mole of
  the element that is, the mass (in grams) that
  contains the Avogadros number of atoms of that
  element.
• gram-atomic mass is molar mass of an element (in
  gram).

7
Atomic Mass Gram-Atomic Mass

• Examples
• Element Atomic mass Gram-atomic mass
• Carbon 12.01 amu 12.01 g/mole
• Oxygen 16.00 amu 16.00 g/mole
• Aluminum 26.98 amu 26.98 g/mole
• Silicon 28.09 amu 28.09 g/mole
• Gold 197.0 amu 197.0 g/mole

8
Molecular Mass and Molar Mass

• Molecular mass the mass of a molecule in amu
• Molar mass the mass of one mole of an element
  or a compound, expressed in grams.
• Examples
• Molecular Mass Molar Mass
• N2 28.02 amu 28.02 g/mole
• H2O 18.02 amu 18.02 g/mole
• C8H18 114.22 amu 114.22 g/mole

9
Calculating Molar Mass

• Calculating the molar mass of sucrose, C12H22O11
• (12 x 12.01 g) (22 x 1.008 g) (11 x 16.00 g)
• 342.3 g/mole
• Molar mass of ammonium hydrogen phosphate,
  (NH4)2HPO4
• (2 x 14.01 g) (9 x 1.008 g) (1 x 30.97 g)
  (4 x 16.00 g)
• 132.06 g/mole

10
Percent Composition of a Compound

• Composition of aluminum sulfate, Al2(SO4)3
• Molar mass of Al2(SO4)3
• (2 x 26.98 g) (3 x 32.06 g) (12 x 16.00 g)
  342.14 g/mole
• Mass percent of Al (53.96 g/342.14 g) x 100
  15.77
• Mass percent of S (96.18 g/342.14 g) x 100
  28.11
• Mass percent of O (192.0 g/342.14 g) x 100
  56.12

11
Formula of Compounds

• Empirical Formula
• A chemical formula that represents a simple whole
  number ratio of the number of moles of elements
  in the compound. Examples MgO, Cu2S, CH2O, etc.
• Molecular Formula
• A formula that shows the actual number of atoms
  of each type in a molecule.
• Examples C4H10, C6H6, C6H12O6.

12
Empirical Formula-1

• Empirical formula from composition
• Example A compound containing carbon, hydrogen,
  and oxygen has the following composition (by mass
  percent) 68.12 C, 13.73 H, and 18.15 O,
  Determine its empirical formula.
• Solution
• Use mass percent to calculate mole and mole ratio
  of CHO
• Mole of C 68.12 g x (1 mol C/12.01 g) 5.672
  mol C
• Mole of H 13.73 g x (1 mol H/1.008 g) 13.62
  mol H
• Mole of O 18.15 g x (1 mol O/16.00 g) 1.134
  mol O
• Divide all moles by mole of O (smallest value) to
  get simple ratio
• 5.672 mol C/1.134 mol O 5 13.62 mol H/1.134
  mol O 12, and
• 1.134 mol O/1.134 mol O 1
• Mole ratio 5C12H1O ? Empirical formula
  C5H12O

13
Empirical Formula-2

• Empirical formula from mass of elements in a
  sample of compound
• Example When 1.96 g of phosphorus is burned,
  4.49 g of a phosphorus oxide is obtained.
  Calculate the empirical formula of the phosphorus
  oxide.
• Solution
• Calculate moles of P and O in the sample and
  obtain a simple mole ratio
• Mole of P 1.96 g P x (1 mol/30.97 g) 0.0633
  mol P
• Mole of O (4.49 g 1.96 g) x (1 mol/16.00 g)
  0.158 mol O
• Divide all moles by mole of P (smaller value) to
  get a simple mole ratio
• 0.0633 mol P/0.0633 1 mol P 0.158 mol
  O/0.0633 2.5 mol O
• Mole ratio 1 mol P to 2.5 mol O, OR 2 mol P to
  5 mol O
• Empirical formula P2O5

