Transcript m5zn_1ed95c16cede0b1

Solutions
Chemical calculations
1- molecular weight (Mwt)
the sum of atomic weights (Awt) of all elements in
the formula.
Examples:
NaCl: Mwt = 23 + 35.5 = 58.5 g/mol
CaBr2 : Mwt = 40 + (2x80) = 200 g/mol
Fe2O3: Mwt = (56x2) + (16x3) = 160 g/mol
glucose C6H12O6:Mwt =(12x6) + (12x1) + (16x6) = 180 g/mol
.
2-Mole-gram-atom calculations
The mole of any objects (ions, atoms, molecules)
= 6 x 1023 Of these objects.
This number is called Avogadro number (given the
symbol NA).
Why did chemists choose this large number for chemical
calculations?
1- The weight of a mole of any element = Atomic weight of
this element, i.e. 1 mole of an element = Awt
Example: 1 mole of oxygen = 16g. 1 mole of sodium = 23g
2- The weight of a mole of any compound= Molecular
weight of this element, i.e. 1 mole of a compound = Mwt
Example: 1 mole of H2O= 18g. 1 mole of H2SO4 = 98g
Calculation of moles (n) in a given weight:
→ m = n x Awt
→ m = n x Mwt
Examples:
1- Calculate the mass of 2 moles calcium. (Ca = 40)
→ m = n x Awt = 2 x 40 = 80 g.
2- Calculate the mass of 0.25 moles of silver. (Ag = 108)
→ m = n x Awt = 0.2 5x 108 = 27 g.
3- Calculate the mass of 0.374 moles of KCN. (KCN =
65)
→ m = n x Mwt = 0.374 x 65 = 24.31 g.
4- Calculate the mass of 100 moles of Al2O3. (Al2O3 =
102)
→ m = n x Mwt = 100 x 102 = 10200 g.
5- Calculate the mass of 0.08 moles of PCl3. (PCl3 =
137.5 )
→ m = n x Mwt = 0.08 x 137.5 = 11 g
3- Percentage weight calculations
The percentage by weight of an element (%wt) in a given formula is
the total weight of this element in the formula divided by the
molecular weight of this compound.
Example:
1- Calculate the %wt by weight of all elements in glucose
C6H12O6 (Mwt = 180).
→ For carbon: %wt C = 6 x 12 / 180 = 0.400 x 100 % = 40 %
For hydrogen: %wt H = 12 x 1 / 180 = 0.0667 x 100 % = 6.67 %
For carbon: %wt C = 16 x 6 / 180 = 0.5334 x 100 % = 53.34 %
2- Calculate the %wt of all elements in NH4NO3 (Mwt = 80).
→ For nitrogen: %wt N = 2 x 14 / 80 = 0.35 x 100 % = 35 %
For hydrogen: %wt H = 4 x 1 / 80 = 0.05 x 100 % = 5 %
For oxygen: %wt C = 16 x 3 / 80 = 0.60 x 100 % = 60 %
The percentage by weight is useful to calculate the mass
of each element in a given mass of a compound. The % of
the element is multiplied by the given mass of the
compound.
m(element) = %wt x m(compound)
Example:
1- Calculate the mass of sulfur in 5 g of SO3. (S = 32, 0 = 16)
→ Mass of sulfur = (32/80) x 5 = 2 g.
2- Calculate the mass of all elements in 8 g of Na3PO4.
(Na = 23, P= 31, 0 = 16)
→ Mwt = (23 x 3) + 31 + (16x4) = 164 g/mol
Mass of sodium = (23x3/164) x 8 = 3.36585 g.
Mass of phosphorus = = (31/164) x 8 = 1.5122 g
Mass of oxygen = = (16x4/164) x 8 = 3.1219 g
Note: the sum of all masses should be equal to 8.0 exactly.
4-Chemical reactions calculations
The chemical reactions occur according to molar
ratios in the balanced chemical equation.
Steps in calculations:
1- Convert mass into moles using molecular weight .
2- Compare moles of reactants and products
3- Obtain moles of unknown substance
4- Convert moles to mass using molecular weight
- Consider the balanced equation:
C + O2
CO2
If 10 g carbon are reacted completely with oxygen, what mass
of
CO2 is produced?
