### ppt unit 4 stoichiometry

```Stoichiometry and Quantitative Analysis
A. Using Mole Ratios
 stoichiometry is the study of the relative
quantities of reactants and products in a
chemical reaction
you can use the number of moles for a given reactant
or product to find the moles for any other reactant or
product
Example
Consider the following chemical reaction:
2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g)
a) Write the ratio for all components of the reaction.
2:7:4:6
b) What amount, in moles, of CO2(g) is formed if 2.50
mol of C2H6(g) reacts?
2 C2H6(g)
n = 2.50 mol
+ 7 O2(g)
 4 CO2(g) + 6 H2O(g)
n = 2.50 mol  4
2
= 5.00 mol
c) What amount, in moles, of O2(g) is required to react
with 10.2 mol of C2H6(g)?
2 C2H6(g)
n = 10.2 mol
+
7 O2(g)
 4 CO2(g) + 6 H2O(g)
n = 10.2 mol  7
2
= 35.7 mol
d) What amount, in moles, of H2O(g) is formed when
100 mmol of CO2(g) is formed?
2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g)
n = 100 mmol
n = 100 mmol  6
4
= 150 mmol
= 0.150 mol
B. Gravimetric Stoichiometry
gravimetric = mass measurements
Steps
1. Write a balanced equation including the states.
Write the information given.
2. Find the moles of the given species using
n=m
M
3. Find the moles of the wanted
mole ratio
wanted
given
species using
4. Calculate mass of the wanted species using
m=nM
Example 1
Iron is produced by the reaction of iron(III) oxide with
carbon monoxide to produce iron and carbon dioxide.
What mass of iron(III) oxide is required to produce
1000 g of iron?
g
w
 2 Fe(s)
1 Fe2O3(s) + 3 CO(g)
+ 3 CO2(g)
m = 1000 g
m=?
M = 55.85 g/mol
M = 159.70 g/mol
n = 17.90… x 1
n=m
2
M
= 8.95… mol
= 1000 g
55.85g/mol
m = nM
= 17.90… mol
= (8.95… mol)(159.70 g/mol)
= 1429.7225 g
= 1430 g
Example 2
The decomposition of the mineral malachite,
Cu2(CO3)(OH)2(s), yields copper(II) oxide, carbon
dioxide and water vapour. What mass of copper(II)
oxide is produced from 1.00 g of malachite?
g
w
 2 CuO(s) + 1 CO2(g) + 1 H2O(g)
1 Cu2(CO3)(OH)2(s)
m=?
m = 1.00 g
M = 79.55 g/mol
M = 221.13 g/mol
n = 0.00452… x 2
n= m
1
M
= 0.00904… mol
=
1.00 g
221.13 g/mol
m = nM
= 0.00452… mol
= (0.00904… mol)(79.55 g/mol)
= 0.7194862 g
= 0.719 g
C. Solution Stoichiometry
use concentration to perform calculations
c=n
v
Example 1
What volume of 14.8 mol/L KOH is needed to react
completely with 1.50 L of 12.9 mol/L sulphuric acid?
w
g
2 KOH(aq) +
1 H2SO4(aq)  1 K2SO4(aq) + 2 H2O(l)
v=?
v = 1.50 L
c= 14.8 mol/L
c = 12.9 mol/L
n = 19.35 x 2
1
= 38.70 mol
v=n
c
= 38.70 mol
14.8 mol/L
= 2.6148649 L
= 2.61 L
n = cv
= 12.9 mol/L x 1.50 L
= 19.35 mol
D. Law of Combining Volumes
you can use the Law of Combining Volumes when the
pressure and temperature conditions are constant for
both the reactants and the products
Example
Consider the following reaction:
N2(g) + 2 O2(g)  N2O4(g)
a) What is the mole ratio for O2(g) and N2O4(g)?
2:1
b) What is the volume ratio for O2(g) and N2O4(g)?
2:1
c) If 1 mol of N2O4(g) is produced, how many moles
of O2(g) must be consumed?
1 mol of N2O4(g)  2
1
= 2 mol of O2(g)
d) If 1 L of N2O4(g) is produced, what volume of
O2(g) must be consumed?