14
Empirical Formula-3

• Empirical formula from data of combustion
  reaction
• Example A compound is composed of carbon,
  hydrogen, and oxygen. When 2.32 g of this
  compound is burned in excess of oxygen, it
  produces 5.28 g of CO2 gas and 2.16 g of water.
  Calculate the composition (in mass percent) of
  the compound and determine its empirical formula.
• Solution
• Find mass of C, H, and O in the sample and then
  calculate their mass percent
• Mass of C 5.28 g CO2 x (12.01 g C/44.01 g CO2)
  1.44 g
• Mass of C (1.44 g C/2.32 g sample) x 100
  62.1
• Mass of H 2.16 g H2O x (2 x 1.008 g/18.02 g
  H2O) 0.24 g
• Mass of H (0.242 g H/2.32 g sample) x 100
  10.4
• Mass of O 2.32 g sample 1.44 g C 0.24 g H
  0.64 g
• Mass of O 100 62.1 C 10.4 H 27.5
• Derive empirical formula from these mass percent
  composition (next slide)

15
Empirical Formula-3

• Empirical formula from data of combustion
  (continued)
• Calculate mole and simple mole ratio from
  calculated mass of each element
• Mole of C 1.44 g C x (1 mol/12.01 g) 0.12 mol
  
• Mole of H 0.242 g x (1 mol/1.008 g) 0.24 mol
• Mole of O 0.64 g x (1 mol/16.00 g) 0.04 mol
• Divide all moles by mole of O (smallest mole) to
  obtain a simple ratio
• 0.12 mol C/0.04 3 mol C 0.24 mol H/0.04 6
  mol H
• 0.04 mol O/0.04 1 mol O
• Simple molar ratio 3 mol C 6 mol H 1 mol O
• Empirical formula C3H6O

16
Molecular Formula

• Molecular formula is derived from empirical
  formula and molecular mass, which should be
  obtained independently
• If empirical formula has the form CxHyOz, then
  molecular formula will be (CxHyOz)n, where n
  (molecular mass/empirical formula mass)
• Example
• A compound has an empirical formula C3H6O and its
  molecular formula is 116.2 u. What is the
  molecular formula of the compound?
• Solution
• Empirical formula mass (2 x 12.01 u) (6 x
  1.008 u) 16.00 u 58.1 u
• Molecular formula (C3H6O)n where n (116.2
  u/58.1 u) 2
• Incorrect molecular formula (C3H6O)2
• Correct molecular formula C6H12O2

17
Chemical Equation 1

• Description of reaction 1
• Iron reacts with oxygen gas and forms solid
  iron(III) oxide
• Identity reactants iron (Fe) and oxygen gas
  (O2) product iron(III) oxide
• Chemical equation Fe(s) O2(g) ? Fe2O3(s)
• Balanced equation 4Fe(s) 3 O2(g) ? 2Fe2O3(s)

18
Chemical Equation 2

• Description of reaction 2
• Propane gas (C3H8) is burned in air (excess of
  oxygen) to form carbon dioxide gas and water
  vapor
• Identity reactants C3H8(g) and O2(g)
• products CO2(g) and H2O(g)
• Equation C3H8(g) O2(g) ? CO2(g) H2O(g)
• Balanced equation
• C3H8(g) 5 O2(g) ? 3CO2(g) 4H2O(g)

19
Chemical Equation 3

• Description of reaction 3
• Ammonia gas (NH3) reacts with oxygen gas to form
  nitrogen monoxide gas and water vapor
• Identity reactants NH3(g) and O2(g) products
  NO(g) and H2O(g)
• Equation NH3(g) O2(g) ? NO(g) H2O(g)
• Balanced eqn. 2NH3(g) 5/2 O2(g) ? 2NO(g)
  3H2O(g)
• A balanced equation should not contain fraction
  as coefficients. Multiply throughout by 2 to get
  rid of the fraction
• Properly balanced equation
• 4NH3(g) 5 O2(g) ? 4NO(g) 6H2O(g)

20
Balancing Chemical Equations

• Rules for balancing equations
• Use smallest integer coefficients in front of
  each reactants and products as necessary
  coefficient 1 need not be indicated
• The formula of the substances in the equation
  MUST NOT be changed.
• Helpful steps in balancing equations
• Begin with the compound that contains the most
  atoms or types of atoms.
• Balance elements that appear only once on each
  side of the arrow.
• Next balance elements that appear more than once
  on either side.
• Balance free elements last.
• Finally, check that smallest whole number
  coefficients are used.

21
Stoichiometry

• Stoichiometry the quantitative relationships
  between one reactant to another, or between a
  reactant and products in a chemical reaction.
• Interpreting balanced equations
• Example C3H8(g) 5 O2(g) ? 3CO2(g) 4H2O(g)
• The equation implies that
• 1 C3H8 molecule reacts with 5 O2 molecules to
  produce 3 CO2 molecules and 4 H2O molecules OR
• 1 mole of C3H8 reacts with 5 moles of O2 to
  produce 3 moles of CO2 and 4 moles of H2O.