→ moles C = 10/12 = 0.8334 mol
1 mol C
1 mol CO2
0.8334 mol C
? mol CO2
Hence moles CO2 = mol C = 0.8334 mol
mass CO2 = 0.8334 X 44 = 36.66 g
- Consider the balanced equation:
Cl2 + H2
2HCl
If 1.5 g of Cl are reacted completely, what mass of HCl is
produced?
→ moles Cl2 = 1.5/35.5 = 0.0423 mol
1 mol Cl2
2 mol HCl
0.0423 mol Cl2
? mol HCl
Hence moles HCl= 2 mol Cl2 = 0.0846 mol
mass CO2 = 0.0846 X 36.5 = 3.088 g
Consider the balanced equation:
NaCl + AgNO3
AgCl + NaNO3
1- What mass of AgNO3 is required to reacts completely with
1.00 g of NaCl ?
→ moles NaCl = 1 / 58.5 = 0.0171 mol
1 mol NaCl
1 mol AgNO3
0.0171 mol NaCl
? mol AgNO3
Hence moles AgNO3 = mol NaCl = 0.0171 mol
mass AgNO3 = 0.0171 x 169 = 2.888 g
2- What mass of AgCl will be produced from this reaction?
→ moles AgCl = moles NaCl = 0.0171 mol
mass AgCl = 0.0171 x 143.5 = 2.454 g
Chemical reactions do not go normally 100 % and the
calculated yield is seldom achieved in the lab
practically unless in ionic reactions which are very fast
and complete.
The following definitions are important:
1- Theoretical yield (YTH): is the yield calculated
theoretically depending on the moles.
2- Practical yield (YPR): the actual yield obtained in the
lab. It is always less than or equal to
3- Percentage yield (Y%): the ratio of practical yield to
the theoretical yield multiplied by 100 %
Examples:
1- in a particular reaction the actual yield was 5 g. the reaction was
expected to give 7.9 g based on calculations. What is the percentage
yield of this reaction?
→ Y% = YPR/YTH X 100 % = [5/7.9] X 100 % = 63.29 %
2- Consider P4 + 5O2
P4O10
If 10 g phosphorus were allowed to burn in enough oxygen, only 18 g
of P4O10 were obtained. What is the percentage yield of P4O10 ?
→ Mwt P4 = 31 x 4 = 124 g / mol
= (131 x 4) + (10 x 16) = 124 + 160 = 284 g/mol
moles P4 = 10/124 = 0.08065 mol
From the equation: 1 mol P4
1 mol P4O10
0.08065 mol P4
? Mol P4O10
hence moles P4 = moles P4O10 = 0.08065 mol
mass P4O10 = 0.08065 x 284 = 22.90 g
Percentage yield of P4O10 ?
Y% = YPR/YTH X 100 % = [18/22.9] X 100 % = 78.6 %
ACTIVITY
1- Consider the balanced equation: 2Na + Cl2
2NaCl
calculate the mass of NaCl resulting from the reaction of 0.85 g
Na with enough chlorine gas.
2- In the reaction: 4Al + 3C
Al4C3
Calculate the mass Al required to react completely with 8.88 g
C. Also calculate the mass of the resulting product.
3- In the reaction: 3NaOH + AlCl3
Al(OH)3 + 3NaCl
a- what mass of NaOH will react completely with 2.4 g AlCl3?
b- what mass of Al(OH)3 will form from the complete
reaction of
11 g AlCl3 with enough NaOH?
4- A reaction gave practically 4.4 g while expected to give 5.00
g theoretically. Calculate the percentage yield.
self-ionization of water, Kw
Pure water is not really pure. The purest
water contains some hydronium ions and
hydroxide ions. These two are formed by
the self-ionization of two water molecules.
This happens rarely. The process is an
equilibrium where the reactants, intact
water molecules, dominate the mixture. At
equilibrium the molarities for the
hydronium ion and hydroxide ion are
equal.
[H3O1+] = [OH1-]
The equation is
H 2 O + H2 O
H3O1+ + OH1-
The equilibrium expression is the normal products over reactants.
K = [H3O1+] [OH1-] / [H2O] [H2O]
The concentration for the water is a constant at
any specific temperature.
