Chemical Quantities

1
Chemical Quantities

• Chapter 9

2
Chemical Stoichiometry

• Stoichiometry - The study of quantities of
  materials consumed and produced in chemical
  reactions.
• Stoichiometry is used to determine how much
  stomach acid an antacid tablet can neutralize.

3
Chemical Equations

• A balanced chemical equation is like a recipe.
  One needs to know what the ingredients are and
  what relative amounts of ingredients are needed
  for both recipes and chemical equations.

4
Chemical Equation

• A representation of a chemical reaction
• C2H5OH(l) __O2(g) ? __CO2(g) __H2O(g)
• reactants products
• Is this equation balanced?
• C2H5OH(l) 3O2(g) ? 2CO2(g) 3H2O(g)

5
Chemical Equation

• C2H5OH (l) 3O2(g) ? 2CO2(g) 3H2O(g)
• Microscopic
• 1 molecule of ethanol reacts with 3 molecules of
  oxygen to produce 2 molecules of carbon dioxide
  and 3 molecules of water.
• Macroscopic
• 1 mole of ethanol reacts with 3 moles of oxygen
  to produce 2 moles of carbon dioxide and 3 moles
  of water.

6
Chemical Equation

• C2H5OH (l) 3O2(g) ? 2CO2(g) 3H2O(g)
• Microscopic
• 1 dozen molecules of ethanol reacts with 3 dozen
  molecules of oxygen to produce 2 dozen molecules
  of carbon dioxide and 3 dozen molecules of water.
• Macroscopic
• 6.02 x 1023 molecules of ethanol reacts with
  3(6.02 x 1023) molecules of oxygen to produce
  2(6.02 x 1023) molecules of carbon dioxide and
  3(6.02 x 1023 ) molecules of water.

7
Chemical Equation

• C2H5OH (l) 3O2(g) ? 2CO2(g) 3H2O(g)
• Macroscopic
• 46.0 g of ethanol reacts with 96.0g of oxygen to
  produce 88.0 g of carbon dioxide and 54.0 g of
  water.
• 142.0 g of reactants 142.0 g of products.
• Atoms and mass are conserved in a chemical
  reaction, but moles and molecules are not!!!

8
Moles Molecules

• __C3H8(g) __O2(g) ---gt __CO2(g) __HOH(g)
• C3H8(g) 5O2(g) ---gt 3CO2(g) 4HOH(g)
• How many moles and molecules of each substance
  are there?

9
Mole Ratio

• Mole ratio -- a conversion factor based upon a
  balanced equation and used to determine relative
  amounts of reactants and products.
• Mole ratios can exist between a reactant and a
  product, between two reactants, or between two
  products.
• C3H8(g) 5O2(g) ---gt 3CO2(g) 4HOH(g)

10
Types of Stoichiometry Problems

• Mole to Mole
• Gram to Mole
• Mole to Gram
• Gram to Molecules
• Molecules to Gram

11
Calculating Masses of Reactants and Products

• 1. Balance the equation.
• 2. Convert mass to moles.
• 3. Set up mole ratios.
• 4. Use mole ratios to calculate moles of desired
  reactant or product.
• 5. Convert moles to grams, if necessary.

12
Mole To Mole Problems

• What number of moles of oxygen would be used in
  burning 5.8 moles of propane, C3H8?
• C3H8(g) 5O2(g) ---gt 3CO2(g) 4HOH(g)
• (5.8 mol C3H8)(5 mol O2/1 mol C3H8)
• 29 mol O2

13
Gram to Mole Gram to Gram

• __Al(s) __I2(s) ---gt __AlI3(s)
• 2Al(s) 3I2(s) ---gt 2AlI3(s)
• How many moles and how many grams of aluminum
  iodide can be produce from 35.0 g of aluminum?

14
Gram to Mole Gram to Gram

• 2Al(s) 3I2(s) ---gt 2AlI3(s)
• (35.0 g Al) (1 mol/26.98 g)(2 mol AlI3/2 mol Al)
  1.30 mol AlI3
• (35.0 g Al) (1 mol/26.98 g)(2 mol AlI3/2 mol
  Al)(407.68 g/1 mol) 529 g AlI3

15
Grams to Molecules

• __LiOH(s) __CO2(g) ---gt __Li2CO3(s) __HOH(l)
• 2LiOH(s) CO2(g) ---gt Li2CO3(s) HOH(l)
• How many molecules of water would be formed from
  1.00 x 103 g of LiOH?
• (1.00 x 103 g LiOH)(1 mol/23.95 g)(1 mol HOH/
• 2 mol LiOH)(6.02 x 1023 molecules/1 mol)
• 1.26 x 1025 molecules HOH

16
Stoichiometric Quantities

• Stoichiometric Quantities -- quantities of
  reactants mixed in exactly the amounts that
  result in their all being used up at the same
  time.
• How often do you think this occurs in reality?
• Almost never!!!!

