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CHAPTER 3. STOICHIOMETRY
Chemical equations describe chemical reactions.
Common symbols:
(g) or ↑ gas;
(l) liquid; (s) or ↓ solid;
(aq) aqueous soln;
∆ heat
BALANCING EQUATIONS
LAW OF CONSERVATION OF MASS tells us that the number of atoms of each
element must be equal on both sides of equation.
Reactants
2H2 + O2
Products
2H2O
→
This is a balanced equation - same number of H and O atoms on both sides.
To balance an equation, we adjust the coefficients - these are numbers in front of
reactant and product molecules.
BALANCING SUGGESTIONS:
1) Change coefficients NOT subscripts.
2) Balance elements in the most complex formula first.
3) Save for last elements present in more than 2 formulas.
4) Balance polyatomic ions as a unit if present on both sides of equation.
5) Make sure you have smallest set of whole number coefficients possible.
A. Mg3N2
+ H2O → Mg(OH)2
Mg3N2
B. Fe(OH)3
+
+ 6H2O → 3Mg(OH)2
NH3
+
2NH3
+ H2SO4 → Fe2(SO4)3 + H2O
2Fe(OH)3
+ 3H2SO4 → Fe2(SO4)3 + 6H2O
TYPES OF REACTIONS
COMBINATION: A + B → C
Two or more reactants form 1 product.
A. N2 + H2 →
NH3
N2 + 3H2 →
B. Mg
2NH3
+ O2 → MgO
Mg + O2 → 2MgO
DECOMPOSITION: C → A + B
1 Reactant breaks down to form 2 or more substances.
A. Air bags - impact cause sodium nitride to ignite, N2 inflates bag:
NaN3(s) →
2NaN3(s) →
Na(s) + N2(g)
2Na(s) + 3N2(g)
B. Metal carbonate decomposes to metal oxide and carbon dioxide:
CaCO3(s) → CaO(s) + CO2(g)
COMBUSTION
ORGANIC COMPOUND (C, H & other nonmetals)
+ O2
→
CO2 + H2O
Examples include burning of gas, candles and glucose, flicking a bic lighter.
A. C3H8 + O2 →
CO2 + H2O
C3H8 + 5O2 →
B. C8H18 + O2 →
3CO2 + 4H2O
CO2 + H2O
2C8H18 + 25O2 → 16CO2 + 18H2O
SINGLE REPLACEMENT- chapter 4
DOUBLE REPLACEMENT – chapter 4
Atomic and Molecular Weights
Masses of atoms are so small that we define the atomic mass unit (amu) to scale up
the numbers
•
The carbon-12 isotope was assigned a mass of exactly 12 amu, masses of other
elements are scaled relative to C-12.
•
1 amu = 1.66054 x 10-24 g; 1 g = 6.02214 x 1023 amu.
Atomic weight of an element is the average atomic mass of all isotopes of an
element. Atomic weight (mass) can be calculated if the % abundance & masses of
the contributing isotopes are known.
E.g. Calculate the atomic mass of Cl.
Isotope
% abundance
35
75.53
17 Cl
37
17
Cl
24.47
0.7553 x 34.96885 = 26.41
0.2447 x 36.96590 = 9.046
35.456
mass of istope (amu)
34.96885
36.96590
⇒
mass of Cl = 35.46 amu
Formula Weight - sum of the masses of all the atoms in a formula. (units = amu)
•
In problems involving masses, round atomic masses to 1st decimal place (tenths
place). E.g. mass of Cl = 35.5 amu
E.g. Calculate Formula Weight of ammonium sulfate: (NH4)2SO4:
2(14.0) + 8(1.0) + 32.1 + 4(16.0) = 132.1 amu
Molecular Weight: Formula weight of a molecular compound. Units – amu
E.g. Calculate the MW of C6H12O6: 6(12.0) + 12(1.0) + 6(16.0) = 180.0 amu
PERCENT COMPOSITION - percent by mass of each element present in a
compound
# atoms x Atomic Weight of element
% element =
x 100
Formula Weight of Compound
E.g. Calculate the % composition of Al2(SO4)3
MM Al2(SO4)3 = 2(27.0) + 3(32.1) + 12(16) = 342.3 a.m.u.
2(27.0)
x 100 =15.8 %;
342.3
12(16.0)
%0=
x 100 = 56.1 %
342.3
% Al =
%S=
3(32.1)
x 100 = 28.1 %
342.3
3.4 The Chemical Mole
The mole is defined as the # of C atoms in exactly 12 grams of Carbon-12:
1 mole C atoms = 6.02x1023 C atoms
•
The mole is a collective quantity. E.g. 1 dozen = 12, 1 pair = 2
1 mole of anything = 6.02 x 1023 units ⇐ Avogadro’s #
units means particles like atoms, ions, molecules, etc.
•
Avogadro's number is a useful conversion factor between moles & particles.
E.g. Calculate the number of moles in 7.67 x1022 atoms Pb.


