10/20/09
Chemical Equations
C6H12O6(aq)
yeast
2 C2H5OH(l) + 2 CO2(g)
glucose
ethanol
carbon dioxide
Conditions may be shown over the arrow. e.g.
Chapter 4: Quantities of Reactants and
Products
heat (Δ)
reflux
catalyst present (yeast)
Physical states are often listed:
(g)
(l)
© 2008 Brooks/Cole
1
Chemical Equations
gas
liquid
(s)
(aq)
solid
aqueous (dissolved in water)
© 2008 Brooks/Cole
2
Chemical Equations
“Mass is neither created nor destroyed in a chemical reaction.”
2 C2H6(g) + 7 O2(g)
4 CO2(g) + 6 H2O(l)
2(30.0)= 60.0 g
7(32.0)= 224.0 g
g
284.0 g
4(44.0)= 176.0 g 6(18.0)= 108.0
CaCO3(s) + 2 HNO3(aq)
Ca(NO3)2(aq) + CO2(g) + H2O(l)
A stoichiometric
coefficient
284.0 g
O2 molar mass
C2H6 molar mass
© 2008 Brooks/Cole
3
Combination Reactions
© 2008 Brooks/Cole
4
Decomposition Reactions
+
+
X
Z
XZ
XZ
X
Z
Often initiated by heat:
Element plus halogen or O2:
2 Mg(s) + O2(g) → 2 MgO(s)
I2(s) + Zn(s) → ZnI2(s)
CaCO3(s)
2 KNO3(s)
800 - 1000°C
heat
CaO(s) + CO2(g)
2 KNO2(s) + O2(g)
Occasionally by shock:
There are other types:
2 SO2(g) + O2(g) → 2 SO3(g)
4 C3H5(NO3)3(l)
12 CO2(g) + 10 H2O(l) + 6 N2(g) + O2(g)
nitroglycerin
© 2008 Brooks/Cole
5
© 2008 Brooks/Cole
6
1
10/20/09
Decomposition Reactions
Displacement Reactions
+
A
+
XZ
AZ
X
Some metals displace another metal from its salt
Fe(s) + CuSO4(aq)
Zn(s) + 2 AgNO3(aq)
FeSO4(aq) + Cu(s)
Zn(NO3)2(aq) + 2 Ag(s)
Some other examples:
F2(g) + 2 LiCl(s)
2 Na(s) + 2 H2O(l)
© 2008 Brooks/Cole
7
Displacement Reactions
2 LiF(s) + Cl2(g)
2 NaOH(aq) + H2(g)
© 2008 Brooks/Cole
8
Exchange Reactions
+
AD
+
XZ
AZ
XD
AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)
Pb(NO3)2(aq) + K2CrO4(aq)
→ PbCrO4(s) + 2 KNO3(aq)
© 2008 Brooks/Cole
9
Balancing Chemical Equations
© 2008 Brooks/Cole
10
Balancing Chemical Equations
Balance : Al + Fe2O3
1.  Write an unbalanced equation with correct
formulas for all substances.
Al2O3 + Fe
2.  Balance the atoms of one element.
a.  Start with the most complex molecule
b.  Change the coefficients in front of the molecules
c.  Do NOT alter the chemical formulas
3.  Balance the remaining elements.
4.  Check the atoms are all balanced.
step 4 – balanced
© 2008 Brooks/Cole
11
© 2008 Brooks/Cole
12
2
10/20/09
Balancing Chemical Equations
Combustion of rocket fuel:
C2H8N2 + N2O4
N2
Balancing Chemical Equations
+ H2O + CO2
NaNO3(s) + H2SO4(aq)
Na2SO4(aq) + HNO3(aq)
Balance Na in Na2SO4
NaNO3(s) + H2SO4(aq)
© 2008 Brooks/Cole
13
Na2SO4(aq) + HNO3(aq)
© 2008 Brooks/Cole
14
The Mole and Chemical Reactions
The Mole and Chemical Reactions
2 C2H6 (g) + 7 O2 (g)
What mass of O2 and Br2 is produced by the reaction
of 25.0 g of TiO2 with excess BrF3?
4 CO2(g) + 6 H2O (l)
2 moles of C2H6 react with 7 moles of O2
2 moles of C2H6 produce 4 moles of CO2
2 mol C2H6 ≡ 7 mol O2
2 mol C2H6 ≡ 4 mol CO2
etc.
3 TiO2(s) + 4 BrF3(l)
3 TiF4(s) + 2 Br2(l) + 3 O2(g)
Notes:
•  Check the equation is balanced!
