Q3U3 Molarity, Solubility, Equilibrium, and Determining
Experimental Yield
 Solutions consist of 2 components
1.
2.


solute: a substance that is dissolved
Solvent: the medium it is dissolved in
Solutions are homogeneous mixtures: constant
composition throughout
Solutes do not have to be solids,

They can be gas


They can be liquid



CO2 dissolves in water, as does O2 (this is what fish breathe)
Ethylene glycol (antifreeze)
The most common solvent is water!
The strength of a solution is important and can be described
in a quantitative way
 The Mole is used to describe the concentration of a
solution
 Or the “strength” of a solution
 A 1 Molar solution contains 1 mole of a solute dissolved in
enough solvent (usually water) to make 1 liter of solution
 Molarity (M) = moles of solute
liter of solution
One mole of NaCl is 58.5g, if 2 moles of NaCl are dissolved in
water to make 1 liter of solution, the solution is a 2M
solution
This number is very consistent and uniform in all 2 molar
solutions of NaCl
 Calculate the Molarity of 1.50 L of solution that
contains 200.0g of MgCl2
Molarity= (200.0g MgCl2) X (1 molMgCl2) = 1.40 mol/L
( 1.50 L )
(95.3 g MgCl)
=1.40 M
 Calculate the molarity of a solution that contains 10.0g
of sodium hydroxide in 5.00 X 102 mL of solution
 Molarity = 10.0 g NaOH X 1 mol NaOH X 1000mL
5.0 X 102 mL 40.0 g NaOH
= 0.500 mol/L = .500 M
1L
 Pg 109 #’s 1-3
1.
2
3
a. 1.00 M
b. 0.25 M
c. 1.21 M
a. 41.6 g
b. 421 g
c. 71.0 g
a. 1.00 L
b. 4.00 L
c. 0.737 L
 Another way to express concentration of solutions
 Defined as the moles of a substance divided by the
moles of solution (sum of moles of solute and moles of
solvent)
 X=
moles solute
(moles solute + moles solvent)
The sum of all of the mole fractions in a solution equals 1
 What are the mole fractions of glucose and water in a
solution made of 7.59g of glucose, dissolved in water?
Molecular mass of glucose is 180.0g, find the moles of
glucose.
mol glucose= (7.59g glucose) (1 mol glucose)= 0.0422 mol
(1)
(180.0 g glucose)
The molecular mass of water is 18.0g, find the moles of water.
mol water = (125 g water) (1 mol water) = 6.94 mol
(18.0 g water)
 Now find the mole fraction of glucose
Mole fraction = mole solute/(mole solute + mol solvent)
=0.0422mol/ (0.0422+6.94)= 0.00605
Now find the mole fraction of water
Mol fraction water = 6.94/6.98 =0.994
If you multiply the mole fraction by 100, you get the mole
percent!
 Calculate the mole fraction of each component in the
following
1. 22.5g CH3CH2OH in 1.0 x 102 g H2O
H2O = 0.919
CH3CH2OH = 0.0807
2. 39.5 g C6H5CH3 toluene, in 1.5 X 102 g C6H6, benzene
C6H5CH3 = 0.183
C6H6 = 0.817
Calculate the molarity of the following solution
1. 500.0mL with 82.0g Ca(NO3)2
Calculate the mass of solute in the following solutions
2. 250.0mL Na2SO4. 7H2O solution that is 2.0 M
3. 1.50 L of KH2PO4 solution that is 0.24 M
How many liters of solution can be made from the following?
4. A 0.10 M solution using 117.0g NaCl
5. A 1.00 M solution using 50.0 g CuSO4 5H2O
6. 0.20 M solution using 200.0g Na2S
1. Molarity of ---500.0mL with 82.0g Ca(NO3)2
(82.0 g/500.mL) (1mol/164.0g) (1000mL/1L) =1.o M
2. Find mass --250.0mL Na2SO4. 7H2O solution that is 2.0 M
(250.0mL/1) (2.0 mol/1L) (268g/1mol) (1L/1000mL) =134g
3. Find mass of -- 1.50 L of KH2PO4 solution that is 0.24 M
(1.50 L/1) (0.24 moles/1L) ( 136g/1mol) = 48.96 g KH2PO4
4. liters solution ? -- a 0.10 M solution using 117.0g NaCl
(117.0g/1) (1 mol/58.5 g) (1L/0.1 mol)= 20.0L
5. liters?- 1.00 M solution using 50.0 g CuSO4 5H2O
(50.0g/1) (1 mol/249.5 g) (1L/1 mol) =0.20 L
6. liters?- 0.20 M solution using 200.0g Na2S
(200.g/1) (1 mol/78.0 g) (1 L/0.2 mol) = 12.8 L
 Previously we have assumed that all reactions have gone to
completion ( react until one of the reactants is used up,
then the reaction stops)
 Reactions go to completion because the elements from the
initial reaction are removed in one of the following 3 ways:
1.