1 mole Pb
 = 0.127 mole Pb
7.67 x1022 atoms Pb 
23
 6.02 x10 atoms Pb 
E.g. How many O atoms are in 0.25 moles C6H12O6
 6.02 x10 23 molecules C 6 H 12 O6 

6 O atoms

 = 9.0x1023 O atoms
0.25 moles C6H12O6 
1 mole C 6 H 12 O6

 1 molecules C 6 H 12 O6 
Molar Mass - mass in grams of 1 mole of substance. (units = g/mol)
• Numerically same as atomic or formula mass, but units are different
Comparison of masses
1 Carbon atom weighs 12 amu
1 H2O molecule weighs 18.0 amu
1 mole of Carbon atoms weighs 12.0 g 1 mole of H2O molecules weighs 18.0 g
MM of C = 12.0 g/mol
MM of H2O = 18.0 g/mol
•
Molar Mass is also a useful conversion factor between grams to moles
E.g. What is the mass of 0.120 mol Cu(NO3)2?
187.5 g Cu(NO 3 ) 2
0.120 mol Cu(NO3)2 x
= 22.5 g Cu(NO3)2
1 mole Cu(NO 3 ) 2
E.g. How many moles are in 12.6 grams of water?
1 mol H 2 O
12.7 g H2O x
= 0.700 mol H2O
18.0 g H 2 O
E.g. How many grams of potassium can be obtained from 57.4 g of potassium
sulfate? potassium sulfate: K+SO42- → K2SO4; MM K2SO4 = 174.3 g/mol
 78.2 g K
57.4 g K2SO4 
 174.3 g K 2 SO 4

 = 25.8 g K

•
•
Can also convert from grams to particles or particles to grams
Use MM & Avogadro’s #: g → moles → particles or particles → moles → grams
1. Calculate the # of molecules in a snowflake containing 4.0 x 10-6 g H2O.
 1 mole H 2 O  6.02 x10 23 molecules H 2 O 
-6
 = 1.3x1017 molecules H2O

4.0 x 10 g H2O 
1 moleH 2 O
 18.0 g H 2 O 

2. Glucose, C6H12O6 contains 4.0 x 1022 C atoms
a. How many H atoms does it contain?

 1 molecule C 6 H 12 O6 
12 H atoms
 = 8.0 x 1022 H atoms

4.0 x 1022 C atoms 
6 C atoms

 1 molecule C 6 H 12 O6 
b. How many grams of glucose does it contain?
 1 molecule C 6 H 12 O6 
1 moleC 6 H 12 O6