This means the equation can be rewritten as:
K[H2O] [H2O] = [H3O1+] [OH1-]
The quantity on the right hand side of the
Equation " K[H2O] [H2O] = Kw "
is formally defined as Kw. The numerical
value for Kw is different at different
temperatures.
At 25oC Kw = 1.0 x 10-14
Kw = K[H2O] [H2O]
Kw = [H3O1+] [OH1-] = 1.0 x 10-14
Kw, pH and buffer calculations
Acidic solutions are those having
[H+] > 1 x 10-7 or [OH-] < 1 x 10-7
Basic solution are those having
[OH-] > 1 x 10-7 or [H+] < 1 x 10-7
A neutral solution must have [H+] = = 7.00
The pH of solution is the negative logarithm of the hydronium
ion concentration. That is:
pH = - log[H+]
Hence if H+ concentration is 1 x 10-3 then pH = 3
Acidic solutions are defined to have pH < 7
Basic (alkaline) solutions are defined to have pH > 7
Neutral solutions are those with pH = 7.00 exactly.
Similarly we can write for the hydroxyl ion:
pOH = - log[OH-]
What is the relation between pH and pOH?
It is found that in all aqueous solutions:
[H+][OH-] = 1 x 10-14 = Kw
Where Kw is called the water dissociation constant. If we
apply the negative logarithm to the above equation we get:
pH + pOH = pKw = 14
For example an acidic solution with pH = 4 must have pOH =
10 and a basic solution with pOH = 2.5 must have pH = 14-2.5
= 11.5
Examples:
Calculate pH of the following solutions:
1- A solution with [H+] = 3.5 x 10-5 M
→ pH = - log [H+] = -log 3.5 x 10-5 = 4.456
2- An HCl solution that has molar concentration = 0.05 M
→
HCl
H+ + ClHCl ionize completely, hence [H+] = [Cl-] = [HCl] = 0.05
pH = - log [H+] = -log 0.05 = 1.30
3- A basic solution with [H+] = 2 x 10-10 M
→ pH = - log [H+] = -log 2 x 10-10 = 9.69
4- A basic solution with [OH-] = 4.4 x 10-4 M
→ pOH = - log [OH-] = -log 4.4 x 10-4 = 3.356
pH = 14 – pOH = 14 – 3.356 = 10.644
5- A solution of KOH with molarity of 1.6 x 10-3
→ KOH ionize completely, hence [OH-] = [K+] = [KOH] = 1.6 x
10-3 M
pOH = - log [OH-] = -log 1.6 x 10-3 = 2.79
pH = 14 – pOH = 14 – 2.79 = 11.21
Definitions
A solution is a homogeneous mixture •
A solute is dissolved in a solvent. •
solute is the substance being dissolved –
solvent is the liquid in which the solute is dissolved –
an aqueous solution has water as solvent –
A saturated solution is one where the •
concentration is at a maximum - no more
solute is able to dissolve.
A saturated solution represents an equilibrium: –
the rate of dissolving is equal to the rate of
crystallization. The salt continues to dissolve, but
crystallizes at the same rate so that there
“appears” to be nothing happening.
Dissolution of Solid Solute
What are the driving forces which cause solutes
to dissolve to form solutions?
1. Covalent solutes dissolve by H-bonding to water or by London dispersion
forces (LDF).
2. Ionic solutes dissolve by dissociation into their ions.
Solution and Concentration
the expressing concentration by:
Percent Composition by Mass (% m/m)
Volume Percent (% v/v)
Mole Fraction (X)
Molarity (M)
Molality (m)
 Normality (N)
* Note that molality is the only concentration unit in which
denominator contains only solvent information rather than
solution.
1. Percent Composition by Mass (%):
This is the mass of the solute divided by
the mass of the solution (mass of solute
plus mass of solvent), multiplied by 100.
mass of the solute / mass of the solution x100%
=
e.g:
Determine the percent
composition by mass of a 100 g salt
solution which contains 20 g salt.
Solution:
20 g NaCl / 100 g solution x 100% =
20% NaCl solution
2. Volume Percent (% v/v):
Volume percent or volume/volume
percent most often is used when
preparing solutions of liquids. Volume
percent is defined as:
v/v % = [(volume of solute)/(volume of
solution)] x 100%
Note that volume percent is relative to
volume of solution, not volume of solvent.