22
Stoichiometric Calculations

• Mole-to-mole relationship
• Example In the following reaction, if 6.0 moles
  of octane, C8H18, is completely combusted in
  excess of oxygen gas, how many moles of CO2 and
  H2O, respectively, will be formed? How many moles
  of O2 does it consumed?
• Reaction 2C8H18(l) 25 O2(g) ? 16CO2(g)
  18H2O(g)
• Calculations
• Mole CO2 formed 6.0 mol C8H18 x (16 mol
  CO2/2mol C8H18) 48 moles
• Mole H2O formed 6.0 mol C8H18 x (18 mol
  H2O/2mol C8H18) 54 moles
• Mole O2 consumed 6.0 mol C8H18 x (25 mol
  O2/2mol C8H18) 75 moles

23
Stoichiometric Calculations

• Mass-to-mole-to-mole-to-mass relationship
• Example-1 In the following reaction, if 690 g of
  octane, C8H18, is completely combusted in excess
  of oxygen gas, how many grams of CO2 are formed?
• Reaction 2C8H18(l) 25 O2(g) ? 16CO2(g)
  18H2O(g)
• Calculation-1
• Moles C8H18 reacted 690 g C8H18 x (1 mol/114.2
  g) 6.0 moles
• Moles CO2 formed 6.0 mol C8H18 x (16 mol CO2/2
  mol C8H18)
• 48 moles CO2
• Mass of CO2 formed 48 mol CO2 x (44.01 g/mol)
  2.1 x 103 g

24
Stoichiometric Calculations

• Mass-to-mole-to-mole-to-mass relationship
• Example-2 In the following reaction, if 690 g of
  octane, C8H18, is completely combusted in excess
  of oxygen gas, how many grams of H2O are formed?
• Reaction 2C8H18(l) 25 O2(g) ? 16CO2(g)
  18H2O(g)
• Calculation-2
• Moles C8H18 reacted 690 g C8H18 x (1 mol/114.2
  g) 6.0 moles
• Moles H2O formed 6.0 mol C8H18 x (18 mol H2O/2
  mol C8H18)
• 54 moles CO2
• Mass of H2O formed 54 mol H2O x (18.02 g/mol)
  970 g

25
Stoichiometric Calculations

• Mass-to-mole-to-mole-to-mass relationship
• Example-3 In the following reaction, if 690 g of
  octane, C8H18, is completely combusted in excess
  of oxygen gas, how many grams of oxygen gas are
  consumed?
• Reaction 2C8H18(l) 25 O2(g) ? 16CO2(g)
  18H2O(g)
• Calculation-3
• Moles C8H18 reacted 690 g C8H18 x (1 mol/114.2
  g) 6.0 moles
• Moles O2 consumed 6.0 mol C8H18 x (25 mol O2/2
  mol C8H18)
• 75 moles O2
• Mass of H2O formed 75 mol O2 x (32.00 g/mol)
  2.4 x 103 g g

26
Stoichiometry Involving Limiting Reactant

• Limiting reactant
• one that got completely consumed in a chemical
  reaction before the other reactants.
• Product yield depends on the amount of limiting
  reactant
• Example for reaction 2Cu(s) S(s) ? Cu2S(s),
  2 moles of copper are required to react
  completely with 1 mole of sulfur, which will
  produce 1 mole of copper(I) sulfide.
• If a reaction is carried out using 1 mole of
  copper and 1 mole of sulfur, then copper will be
  the limiting reactant and sulfur is in excess.
  Only 0.5 mole of copper(I) sulfide is obtained.

27
Reactions Involving Limiting Reactant

• In the reaction N2(g) 3H2(g) ? 2NH3(g)
• Which is the limiting reactant if 20.0 g of H2
  is reacted with 80.0 g of N2? How many grams of
  NH3 will be formed?

28
Theoretical, Actual and Percent Yields

• Theoretical yield
• yield of product calculated based on the
  stoichiometry of balanced equation and amount of
  limiting reactant (assuming the reaction goes to
  completion and the limiting reactant is
  completely consumed).
• Actual Yield
• Yield of product actually obtained from
  experiment
• Percent Yield (Actual yield/Theoretical yield)
  x 100