4.0 x 1022 C atoms 
23
6 C atoms

 6.02 x 10 molecule C 6 H 12 O6
 180.0 g C 6 H 12 O6 

 =
 1 mole C 6 H 12 O6 
1.99 g C6H12O6
EMPIRICAL & MOLECULAR FORMULAS
Empirical formula can be determined from % composition of elements in compound
Steps to find Empirical formula
1) Assume 100 g (change % signs to grams symbols)
2) Convert g to moles (divide by MM of element)
3) Divide by the smallest # of moles
4) Convert to whole #'s
• Round if close to whole #: e.g. 8.9 → 9
• Or if not close to whole #, round to closest fraction & multiply each mole ratio by
smallest integer that will yield a whole number. e.g. 2.5 = 5/2 x 2 = 5
E.g. Determine the empirical formula of a compound with 62.1% C, 5.21% H,
12.1% N and 20.7% O.
1 mol C
= 5.18 mol C/0.864 = 6 x 2 = 12
12.0 g C
1 mol H
5.21 g H
= 5.21 mol H/0.864 = 6.03 → 6 x 2 =12
1.0 g H
1 mol N
12.1 g N
= 0.864 mol N/0.864 = 1 x 2 = 2
14.0 g N
1 mol O
3
= 1.29 mol O/0.864 = 1.49 → 1.5 = x 2 = 3
20.7 g O
16.0 g O
2
Formula = C12H12N2O3
62.1 g C
Molecular formulas give the actual numbers of atoms in a molecule. The
molecular formula is a multiple of the empirical formula. E.g. CH4, H2O2, C2H4
Steps to find Molecular Formula:
1) find empirical formula
2) calculate MM of empirical formula
3) Determine n, where n = number of empirical formula units
MM (molecular formula )
=n
MM (empirical formula )
4) Multiply each subscript in formula by n
E.g. Determine the empirical formula and molecular formula for nicotine:
% composition: 74.1% C, 8.6% H, 17.3% N; molar mass = 160 g
1 mol C
= 6.18/1.24 = 4.98 → 5
12.0 g C
1 mol H
8.6 g H
= 8.6/1.24 = 6.94 → 7
1.0 g H
1 mol N
= 1.24/1.24 = 1
17.3 g N
14.0 g N
Empirical Formula = C5H7N
MM= 5(12.0) + 7(1.0) + 14.0 = 81.0 g
160 g
= 2 → multiply subscripts by 2
81 g
Molecular Formula: C10H14N2
74.1 g C
3.6 Quantitative Information
Stoichiometry problems are based on quantitative relationships between the
different substances involved in a chemical reaction.
•
A balanced chemical equation is a source of several conversion factors
e.g.
PCl5
+
4H2O
→
H3PO4
+
5 HCl
1 molecule PCl5 + 4 molecules H2O → 1 molecule H3PO4 + 5 molecules HCl
1 mole PCl5 + 4 mole H2O
→ 1 mole H3PO4 + 5 mole HCl
1 mole PCl 5
5 mole HCl
e.g. mole to mole conversion factors:
4 moles H 2 O
1 mole H 3 PO4
Problem. In the above reaction, how many moles of HCl are formed from 6.0 moles
of H2O?
5 mole HCl
6.0 moles H2O
= 7.5 moles HCl
4 moles H 2 O
•
In actual lab experiments we measure mass (in g) of reactants and products, so
its more useful to know how to convert from grams of 1 substance to grams of
another. To do this we combine g → mole and mole → mole conversion factors.
BASIC STOICHIOMETRY PATH:
gA↔
moles A ↔ moles B
⇑
Balanced Equation
MM - Periodic Table
↔ grams B
Steps:
1) g A ↔ mole A (divide by MM of A)
2) mole A ↔ mol B (Use mole ratios from equation)
3) mole B ↔ g B (multiply by MM of B)
¾Important to include units & formulas - units cancel except wanted units
1. Given the following reaction:
3H2 + N2 → 2NH3
A. How many moles of NH3 are formed from 56.0 g of N2?
g N2 → moles N2 → mole NH3
 1 mol N 2
56.0 g N2 
 28.0 g N 2
 2 mol NH 3 