For example:
Solution is about 12% v/v ethanol.
This means there are 12 ml ethanol
for every 100 ml of H2O. It is
important to realize liquid and gas
volumes are not necessarily
additive. If you mix 12 ml of ethanol
and 100 ml of H2O, you will get less
than 112 ml of solution.
3. Mole Fraction (X)
This is the number of moles of a
compound divided by the total number of
moles of all chemical species in the
solution.
the number of moles
Mole Fraction (X) =
the total number of moles
Keep in mind, the sum of all mole
fractions in a solution always equals 1.
Example:
What are the mole fractions of the components of the
solution formed when 92 g glycerol is mixed with 90 g
water? (molecular weight water = 18; molecular
weight of glycerol = 92)
Solution:
90 g water = 90 g x 1 mol / 18 g = 5 mol water
92 g glycerol = 92 g x 1 mol / 92 g = 1 mol glycerol
total mol = 5 + 1 = 6 mol
xwater = 5 mol / 6 mol = 0.833
xglycerol = 1 mol / 6 mol = 0.167
It's a good idea to check your math by making sure
the mole fractions add up to 1:
xwater + xglycerol = .833 + 0.167 = 1.000
4. Molarity (M)
Molarity is probably the most commonly
used unit of concentration.
It is the number of moles of solute per liter
of solution (not necessarily the same as the
volume of solvent!).
Molar Concentration
calculations
The molar concentration (Molarity) of solutions is the
number of moles of solute per liter of solution. That is:
Since n = m/Mwt , substituting in the above formula:
and
m = Mwt x M x
V(L)
Example:
What is the molarity of a solution
made when water is added to 11 g
CaCl2 to make 100 mL of solution?
Solution:
11 g CaCl2 / (110 g CaCl2 / mol
CaCl2) = 0.10 mol CaCl2
100 mL x 1 L / 1000 mL = 0.10 L
molarity = 0.10 mol / 0.10 L
molarity = 1.0 M
Examples:
1- Calculate the molar concentration of a solution made by:
A- dissolving 3 g MgBr2 in 500 mL aq. solution.
→ M = m/Mwt x V(L) = 3/(184 x 0.5) = 0.0326 M
B- dissolving 1.11 g FeSO4 in 75 mL aq. solution.
FeSO4 = 56 + 32 + 64 = 154 g/mol
→ M = m/Mwt x V(L) = 1.11 / (154 x 0.075) = 0.0961 M
2- What is the molarity of a 23 mL aqueous solution of NH4CN
containing 0.023 g solute?
NH4CN = 14 + 4 + 12 + 14 = 44 g/mol
→ M = m/Mwt x V(L) = 0.023 / (44 x 0.023) = 0.0227 M
3- if 15 g of K3PO4 were dissolved in water and the volume
completed to 2.5 L, calculate the molar concentration.
K3PO4 = (39 x 3) + 31 + (16 x 4) = 212g/mol
→ M = m/Mwt x V(L) = 15 / (212 x 2.5) = 0.0283 M
5. Molality (m)
Molality is the number of moles of solute per
kilogram of solvent.
Molality (m) = moles of solute / kilogram of solvent
Because the density of water at 25°C is about 1
kilogram per liter, molality is approximately
equal to molarity for dilute aqueous solutions at
this temperature.
Example:
What is the molality of a solution of 10 g
NaOH in 500 g water?
Solution:
10 g NaOH / (40 g NaOH / 1 mol NaOH) =
0.25 mol NaOH
500 g water x 1 kg / 1000 g = 0.50 kg water
molality = 0.25 mol / 0.50 kg
molality = 0.05 M / kg
molality = 0.50 m
6. Normality (N)
Normality is equal to the gram equivalent
weight of a solute per liter of solution. A
gram equivalent weight or equivalent is a
measure of the reactive capcity of a given
molecule. Normality is the only
concentration unit that is reaction
dependent.
Normality (N)=
gram equivalent weight of a solute/ liter of solution
Example:
1 M sulfuric acid (H2SO4) is 2 N for acidbase reactions because each mole of
sulfuric acid provides 2 moles of H+ ions.
On the other hand, 1 M sulfuric acid is 1
N for sulfate precipitation, since 1 mole
of sulfuric acid provides 1 mole of sulfate
ions.