1 L of N2O4(g)  2
1
= 2 L of O2(g)
e) If 2.5 L of N2(g) is consumed, what volume of
O2(g) must be consumed?
2.5 L of N2(g)  2
1
= 5.0 L of O2(g)
E. Gas Stoichiometry
if the other information is given for the chemicals
in the reaction (eg. mass), use ideal gas law to
perform calculations
PV = nRT
Example 1
If 300 g of propane burns in a gas barbeque, what
volume of oxygen at SATP is required for the
reaction?
w
g
 3 CO2(g) + 4 H2O(g)
1 C3H8(g)
+
5 O2(g)
m = 300 g
M = 44.11 g/mol
n=m
M
= 300 g
44.11 g/mol
= 6.80… mol
v=?
P = 100.000 kPa
T = 298.15 K
R = 8.314 kPaL/molK
n = 6.80… mol x 5
1
= 34.0… mol
PV = nRT
(100.000kPa)V = (34.0… mol)(8.314)(298.15K)
V = 842.94…L
= 843 L
Chapter 8: Applications of Stoichiometry
8.1 Limiting and Excess Reagents
let’s make double burgers…
1 bun + 2 meat patties  1 double burger
2 buns + 4 meat patties  2 double burgers
2 buns + 2 meat patties  1 double burger
excess
limiting
2 buns + one million meat patties  2 double burgers
limiting
excess
limiting reagent = the reactant that determines
how much product can be formed
in a
reaction
excess reagent = the reactant that is present in
larger quantities than necessary
Steps
1. Write the balanced chemical equation,
including states.
2. Calculate the number of moles of each reactant
using
n=m
M
3. “Even out ” the number of moles for each
reactant by dividing by the coefficient in the
balanced equation and compare…the one with the
lowest number of moles is the limiting reagent
4. Use limiting reagent moles to calculate the
Example 1
When 80.0 g copper and 25.0 g of sulphur react, which
reactant is limiting and what is the maximum amount of
copper(I) sulphide that can be produced?
g
16 Cu(s)
+
m = 80.0 g
M = 63.55 g/mol
n = 80.0 g
63.55g/mol
= 1.25… mol
n/16 = 0.0786…mol
\ limiting

1 S8(s)
m = 25.0 g
M = 256.56 g/mol
n = 25.0 g
256.56g/mol
= 0.0974… mol
w
8Cu2S(s)
m=?
M = 159.17 g/mol
n = 1.25…mol 8/16
= 0.629… mol
n/1 = 0.0974… mol
\ excess
m = (0.629…mol )  (159.17 g/mol)
= 100.17… g
= 100 g
Example 2
You are supplied with 9.00 g of KCl and 6.50 g of
AgNO3. What is the mass of the precipitate formed
when these two chemicals react?
1 KCl(aq) +
m = 9.00 g
M = 74.55 g/mol
n = 9.00 g
74.55g/mol
=0.120… mol
n/1 = 0.120…mol
\ excess
1 AgNO3(aq) 1 KNO3(aq) + 1 AgCl(s)
m = 6.50 g
M = 169.88 g/mol
n = 6.50 g
169.88g/mol
= 0.0382… mol
m=?
M = 143.32 g/mol
n =0.0382… mol 1/1
n/1 = 0.0382… mol
\ limiting
m = (0.0382…mol )  (143.32 g/mol)
= 5.48 g
Example 3
A 200 mL sample of a 0.221 mol/L mercury (II)
chloride solution reacts with 100.0 mL of a
0.500 mol/L solution of sodium sulphide. What is
the mass of the precipitate formed?
1 HgCl2(aq) + 1 Na2S(aq) 2 NaCl(aq) + 1 HgS(s)
C = 0.221 mol/L
V = 0.200 L
n = CV
=(0.221 mol/L)
 (0.200 L)
=0.0422… mol
n/1 = 0.0422…mol
\ limiting
C = 0.500 mol/L
V = 0.1000 L
n = CV
=(0.500 mol/L)
 (0.1000 L)
=0.0500… mol
m=?
