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Stoichiometry
(Mass Relationships in Chemical Reactions)
Chapter 3
Stoichiometric Amounts: The exact molar amounts of
reactants and products that appear in the balanced chemical
equation.
Dr. Sa’ib Khouri
AUM- JORDAN
Chemistry
By Raymond Chang
2
Carbon-12 (6 protons and 6 neutrons):
1 atom 12C “weighs” 12 amu
i.e. amu is 1/12 the mass of carbon atom
By experiments: H = 1.008 amu
O = 16.00 amu
Fe = 55.85 amu
Atomic mass is the mass of an atom in
atomic mass units (amu)
Micro World
atoms & molecules
Macro World
grams
3
The average atomic mass is the weighted
average of all of the naturally occurring
isotopes of the element.
4
How?
Carbon:
98.90% 12C (12.00000 amu)
1.10% 13C (13.00335 amu)
98.90 x 12.00000 + 1.10 x 13.00335
100
= 12.01 amu
Average atomic mass of carbon:
Look to example 3.1 on copper
5
Average atomic mass (6.941)
6
In the SI system, the mole (mol) is the amount of a
substance that contains as many elementary entities
(atoms, molecules, or other particles) as there are
atoms in exactly 12 g (or 0.012 kg) of 12C isotope
1 mol = NA = 6.0221415 x 1023
NA: Avogadro’s number (Italian scientist Amedeo Avogadro)
Dozen = 12 items Pair = 2 items
The mole: A unit to count numbers of particles
Avogadro’s Number and Molar mass of an element
7
Molar mass (M) is the mass (in grams or kilograms)
of one mole of units (such as atoms or molecules)of a
substance.
1 mol 12C atoms = 6.022 x 1023 atoms = exactly 12 g
1 mol C-12 atoms = 12.00 g 12C
1 mol lithium atoms = 6.941 g of Li
1 mol sodium atoms = 22.99 g Na
For any element
atomic mass (amu) = molar mass (grams)
8
One Mole of:
C S
Cu Fe
Hg
12.01 g
32.07 g
200.6 g
55.85 g63.55 g
9
therefore
1 amu = 1.661 x 10-24 g or 1 g = 6.022 x 1023 amu
1 12C atom
12 amu
x
12 g
6.022 x 1023 12C atoms
=
1.661 x 10-24 g
1 amu
What is the mass per gram of amu?
10
x 6.022 x 1023 Na atoms
1 mol Na
=
e.g.
How many atoms are in 0.551 g of sodium (Na) ?
1 mol Na = 22.99 g Na
1 mol Na = 6.022 x 1023 Na atoms
0.551 g Na 1 mol Na
22.99 g Na
x
1.44 x 1022 Na atoms
11
Molecular mass
is the sum of the atomic masses (in amu) in the molecule
1S 32.07 amu
2O + 2 x 16.00 amu
SO2 64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g
SO2
e.g.
H.W. What is the molecular mass (amu) and molar
mass (g) of caffeine (C8H10N4O2)
12
e.g.
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol H = 6.022 x 1023 H atoms
5.82 x 1024 atoms H
1 mol C3H8O molecules = 8 mol H atoms
72.5 g C3H8O
1 mol C3H8O
60 g C3H8O
x
8 mol H atoms
1 mol C3H8O
x
6.022 x 1023 H atoms
1 mol H atoms
x =
HW. How many H atoms are present in 43.8 g urea
[(NH2)2CO] M of urea is 60.06 g
13
Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound
(like NaCl and MgO)
1Na 22.99 amu
1Cl + 35.45 amu
NaCl 58.44 amu
For any ionic compound
formula mass (amu) = molar mass (grams)
1 formula unit NaCl = 58.44 amu
1 mole NaCl = 58.44 g NaCl
NaCl
14
e.g. What is the formula mass (amu) and molar mass
(grams) of Ca3(PO4)2 ?
1 formula unit of Ca3(PO4)2
3 Ca 3 x 40.08
2 P 2 x 30.97
8 O + 8 x 16.00
310.18 amu
The molar mass = 310.18 g
15
Percent composition (% by mass) of an element in a
compound
=
n x molar mass of element
molar mass of compound
x 100%
n is the number of moles of an element in 1 mole
of a compound that contains of this element
C2H6O
%C =
2 x (12.01 g)
46.07 g
x 100% = 52.14%
%H =
6 x (1.008 g)
46.07 g
x 100% = 13.13%
%O =
1 x (16.00 g)
46.07 g
x 100% = 34.73%
52.14% + 13.13% + 34.73% = 100.0%
e.g.