2.
3.
Formation of a gas
Formation of a slightly ionized substance
Formation of a precipitate
Experimentation has proved that, in some cases, not all of
the reactants are converted to the product, no matter
how much time is given.
These reactions are reversible!
 Equilibrium reactions do not go to completion
 As the reaction proceeds, the reactant concentrations
decrease as the product is formed.
 However at a certain point, the concentrations level off.
 There is no additional product formed, or additional reactant
consumed
 The concentration of reactants and products no longer
change, A state of Chemical equilibrium has been
established.
 Ex. N2 (g) + 3H2 (g) <----> 2 NH3 (g) + energy
A reversible chemical reaction is in equilibrium when :
 the rates of the opposing reactions are equal and
 overall concentrations remain constant
 Le Chatelier’s Principle: If a system in equilibrium is
subjected to a disturbance, the equilibrium will shift in
an attempt to reduce the disturbance and regain
equilibrium
 Disturbances or outside influences can consist of the
following:
 Concentration change
 Pressure change
 Temperature change
 ALSO> If more reactant is added to the system in
equilibrium, the reaction shifts right and more product
is formed
 Concentration Change Example: Ammonia
equilibrium equation
N2 (g) + 3H2 (g) ----- 2NH3 (g) + energy
 Adding N2 (g) disturbs the system, the reaction shifts
right
 This can be corrected through the consumption of N2,
so the reaction shifts right and more NH3 is made.
 If H2 (reactant) is removed, the reaction shifts left
 The disturbance is relieved by producing more H2 and
N2, consuming NH3
 Pressure Change Example: Ammonia equilibrium equation
N2 (g) + 3H2 (g) ----- 2NH3 (g) + energy
NOTE! Pressure Change Only Affects Gaseous equilibrium
systems!
 If Pressure on reactant gas is increasing, this
causes the reaction to shift towards the side with
the least volume.
 In the above reaction, the shift would be to the right because
this would make 4 particles (N2(g) + 3H2) combine to make
only 2 particles (NH3)
 This would reduce particle collisions, this reduces the pressure,
thereby relieving the disturbance
 Temperature Change Example: Ammonia equilibrium
equation
N2 (g) + 3H2 (g) ----- 2NH3 (g) + energy
A temperature increase will cause the reaction to shift
in the direction of the endothermic reaction
 In the example, the left to right reaction is exothermic
produces heat
 A rise in temp will therefore shift the reaction toward the
left (endothermic reaction)!
 Using up the heat, corrects the disturbance
 The Following disturbances do not cause equilibrium
shifts
 Catalysts, which speed up the reaction by lowering the
activation energy, DO NOT CAUSE AN EQUILIBRIUM
SHIFT!
 They speed up the reaction but do not increase the amount of
product produced
 Inhibitors, which slow down reactions, also do not cause a
shift.
 They cause the reaction to take longer, but do not reduced the
amount of product produced.
 For the following gaseous equilibrium reactions, indicate
what happens to the equilibrium position when the given
disturbance or condition change occurs. (shift left, right or
no shift)
a. Remove NH3 gas
b. Decrease pressure
N2 + 3H2  2NH3 + energy
c. Decrease temperature
d. add a catalyst
CO2 + H2 + energy  CO + H2O
e. Increase SO2 concentration
f. increase temperature
2SO2 + O2  2 SO3 + energy
g. Increase temperature
h. increase CO concentration
CO2 + C + energy  2CO
 For the following gaseous equilibrium reactions,
indicate what happens to the equilibrium position
when the given disturbance or condition change
occurs. (shift left, right or no shift)
i.