 = 4.00 mole NH3
 1 mol N 2 
B. How many grams of N2 are required to react with 9.0 moles of H2?
moles H2 → moles N2 → g N2
 1 mole N 2
9.0 moles H2 
 3 mole H 2
 28.0 g N 2

 1 mole N 2

 = 84.0 g N2

2. For this unbalanced reaction: C2H6 +
O2 →
H2O +
CO2
How many grams of water will be produced during the combustion of 90.0 g C2H6?
Balance the reaction first!
2C2H6 +
7O2 →
g C2H6 →
moles C2H6 →
 1 mole C 2 H 6
90.0 g C2H6 
 30.0 g C 2 H 6
6H2O
+
4CO2
moles H2O
→
 6 mole H 2 O  18.0 g H 2 O 


 =
mole
C
H
mole
H
O
2
1
2
6 
2


g H2O
162 g H2O
3. How many pounds of CO2 are formed during the combustion of 1 gallon of
gasoline, C8H18. The density of gasoline is 0.70 g/ml. (1 gallon = 3.785 L, 1 lb
= 454 g)
2C8H18 + 25O2 → 16CO2 + 18H2O
 3.785 L  1000 mL  0.70g 
3


1 gallon 
 = 2.6 x10 g C8H18
 1 gallon  1 L  mL 
 1 mol C8 H 18  16 mol CO2  44.0 g CO2 


 = 8.0 x10 3 g CO2
2.6x103 g C8H18 
g
C
H
mol
C
H
mol
CO
114
2
8
18 
8
18 
2 

 1lb 
 = 18 lb CO2 ⇐ results in global warming
8.0 x 103 g CO2 
 454 g 
3.7 Limiting Reactants
In many chemical reactions, the reactants are not present in the precise ratios
specified by the balanced chemical reaction. One or more of the reactants is often
present in excess and remain after completion of the reaction.
The limiting reactant is the reactant that is completely used up.
Hot dog Analogy:
You have 2 packages of hot dogs (10 hot dogs/package)
& 2 packages of buns (8 buns/package)
How many hot dogs can you make? 16 hot dogs with buns
E.g. Four molecules of H2 react with 3 molecules of O2. How many molecules of
water can form? Which is the limiting reactant?
2H2O
2H2 + O2 →
Answer: 4 water molecules, H2 is LR – completely used up
Learning Check. How many moles of water can form if 11 moles of O2 react with
25 moles of H2?
Answer: 11 moles O2
Smallest Amount Method
1) convert grams of each reactant to grams of product
(g A → mol A → mol B → g B)
2) compare grams of product- smaller amount of product is the amount of product
that can form
3) limiting reactant is the reactant that yields smallest amount of product
E.g. Magnesium reacts with nitrogen to form magnesium nitride. How many grams
of Mg3N2 can be made through the reaction of 35.0 g of Mg with 15.0 g of N2?
3Mg + N2 →
Mg3N2
 1 mol Mg  1 mol Mg 3 N 2  100.9 g Mg 3 N 2


35.0 g Mg 
 24.3 g Mg  3 mol Mg  1 mol Mg 3 N 2
 1 mol N 2
15.0 g N2 
 28.0 g N

2
 1 mol Mg 3 N 2  100.9 g Mg 3 N 2


 1 mol N 2  1 mol Mg 3 N 2

 = 48.4 g Mg3N2


 = 54.1 g Mg3N2

Theoretical yield = 48.4 g Mg3N2; Limiting reactant is Mg
Theoretical yield – the amount of product calculated to form by the complete
conversion of the reactants.
Actual yield – the amount of product obtained in a real lab experiment. It is
always less than the calculated theoretical yield due to incomplete reactions,
competing side reactions, spillage, etc.
Actual Yield
x 100%
Theoretical Yield
% yield should be less than 100%
Percent Yield =
•