16
Percent Composition and Empirical Formulas
e.g.
Determine the empirical formula of a
compound that has the following
percent composition by mass:
K 24.75, Mn 34.77, O 40.51 percent.
nK = 24.75 g K x = 0.6330 mol K
1 mol K
39.10 g K
nMn = 34.77 g Mn x = 0.6329 mol Mn
1 mol Mn
54.94 g Mn
nO = 40.51 g O x = 2.532 mol O
1 mol O
16.00 g O
Suppose total mass is 100 gram
17
Percent Composition and Empirical Formulas
K : ~~ 1.0
0.6330
0.6329
Mn :
0.6329
0.6329
= 1.0
O : ~~ 4.0
2.532
0.6329
nK = 0.6330, nMn = 0.6329, nO = 2.532
KMnO4
Divide all by the smallest one
18
Soln.
Moles of N= 87.5/14 = 6.25
Moles of H= 12.5/1 = 12.5
A compound with a percent composition by mass of 87.5% N and
12.5% H, What is the empirical formula of this compound?
A) NH2
B) N2H3
C) NH
D) N2H2
E) N2H
e.g.
N6.25 H12.5
19
g CO2 mol CO2 mol C g C
g H2O mol H2O mol H g H
g of O = g of sample – (g of C + g of H)
e.g. Combustion of 11.5 g of ethanol
Produced 22.0 g CO2 and 13.5 g H2O
6.0 g C = 0.5 mol C
1.5 g H = 1.5 mol H
4.0 g O = 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
Experimental Determination of Empirical Formulas
Determination of Molecular Formulas
To calculate the actual molecular formula we
must know the approximate molar mass of the
compound in addition to its empirical formula.
23
3 ways of representing the reaction between H2 and O2 to form H2O
Chemical reaction: a process in which a substance (or
substances) is changed into one or more new substances
A chemical equation uses chemical symbols to show
what happens during a chemical reaction
reactants products
Chemical reactions and chemical equations
24
How to “Read” Chemical Equations
2 Mg + O2 2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
25
Balancing Chemical Equations
1. 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.
e.g.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2 CO2 + H2O
2. 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.
2C2H6 NOT C4H12
26
3. Start by balancing those elements that appear in
only one reactant and one product.
C2H6 + O2 CO2 + H2O start with C or H but not O
2 carbon
on left
1 carbon
on right
multiply CO2 by 2
C2H6 + O2 2CO2 + H2O
6 hydrogen
on left
2 hydrogen
on right
multiply H2O by 3
C2H6 + O2 2CO2 + 3H2O
27
4. Balance those elements that appear in two or
more reactants or products.
2 oxygen
on left
4 oxygen
(2x2)
C2H6 + O2 2CO2 + 3H2O
+ 3 oxygen
(3x1)
multiply O2 by
7
2
= 7 oxygen
on right
C2H6 + O2 2CO2 + 3H2O7
2
remove fraction
multiply both sides by 2
2C2H6 + 7O2 4CO2 + 6H2O
28
5. Check to make sure that you have the same
number of each type of atom on both sides of the
equation.
2C2H6 + 7O2 4CO2 + 6H2O
Reactants Products
4 C
12 H
14 O
4 C
12 H
14 O
4 C (2 x 2) 4 C
12 H (2 x 6) 12 H (6 x 2)
14 O (7 x 2) 14 O (4 x 2 + 6)
29
1. Write a balanced chemical equation
2. Convert quantities of known substances into moles
3. Use stoichiometric coefficients in a balanced equation as
the number of moles of each substance.
4. Convert moles of required quantity into desired units
Amounts of Reactants and Products
The procedure for
calculating the amounts
of reactants or products
in a reaction using the
mole method.
30
e.g.