Decrease pressure
j. remove N2O4
N2O4 + energy  2 NO2
k. Increase H2 concentration
l. increase pressure
H2 + Cl2  2HCl + energy
m. Decrease O2 concentration
N2 + O2 + energy  2NO
n. add catalyst
 A. right
 B. left , gas will fill space
 C. left, produces heat (exothermic)
 D. no change, increases speed but does not change volume
 E. right
 F. left, toward endothermic
 G. right, toward endothermic
 H. left, consume excess
 I. right, increase volume
 J. left
 K. right
 L. no change, particle number is not changed either way
 M. left
 N. no change
 The submicroscopic interactions that occur between
water molecules and various solute particles determine
the extent to which water is able to dissolve a solute.
 Ionic compounds
 Like salt in water
 Dissociation: the process by which changed particles in an ionic
solid separate from one another
 Produces ions
 Covalent compounds
 Sugar
 No dissociation, just separation (neutral particles)
 Both cases: interparticle attractive forces between the
solvent and the solute particles, overcome attractive
forces between the solute particles
 Like dissolves like: dissolving occurs when similarities
occur between solvent and solute
 Polar and polar
 Non-polar and non-polar
 Saturated solution: a solution that holds the maximum
amount of solute per solution, the solution can not
dissolve any more solute
 Unsaturated solution: The amount of solute dissolved
is less than the amount of solute than can be dissolved,
the solvent can dissolve more solute
 Supersaturated solution: Unstable solutions which
contain more dissolved solute than they can usually
hold. Conditions were manipulated to get additional
solute dissolved, temporarily.
 Ex fudge making, more sugar is dissolved by heating the
mixture, then slow cooling with constant stirring keeps
the sugar trapped in small crystals in the fudge
 Temperature effects solubility!
 The solubility of most SOLID solutes
generally increases with increasing
temperature
 The solubility of GASES always decreases
with increased temperature
 The amount of increase or decrease
depends on the individual solutes
 Many reactants continue until one of the reactants is
consumed, the reactant used up first is called the limiting
reactant.
 This is used only in non-reversible reactions!
 Limiting reactant problems are solved by comparing the
moles of each reactant.
 Steps
1.
2.
3.
4.
Write a balanced equation
Change given to moles
Determine moles produced by given quantity, for each
Determine predicted quantities produced using the lesser
amount produced (limiting reaction)
 Example: If 40.0g H3PO4 react with 60.0g MgCO3.
calculate the volume of CO2 produced at STP.
2H3PO4 + 3MgCO3 Mg3(PO4)2 + 3CO2 +3H2O
Step 2: Change Grams to moles
Mol H3PO4 = 40g H3PO4 1mol H3PO4 = 0.408 mol
1
98.0 g H3PO4
Mol MgCO3 = 60.0 g MgCO3 1mol MgCO3 = 0.712 mol
1
84.3g MgCO3
 Step 3 :determine moles of CO2 which will be
produced by each reactant(molar ratio from balanced
equation)
Mol CO2 = 0.408 mol H3PO4 3 mol CO2 = 0.612 mol
1
2 mol H3PO4
Mol CO2 = 0.712 mol MgCO3 3 mol CO2 =0.712 mol
1
3 mol MgCO3
The limiting reactant produces the lesser amount of
product- H3PO4
Step 4 :Use the limiting reactant to complete the problem
(remember molar volume is 22.4L/mole at STP)
Volume CO2 = 0.612 mol CO2 22.4 L = 13.7 L at STP
1
1 mol CO2
4og of H3Po4 will produce 13.7 L of CO2 at STP
 If 20.og of NaOH react with 30.0g of H2SO4 to
produce Na2SO4, which reactant is limiting?
2NaOH + H2SO4  Na2SO4 +2H2O
30 g H2SO4 1 mol H2SO4 = 0.30 mol H2SO4
1
98 g H2SO4
20g NaOH 1mol NaOH = 0.5 mol NaOH
1
40.0g NaOH
0.30 mol H2SO4 1 mol Na2SO4 = 0.3 mol Na2SO4 prod
1
1 mol H2SO4
0.5 mol NaOH 1 mol Na2SO4 = 0.25 mol Na2SO4 prod
1
2 mol NaOH
NaOH is limiting reactant, 0.25 mol Na2SO4 prod
 If 5.00g of copper metal react with a solution containing
20.0g of AgNO3 to produce silver metal, which reactant is
limiting?