•  Stoichiometric ratios:
2 mol C2H6
=1
7 mol O2
3TiO2 ≡ 3O2 ; 3TiO2 ≡ 2Br2 ; and many others
7 mol O2
2 mol C2H6 =1
© 2008 Brooks/Cole
15
© 2008 Brooks/Cole
16
The Mole and Chemical Reactions
The Mole and Chemical Reactions
What mass of O2 and Br2 is produced by the reaction of 25.0g of TiO2 with
excess BrF3? 3 TiO2(s) + 4 BrF3(l)
3 TiF4(s) + 2 Br2(l) + 3 O2 (g)
What mass of O2 and Br2 is produced by the reaction of 25.0g of TiO2 with
excess BrF3? 3 TiO2(s) + 4 BrF3(l)
3 TiF4(s) + 2 Br2(l) + 3 O2(g)
Mass of O2 produced = nO2 (mol. wt. O2)
nTiO2 = mass TiO2 / FM TiO2
= 0.3130 mol x 32.00 g/mol
= 10.0 g
1 mol
= 25.0 g x 79.88 g = 0.3130 mol TiO2
3 mol TiO2 ≡ 3 mol O2
0.3130 mol TiO2
© 2008 Brooks/Cole
nBr2 = 0.3130 mol TiO2
3 mol O2
= 0.3130 mol O2
3 mol TiO2
2Br2
= 0.2087 mol Br2
3 TiO2
Mass of Br2 = 0.2087 mol 159.81 g = 33.4 g Br2
mol Br2
17
© 2008 Brooks/Cole
18
3
10/20/09
Practice Problem 4.8
Practice Problem 4.8
The purity of Mg can be found by reaction with excess HCl
(aq), evaporating the water from the resulting solution and
weighing the solid MgCl2 formed.
Mg(s) + 2 HCl(aq)
MgCl2(aq) + H2(g)
Calculate the % Mg in a 1.72-g sample that produced 6.46 g of
MgCl2 when reacted with excess HCl.
FW of MgCl2
= 24.31 + 2(35.45) = 95.21 g/mol
nMgCl2 = 6.46 g MgCl2 1 mol = 0.06785 mol MgCl2
95.21 g
More difficult – What should you calculate?
Mg required:
– Express as a % of the original mass.
© 2008 Brooks/Cole
19
Practice Problem 4.8
0.06785 mol MgCl2
1 Mg
1 MgCl2
© 2008 Brooks/Cole
20
Reactions with Reactant in Limited Supply
Given 10 slices of cheese and 14 slices
of bread. How many sandwiches can
you make?
0.06785 mol Mg x 24.31 g = 1.649 g Mg
1 mol
Given 1.72 g of impure Mg.
Purity (as mass %) =
Balanced equation
1 cheese + 2 bread
1.649 g x 100% = 95.9 %
1.72 g
© 2008 Brooks/Cole
1 sandwich
1 cheese ≡ 2 bread
1 cheese ≡ 1 sandwich
2 bread ≡ 1 sandwich
21
Reactions with Limited Reactants
© 2008 Brooks/Cole
22
Reactions with Limited Reactants
Two methods can be used:
•  Yes = Your choice is the limiting reactant.
•  No = Another reactant is limiting.
10 cheese x
14 bread x
1 sandwich
1 cheese
1 sandwich
2 bread
Bread is limiting. It will
be used up first
© 2008 Brooks/Cole
= 10 sandwiches
= 7 sandwiches
Correct answer
23
© 2008 Brooks/Cole
24
4
10/20/09
Reactions with Limited Reactants
Reactions with Limited Reactants
How much water will be produced by the combustion
of 25.0 g of H2 in the presence of 100. g of O2?
…base all other calculations on the limiting reactant.
Write a balanced equation:
2 H2(g) + O2(g)
Started with 10 cheese. Cheese remaining
10 – 7 = 3 slices
© 2008 Brooks/Cole
25
Reactions with Limited Reactants
2 H2O(l)
nH2 = 25.0 g
1 mol H2
= 12.40 mol H2
2.016 g
nO2 = 100. g
1 mol O2
= 3.125 mol O2
32.00 g
© 2008 Brooks/Cole
26
Reactions with Limited Reactants
How much water will be produced?
Using H2
12.40 mol H2 (2H2O /2H2 ) = 12.40 mol H2O
Using O2
3.125 mol O2 (2 H2O /1 O2 ) = 6.250 mol H2O
H2O formed: 3.125 mol O2 (2H2O/1O2) = 6.250 mol.
= 113. g
© 2008 Brooks/Cole
27
© 2008 Brooks/Cole
28
Reactions with Limited Reactants
Reactions with Limited Reactants
Consider :
4 NH3(g) + 5 O2(g)
4 NO(g) + 6 H2O(g)
If 374 g of NH3 and 768 g of O2 are mixed, what mass
of NO will form?