Methanol (CH3OH) burns in air according
to the equation:
2CH3OH + 3O2 2CO2 + 4H2O
If 209 g of methanol are used up in the combustion, what
mass of water is produced?
grams CH3OH moles CH3OH moles H2O grams H2O
209 g CH3OH
1 mol CH3OH
32.0 g CH3OH
x
4 mol H2O
2 mol CH3OH
x
18.0 g H2O
1 mol H2O
x = 235 g H2O
Jelled Methanol
33
Limiting Reagents:
2NO + O2 2NO2
NO is the limiting reagent
O2 is the excess reagent, i.e. presents
in a quantity greater than necessary to
react with the quantity of the limiting reagent.
Limiting reagent is the reactant
used up first in a reaction.
e.g.
Limiting Reagentexcess
34
e.g.
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3 Al2O3 + 2Fe
Calculate the mass (in grams) of Al2O3
formed.
g Al mol Al mol Fe2O3 needed g Fe2O3 needed
OR
g Fe2O3 mol Fe2O3 mol Al needed g Al needed
124 g Al
1 mol Al
27.0 g Al
x
1 mol Fe2O3
2 mol Al
x
160. g Fe2O3
1 mol Fe2O3
x = 367 g Fe2O3
Start with 124 g Al need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
35
Cont.
Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al mol Al mol Al2O3 g Al2O3
124 g Al
1 mol Al
27.0 g Al
x
1 mol Al2O3
2 mol Al
x
102. g Al2O3
1 mol Al2O3
x = 234 g Al2O3
2Al + Fe2O3 Al2O3 + 2Fe
At this point, all the Al is consumed and Fe2O3 remains in excess.
H.W. How much of the excess reagent (Fe2O3) is left at the
end of the reaction.
36
Theoretical yield is the amount of product that would
result if all the limiting reagent reacted (maximum
obtainable yield)
Actual yield or the amount of product actually
obtained from a reaction is almost always less than
the theoretical yield.
% yield =
actual yield
theoretical yield
x 100%
Reaction Yield
Percent yield is the proportion of the actual yield to
the theoretical yield and determines how efficient a
given reaction is.
46
H.W.
When octane (C8H18) is burned in a particular internal
combustion engine, the yield of products (carbon
dioxide and water) is 93%. What mass of carbon
dioxide will be produced in this engine when 15.0 g of
octane is burned with 15.0 g of oxygen gas?
A) 43 g
B) 21 g
C) 13 g
D) 12 g
E) 54 g

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Stoichiometry Calculations for Combustion of Octane

  • 1. Stoichiometry (Mass Relationships in Chemical Reactions) Chapter 3 Stoichiometric Amounts: The exact molar amounts of reactants and products that appear in the balanced chemical equation. Dr. Sa’ib Khouri AUM- JORDAN Chemistry By Raymond Chang
  • 2. 2 Carbon-12 (6 protons and 6 neutrons): 1 atom 12C “weighs” 12 amu i.e. amu is 1/12 the mass of carbon atom By experiments: H = 1.008 amu O = 16.00 amu Fe = 55.85 amu Atomic mass is the mass of an atom in atomic mass units (amu) Micro World atoms & molecules Macro World grams
  • 3. 3 The average atomic mass is the weighted average of all of the naturally occurring isotopes of the element.
  • 4. 4 How? Carbon: 98.90% 12C (12.00000 amu) 1.10% 13C (13.00335 amu) 98.90 x 12.00000 + 1.10 x 13.00335 100 = 12.01 amu Average atomic mass of carbon: Look to example 3.1 on copper
  • 6. 6 In the SI system, the mole (mol) is the amount of a substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in exactly 12 g (or 0.012 kg) of 12C isotope 1 mol = NA = 6.0221415 x 1023 NA: Avogadro’s number (Italian scientist Amedeo Avogadro) Dozen = 12 items Pair = 2 items The mole: A unit to count numbers of particles Avogadro’s Number and Molar mass of an element
  • 7. 7 Molar mass (M) is the mass (in grams or kilograms) of one mole of units (such as atoms or molecules)of a substance. 1 mol 12C atoms = 6.022 x 1023 atoms = exactly 12 g 1 mol C-12 atoms = 12.00 g 12C 1 mol lithium atoms = 6.941 g of Li 1 mol sodium atoms = 22.99 g Na For any element atomic mass (amu) = molar mass (grams)
  • 8. 8 One Mole of: C S Cu Fe Hg 12.01 g 32.07 g 200.6 g 55.85 g63.55 g
  • 9. 9 therefore 1 amu = 1.661 x 10-24 g or 1 g = 6.022 x 1023 amu 1 12C atom 12 amu x 12 g 6.022 x 1023 12C atoms = 1.661 x 10-24 g 1 amu What is the mass per gram of amu?