Cu + 2AgNO3  Cu(NO3)2 + 2Ag
Answer: AgNO3, 0.118 mol Ag
5.0 g copper  moles
20.0 g AgNO3  moles
Moles and mole ratio # of moles produced
1 mol Cu : 2 mol Ag
2 mol AgNO3 : 2 mol Ag
Smallest mole amount produced is limiting reactant
 Remember
 The amount of product produced by a chemical
reaction, as predicted by stoichiometry is called
Theoretical Yield
 when actual yield, produced via experimentation, is
different than theoretical yield, the efficiency of the
reaction is expressed by determining percent yield.
Percent yield = (actual yield
X 100
(Theoretical yield)
This yield can be affected by many different things
Example 1.
 A student conducts a single displacement reaction
that produces 2.755 grams of copper.
 Mathematically he determines that 3.150 grams of
copper should have been produced.
 Calculate the student's percentage yield.
 Solve:
actual amount of product: 2.755 g
expected amount of product: 3.150 g
actual amount of product
percentage yield = --------------------------------------- x 100
expected amount of product
2.755g
percentage yield = --------------- x 100
3.150g
 percentage yield = 87.4603174 % = 87.46 %
Example 2.
 A student completely reacts 5.00g of magnesium
with an excess of oxygen to produce magnesium
oxide.
 Analysis reveals 8.10 g of magnesium oxide.
 What is the student's percentage yield?
In this problem, you are given the actual amount of
the product, but you are not given the expected
amount of the product.
 The second mass shown, 5.00g, is the mass of one
of the reactants.
 In order to determine the expected amount of the
product in this problem, you must begin with a
mass-mass problem.
 First, write a balanced chemical equation for the
reaction:
2Mg(s) + O2(g) ----> 2MgO(s)
 Now, label the given and the unknown:
given
unknown
2Mg(s) + O2(g) ----> 2MgO(s)
5.00 g
Xg
 Change the mass given to moles by dividing by the
molar mass of Mg.
mass given: 5.00g
molar mass of Mg: 24.3 g
5.00g
Number of moles of Mg = ----------24.3 g/mole
Number of moles of Mg = 0.206 moles
 Compare the molar ratio between the given and the
unknown to determine the number of moles produced.
Coefficient of Mg; 2
Coefficient of MgO:2
# of moles of Mg: = 0.206 moles
Number of moles of MgO: = ?
number of moles of given
number of moles of unknown
-------------------------------- = -------------------------------------coefficient of given
coefficient of unknown
0.206 moles X moles
-------------- = ------------2
2
Number of moles of MgO produced = 0.206 moles
 Now, change the number of moles of MgO
produced to the mass by multiplying by the
molar mass of MgO.
# of moles of MgO = 0.206 moles
Molar mass of MgO = 40.3g/mole
mass = # of moles x molar mass
mass of MgO = 0.206 moles x 40.3 g/mole
mass of MgO = 8.30 g
Now, you are ready to solve the percentage yield
problem.
actual mass of MgO produced = 8.10 g
expected mass of MgO = 8.30 g
actual amount of product
percentage yield = expected amount of product X 100
8.10 g
percentage yield = ------------ x 100
8.30 g
percentage yield = 97.6 %
 A student completely reacts 6.00g of magnesium
with an excess of oxygen to produce magnesium
oxide.
 Analysis reveals 8.50 g of magnesium oxide.
 What is the student's percentage yield?
2Mg(s) + O2(g) ----> 2MgO(s)
 Change the mass given to moles
mass given: 6.00g
molar mass of Mg: 24.3 g
6.00g
Number of moles of Mg = ----------24.3 g/mole
Number of moles of Mg = 0.247 moles
Compare the molar ratio:

number of moles of given
number of moles of unknown
-------------------------------- = -------------------------------------coefficient of given
coefficient of unknown
0.247 moles X moles
-------------- = ------------2
2
Number of moles of MgO produced = 0.247 moles
mass = # of moles x molar mass
mass of MgO = 0.247 moles x 40.3 g/mole
mass of MgO = 9.95 g
8.50 g
percentage yield = ------------ x 100
9.95 g
percentage yield = 85.4 %