17
Limiting Reactant

• The limiting reactant is the reactant that is
  consumed first, limiting the amounts of products
  formed.
• Almost all stoichiometric situations are of the
  limiting reactant type.
• The reactants that are left over and unreacted
  are said to be in excess.

18
Figure 9.1 A mixture of 5CH4 and 3H20 molecules
undergoes the reaction CH4(g) H20(g) ---gt 3H2
CO(g)
19
Double Cheeseburger Problem

• At the local Burger Barn a worker finds the
  following inventory
• 22 hamburger patties
• 15 hamburger buns
• 7 slices of onion
• 18 slices of cheese
• How many double cheeseburgers with onion and
  cheese can be made to sell?

20
Double Cheeseburger Problem

• 2 h.b. patties 1 h.b. bun 2 slices cheese
• 1 slice onion ---gt 1 double cheeseburger
• What is the limiting reactant?
• Onion
• How many double cheeseburgers with onion and
  cheese can be made to sell?
• 7 double cheeseburgers

21
Double Cheeseburger Problem

• 2 h.b. patties 1 h.b. bun 2 slices cheese
• 1 slice onion ---gt 1 double cheeseburger
• What materials are in excess and by how much?
• 8 hamburger patties
• 8 hamburger buns
• 4 slices cheese

22
Solving a Stoichiometry Problem

• 1. Balance the equation.
• 2. Convert masses to moles.
• 3. Determine which reactant is limiting.
• 4. Use moles of limiting reactant and mole ratios
  to find moles of desired product.
• 5. Convert from moles to grams.

23
Limiting Reactant Problem

• If 56.0 g of Li reacts with 56.0 g of N2, how
  many grams of Li3N can be produced?
• __Li(s) __N2(g) ---gt __Li3N(s)
• 6 Li(s) N2(g) ---gt 2 Li3N(s)
• (56.0 g Li) (1 mol/6.94g)(1 mol N2/6 mol Li)
  (28.0 g/1 mol) 37.7 g N2
• Since there were 56.0 g of N2 and only 37.7 g
  used, N2 is the excess and Li is the Limiting
  Reactant.

24
Limiting Reactant Problem

• 6 Li(s) N2(g) ---gt 2 Li3N(s)
• (56.0 g Li)(1 mol/6.94g)(2 mol Li3N/6 mol Li)
  (34.8 g/1 mol) 93.6 g Li3N
• How many grams of nitrogen are left?
• 56.0g N2 given - 37.7 g used 18.3 g excessN2

25
Double Cheeseburger Yield

• At the local Burger Barn a worker finds the
  following inventory
• 22 hamburger patties
• 15 hamburger buns
• 7 slices of onion
• 18 slices of cheese
• We found that seven double cheeseburgers could be
  made from these ingredients.

26
Double Cheeseburger Yield

• If a worker eats one slice of onion, how many
  double cheeseburgers can actually be made?
• 6 double cheeseburgers
• The number of cheeseburgers that could have been
  made (7) is the theoretical yield.
• The number of cheeseburgers that actually were
  made (6) is the actual yield.

27
Double Cheeseburger Yield

• Yield 85.7

28
Yield

• Values calculated using stoichiometry are always
  theoretical yields!
• Values determined experimentally in the
  laboratory are actual yields!

29
Limiting Reactant Yield

• If 68.5 kg of CO(g) is reacted with 8.60 kg of
  H2(g), what is the theoretical yield of methanol
  that can be produced?
• __H2(g) __CO(g) ---gt __CH3OH(l)
• 2 H2(g) CO(g) ---gt CH3OH(l)
• (68.5 kg CO)(1 mol/28.0 g)(2 mol H2/1 mol CO)
• (2.02 g/1mol) 9.88 kg H2

30
Limiting Reactant Yield

• 2 H2(g) CO(g) ---gt CH3OH(l)
• Since only 8.60 kg of H2 were provided, the H2 is
  the limiting reactant, and the CO is in excess.
• (8.60 kg H2)(1000 g/1 kg)(1 mol/2.02 g)(1 mol
  CH3OH/2 mol H2)(32.0 g/1 mol) 6.85 x 104 g CH3OH

31
Limiting Reactant Yield

• 2 H2(g) CO(g) ---gt CH3OH(l)
• If in the laboratory only 3.57 x 104 g of CH3OH
  is produced, what is the yield?

Yield 52.1