  • 10. 10 x 6.022 x 1023 Na atoms 1 mol Na = e.g. How many atoms are in 0.551 g of sodium (Na) ? 1 mol Na = 22.99 g Na 1 mol Na = 6.022 x 1023 Na atoms 0.551 g Na 1 mol Na 22.99 g Na x 1.44 x 1022 Na atoms
  • 11. 11 Molecular mass is the sum of the atomic masses (in amu) in the molecule 1S 32.07 amu 2O + 2 x 16.00 amu SO2 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2 e.g. H.W. What is the molecular mass (amu) and molar mass (g) of caffeine (C8H10N4O2)
  • 12. 12 e.g. How many H atoms are in 72.5 g of C3H8O ? 1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O 1 mol H = 6.022 x 1023 H atoms 5.82 x 1024 atoms H 1 mol C3H8O molecules = 8 mol H atoms 72.5 g C3H8O 1 mol C3H8O 60 g C3H8O x 8 mol H atoms 1 mol C3H8O x 6.022 x 1023 H atoms 1 mol H atoms x = HW. How many H atoms are present in 43.8 g urea [(NH2)2CO] M of urea is 60.06 g
  • 13. 13 Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound (like NaCl and MgO) 1Na 22.99 amu 1Cl + 35.45 amu NaCl 58.44 amu For any ionic compound formula mass (amu) = molar mass (grams) 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaCl NaCl
  • 14. 14 e.g. What is the formula mass (amu) and molar mass (grams) of Ca3(PO4)2 ? 1 formula unit of Ca3(PO4)2 3 Ca 3 x 40.08 2 P 2 x 30.97 8 O + 8 x 16.00 310.18 amu The molar mass = 310.18 g
  • 15. 15 Percent composition (% by mass) of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of an element in 1 mole of a compound that contains of this element C2H6O %C = 2 x (12.01 g) 46.07 g x 100% = 52.14% %H = 6 x (1.008 g) 46.07 g x 100% = 13.13% %O = 1 x (16.00 g) 46.07 g x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0% e.g.
  • 16. 16 Percent Composition and Empirical Formulas e.g. Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent. nK = 24.75 g K x = 0.6330 mol K 1 mol K 39.10 g K nMn = 34.77 g Mn x = 0.6329 mol Mn 1 mol Mn 54.94 g Mn nO = 40.51 g O x = 2.532 mol O 1 mol O 16.00 g O Suppose total mass is 100 gram
  • 17. 17 Percent Composition and Empirical Formulas K : ~~ 1.0 0.6330 0.6329 Mn : 0.6329 0.6329 = 1.0 O : ~~ 4.0 2.532 0.6329 nK = 0.6330, nMn = 0.6329, nO = 2.532 KMnO4 Divide all by the smallest one
  • 18. 18 Soln. Moles of N= 87.5/14 = 6.25 Moles of H= 12.5/1 = 12.5 A compound with a percent composition by mass of 87.5% N and 12.5% H, What is the empirical formula of this compound? A) NH2 B) N2H3 C) NH D) N2H2 E) N2H e.g. N6.25 H12.5
  • 19. 19 g CO2 mol CO2 mol C g C g H2O mol H2O mol H g H g of O = g of sample – (g of C + g of H) e.g. Combustion of 11.5 g of ethanol Produced 22.0 g CO2 and 13.5 g H2O 6.0 g C = 0.5 mol C 1.5 g H = 1.5 mol H 4.0 g O = 0.25 mol O Empirical formula C0.5H1.5O0.25 Divide by smallest subscript (0.25) Empirical formula C2H6O Experimental Determination of Empirical Formulas
  • 20. Determination of Molecular Formulas To calculate the actual molecular formula we must know the approximate molar mass of the compound in addition to its empirical formula.