M = 232.66 g/mol
n =0.0422…mol 1/1
n/1 = 0.0500… mol
m = (0.0422…mol )  (232.66 g/mol)
=10.3 g
8.2 Predicted and Experimental Yield
the expected amount of product is called the
predicted or theoretical yield
the quantity of the product actually obtained is
called the experimental or actual yield
it is extremely rare for the predicted and
experimental yield to be the same
factors affecting the experimental yield include:
1. two chemicals can react to give different
products …called competing reactions
eg) C(s) and O2(g) can react to form CO2(g) or CO(g)
2. reaction is very slow
3. collection and transfer methods
4. reactant or product purity
5. reaction doesn’t go to completion
ideally, percent yield should be as close to 100%
as possible
percent yield is calculated as follows:
% yield = actual yield
 100
predicted yield
we can also calculate our % error for the
experiment, which tells us how far we are from the
theoretical yield
the closer to zero the percent error is, the better
the experiment
% error = actual yield – predicted yield
predicted yield
 100
Example 1
Calculate the % error and % yield for the
following:
predicted mass of ppt = 6.20 g
actual mass of ppt = 7.12 g
% error = 7.12 g - 6.20 g x 100
6.20 g
= 14.8 %
% yield = 7.12 g x 100
6.20 g
= 115 %
Example 2
Calculate the % error and % yield for the
following:
predicted mass of ppt = 100 g
actual mass of ppt = 93.5 g
% error = 93.5 g - 100 g x 100
100 g
= -6.50 %
% yield = 93.5 g x 100
100 g
= 93.5 %
8.3 Acid-Base Titration
A. Titrations
 in solution stoichiometry, sometimes you don’t have
enough information to solve the problem on paper
eg) 10 mL of acetic acid reacts with a 0.202 mol/L
NaOH solution. What is the concentration of the
acetic acid?
CH3COOH(aq) + NaOH(aq)  H2O(l) + NaCH3COO(aq)
c = 0.202 mol/L
x mol/L
v = 0.0100 L v = ??? *****
**** you need this volume in order to solve the
problem
 a titration is a procedure used to find the
volume of substances so you can calculate
concentration
 standardization is when a solution of known
concentration, a standard solution , is reacted
with a solution of unknown concentration
 both strong acids and strong bases
need to be
standardized since their concentrations will
change over time
 a solution called the titrant is
transferred from a precisely marked
tube called a burette to a flask
containing the sample and an
indicator
 an indicator (eg. methyl orange,
bromothymol blue) is used
because a sudden change in
colour indicates the
completion of the reaction
 the endpoint is the point where the
titrant reacts
completely with the sample
 equivalence point is the volume needed to reach
the endpoint
 you need a minimum of 3 trials within 0.20 mL of
each other to ensure results are accurate
Here is a website to view a virtual
titration:
Virtual Titration
(Click on the “View Tutorial” beside Section 16.716.8 Strong Acid and Strong Base Titration)
Example
A 10.00 ml sample of HCl(aq) was titrated with a
standardized solution of 0.685 mol/L NaOH(aq).
Bromothymol blue indicator was used and it changes
from yellow to blue at the endpoint. What is the
concentration of the HCl(aq)?
Note: HCl(aq) “is titrated with” NaOH(aq)
titrant in burette
Data Table
Trial
1
overshoot
Final Volume
(mL)
Initial
Volume (mL)
Volume
NaOH (mL)
Endpoint
Colour
2
3
4
average volume = (
+
+
)
3
=
mL
1 HCl(aq)
+ 1 NaOH(aq)→ 1 H2O(l) + 1 NaCl(aq)
x mol/L
C = 0.685 mol/L
V = 10.00 mL
V=
mL
= 0.01000 L
=
L
n=
n=
=
mol
C=n
V
=
=
B. Titration Curves
 a plot of the pH vs. the volume of titrant added
is called a pH titration curve
 titration curves are S-shaped
 when a strong monoprotic acid is titrated with a
strong monoprotic base, the equivalence point
will always have a pH of 7 (at 25C)
 the first point on the curve is always the pH of the
sample
 the pH changes very gradually at first…this is
called the buffer region
 as the endpoint is approached, the pH changes
very rapidly
 the overtitration is asymptotic to the pH of the
titrant
Strong Acid Titrated with Strong Base
14
pH
7

0
Strong Base Titrated with Strong Acid
14
pH

7
0