Dilution
Suppose you have 0.500
M sucrose stock solution.
How do you prepare 250
mL of 0.348 M sucrose
solution ?
When a solution is diluted,
solvent is added to lower its
concentration.
The amount of solute remains
constant before and after the
dilution:
moles BEFORE = moles AFTER
C1V1 = C2V2
Concentration
0.500 M
Sucrose
250 mL of 0.348 M sucrose
A bottle of 0.500 M standard
sucrose stock solution is in the
lab.
Give precise instructions to your
assistant on how to use the stock
solution to prepare 250.0 mL of a
0.348 M sucrose solution.
Example:
How many millilieters of 5.5 M NaOH are
needed to prepare 300 mL of 1.2 M NaOH?
Solution:
5.5 M x V1 = 1.2 M x 0.3 L
V1 = 1.2 M x 0.3 L / 5.5 M
V1 = 0.065 L
V1 = 65 mL
So, to prepare the 1.2 M NaOH solution, you
pour 65 mL of 5.5 M NaOH into your
container and add water to get 300 mL final
volume.
3 Stages of Solution Process
• Separation of Solute
– must overcome Intermolecular Forces (IMF) or
ion-ion attractions in solute
– requires energy, ENDOTHERMIC ( + DH)
• Separation of Solvent
– must overcome Intermolecular Forces (IMF) of
solvent particles
– requires energy, ENDOTHERMIC (+ DH)
• Interaction of Solute & Solvent
– attractive bonds form between solute particles
and solvent particles
– “Solvation” or “Hydration” (where water =
solvent)
– releases energy, EXOTHERMIC (- DH)
Factors Affecting Solubility
1. Nature of Solute / Solvent. - Like dissolves like (IMF)
2. Temperature i) Solids/Liquids- Solubility increases with Temperature
Increase K.E. increases motion and collision between solute /
solvent.
ii) gas - Solubility decreases with Temperature
Increase K.E. result in gas escaping to atmosphere.
3. Pressure Factor i) Solids/Liquids - Very little effect
Solids and Liquids are already close together, extra pressure will
not
increase solubility.
ii) gas - Solubility increases with Pressure.
Increase pressure squeezes gas solute into solvent.
Definition of Chemical Equilibrium:
Chemical equilibrium applies to
reactions that can occur in both
directions. In a reaction such as:
CH4(g) + H2O(g)
CO(g) + 3H2(g)
The reaction can happen both ways.
So after some of the products are created
the products begin to react to form the
reactants.
At the beginning of the reaction, the rate
that the reactants are changing into the
products is higher than the rate that the
products are changing into the reactants.
Therefore, the net change is a higher
number of products.
Even though the reactants are constantly
forming products and vice-versa the amount
of reactants and products does become
steady.
When the net change of the products and
reactants is zero the reaction has reached
equilibrium.
The equilibrium is a dynamic equilibrium.
The definition for a dynamic equilibrium is
when the amount of products and reactants
are constant. (They are not equal but
constant. Also, both reactions are still
occurring.)
Equilibrium Constant
To determine the amount of each
compound that will be present at
equilibrium you must know the
equilibrium constant.
To determine the equilibrium constant
you must consider the following equation:
aA + bB
cC + dD
Use the equation to determine the
equilibrium constant (Kc).
For example:
Determining the equilibrium constant of
the following equation can be
accomplished by using the Kc equation.
Using the following equation, calculate
the equilibrium constant.
N2(g) + 3H2(g)
2NH3(g)
N2(g) + 3H2(g)
2NH3(g)
A one-liter vessel contains 1.60 moles
NH3, .800 moles N2, and 1.20 moles of H2.
What is the equilibrium constant?
Le Chatelier's Principle
Le Chatelier's principle states that when
a system in chemical equilibrium is
disturbed by a change of temperature,
pressure, or a concentration, the system
shifts in equilibrium composition in a
way that tends to counteract this change
of variable.
The three ways that Le Chatelier's
principle says you can affect the outcome
of the equilibrium are as follows:
1- Changing concentrations by adding or
removing products or reactants to the
reaction vessel.
2- Changing partial pressure of gaseous
reactants and products.
3- Changing the temperature.
THE END