4 NH3(g) + 5 O2(g)
Mol available: 21.96
24.00
From NH3
NO formed: 21.96 mol NH3 4 NO
4 NH3
nNH3 = 374 g 1 mol = 21.96 mol
17.03 g
nO2 = 768g
© 2008 Brooks/Cole
1 mol
32.00 g
4 NO(g) + 6 H2O(g)
= 21.96 mol NO
4 NO
5 O2
= 24.00 mol
29
© 2008 Brooks/Cole
30
5
10/20/09
Reactions with Limited Reactants
Reactions with Limited Reactants
Mass of NO formed?
What mass of MgI2 is made by the reaction of 75.0 g
of Mg with 75.0 g of I2?
Mg + I2 → MgI2
4 NH3(g) + 5 O2(g)
21.96 mol
24.00 mol
4 NO(g) + 6 H2O(g)
19.20 mol
•  Calculate moles
  75.0 g of Mg = 75.0g/(24.31 g mol-1) = 3.085 mol Mg
  75.0 g of I2 = 75.0g/(253.9 g mol-1) = 0.2955 mol I2
NO formed = 19.20 mol NO
Mass of NO = 19.20 mol NO x 30.01g = 576 g NO
1 mol
© 2008 Brooks/Cole
31
Percent Yield
© 2008 Brooks/Cole
32
Percent Yield
Few reactions have 100% yield.
Actual yield
x 100%
Theoretical yield
% yield =
© 2008 Brooks/Cole
33
© 2008 Brooks/Cole
34
Percent Yield
Percent Yield
You heat 2.50 g of copper with an excess of sulfur
and synthesize 2.53 g of copper(I) sulfide
16 Cu(s) + S8(s)
8 Cu2S(s)
What was the percent yield for your reaction?
Heat 2.50 g of Cu with excess S8 and make 2.53 g of copper(I) sulfide:
16 Cu(s) + S8(s) → 8 Cu2S(s). What was the %-yield for your reaction?
nCu used:
2.50 g
= 0.01967 mol Cu2S
159.2 g
= 3.131 g Cu2S
1 mol
1 mol
= 0.03934 mol Cu
63.55g
16 mol Cu used ≡ 8 mol Cu2S made
2.53 g
3.131 g
8 Cu2S
16 Cu
© 2008 Brooks/Cole
35
© 2008 Brooks/Cole
36
6
10/20/09
Atom Economy
Percent Composition & Empirical Formulas
Examines the fate of all starting-material atoms.
% atom economy =
atomic mass of atoms in useful product(s)
x 100%
atomic mass of all reactants used
Sample in
a furnace
O2
H 2O
absorber
CO2
absorber
Mg(ClO4)2
NaOH
C and H are converted to CO2 & H2O.
•  Both are trapped and the weight gain measured.
•  Other elements (N, O …) with other traps or by mass
difference.
© 2008 Brooks/Cole
37
Percent Composition & Empirical Formulas
38
Percent Composition & Empirical Formulas
Vitamin C contains C, H & O only. Combustion of 1.000 g of vitamin C
produced 1.502 g of CO2 and 0.409 g of H2O.
Vitamin C contains C, H & O only. 1.502 g of CO2
and 0.409 g of H2O are produced when 1.000 g is
burned in O2. Given the molar mass of vitamin C is
176.12 g/mol, find its empirical & molecular formula.
0.409 g H2O
2.0158 g H
18.015 g H2O
= 0.04577 g H
mass of O = sample mass – (mass of C + mass of H)
1.502 g CO2 1 mol CO2 1 mol C 12.011 g C = 0.4099 g C
44.009 g CO21 mol CO2 1 mol C
1.502 g CO2
© 2008 Brooks/Cole
12.011 g C
= 0.4099 g C
44.009 g CO2
© 2008 Brooks/Cole
39
© 2008 Brooks/Cole
40
Percent Composition & Empirical Formulas
Percent Composition & Empirical Formulas
Convert to moles:
Find the mole ratio (divide by smallest…):
0.4099 g C = 0.03413 mol C
12.011 g/mol
0.04577 g H = 0.04541 mol H
1.0079 g/mol
0. 544 g O
= 0.0340 mol O
15.999 g/mol
C
0.03413 / 0.0340 = 1.00
H
0.04541 / 0.0340 = 1.34
O
0.0340 / 0.0340 = 1.00
Close to 1 : 1⅓ : 1 (C : H : O)
Multiply by 3 to get an integer ratio (3 : 4 : 3)
© 2008 Brooks/Cole
41
© 2008 Brooks/Cole
42
7
10/20/09
Percent Composition & Empirical Formulas
…Vitamin C contains C, H & O only… … molar mass of vitamin C is
176.12 g/mol, find its empirical and molecular formula.
Empirical formula = C3H4O3
Empirical mass = 3(12) + 4(1) + 3(16) g = 88 g
Molar mass = 176.12 g
Molar mass ≈ 2 x (empirical mass)
© 2008 Brooks/Cole
43
8