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  • 23. 23 3 ways of representing the reaction between H2 and O2 to form H2O Chemical reaction: a process in which a substance (or substances) is changed into one or more new substances A chemical equation uses chemical symbols to show what happens during a chemical reaction reactants products Chemical reactions and chemical equations
  • 24. 24 How to “Read” Chemical Equations 2 Mg + O2 2 MgO 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO NOT 2 grams Mg + 1 gram O2 makes 2 g MgO
  • 25. 25 Balancing Chemical Equations 1. 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. e.g. Ethane reacts with oxygen to form carbon dioxide and water C2H6 + O2 CO2 + H2O 2. 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. 2C2H6 NOT C4H12
  • 26. 26 3. Start by balancing those elements that appear in only one reactant and one product. C2H6 + O2 CO2 + H2O start with C or H but not O 2 carbon on left 1 carbon on right multiply CO2 by 2 C2H6 + O2 2CO2 + H2O 6 hydrogen on left 2 hydrogen on right multiply H2O by 3 C2H6 + O2 2CO2 + 3H2O
  • 27. 27 4. Balance those elements that appear in two or more reactants or products. 2 oxygen on left 4 oxygen (2x2) C2H6 + O2 2CO2 + 3H2O + 3 oxygen (3x1) multiply O2 by 7 2 = 7 oxygen on right C2H6 + O2 2CO2 + 3H2O7 2 remove fraction multiply both sides by 2 2C2H6 + 7O2 4CO2 + 6H2O
  • 28. 28 5. Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C2H6 + 7O2 4CO2 + 6H2O Reactants Products 4 C 12 H 14 O 4 C 12 H 14 O 4 C (2 x 2) 4 C 12 H (2 x 6) 12 H (6 x 2) 14 O (7 x 2) 14 O (4 x 2 + 6)
  • 29. 29 1. Write a balanced chemical equation 2. Convert quantities of known substances into moles 3. Use stoichiometric coefficients in a balanced equation as the number of moles of each substance. 4. Convert moles of required quantity into desired units Amounts of Reactants and Products The procedure for calculating the amounts of reactants or products in a reaction using the mole method.
  • 30. 30 e.g. Methanol (CH3OH) burns in air according to the equation: 2CH3OH + 3O2 2CO2 + 4H2O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH3OH moles CH3OH moles H2O grams H2O 209 g CH3OH 1 mol CH3OH 32.0 g CH3OH x 4 mol H2O 2 mol CH3OH x 18.0 g H2O 1 mol H2O x = 235 g H2O Jelled Methanol
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  • 33. 33 Limiting Reagents: 2NO + O2 2NO2 NO is the limiting reagent O2 is the excess reagent, i.e. presents in a quantity greater than necessary to react with the quantity of the limiting reagent. Limiting reagent is the reactant used up first in a reaction. e.g. Limiting Reagentexcess
  • 34. 34 e.g. In one process, 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe2O3 Al2O3 + 2Fe Calculate the mass (in grams) of Al2O3 formed. g Al mol Al mol Fe2O3 needed g Fe2O3 needed OR g Fe2O3 mol Fe2O3 mol Al needed g Al needed 124 g Al 1 mol Al 27.0 g Al x 1 mol Fe2O3 2 mol Al x 160. g Fe2O3 1 mol Fe2O3 x = 367 g Fe2O3 Start with 124 g Al need 367 g Fe2O3 Have more Fe2O3 (601 g) so Al is limiting reagent
  • 35. 35 Cont. Use limiting reagent (Al) to calculate amount of product that can be formed. g Al mol Al mol Al2O3 g Al2O3 124 g Al 1 mol Al 27.0 g Al x 1 mol Al2O3 2 mol Al x 102. g Al2O3 1 mol Al2O3 x = 234 g Al2O3 2Al + Fe2O3 Al2O3 + 2Fe At this point, all the Al is consumed and Fe2O3 remains in excess. H.W. How much of the excess reagent (Fe2O3) is left at the end of the reaction.
  • 36. 36 Theoretical yield is the amount of product that would result if all the limiting reagent reacted (maximum obtainable yield) Actual yield or the amount of product actually obtained from a reaction is almost always less than the theoretical yield. % yield = actual yield theoretical yield x 100% Reaction Yield Percent yield is the proportion of the actual yield to the theoretical yield and determines how efficient a given reaction is.
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  • 46. 46 H.W. When octane (C8H18) is burned in a particular internal combustion engine, the yield of products (carbon dioxide and water) is 93%. What mass of carbon dioxide will be produced in this engine when 15.0 g of octane is burned with 15.0 g of oxygen gas? A) 43 g B) 21 g C) 13 g D) 12 g E) 54 g