Transcript Lab # 1.ppt

Introduction
Review of fundamental concept
Formula weight
It is assumed that you can calculate the formula or
molecular weights of compound from respective atomic
weight of the elements forming these compounds.
The formula weight of the substance is the sum of the
atomic weight of the elements from which this substance is
formed from.
Find formula weight of CaSO4.7H2O
Element
Atomic weight
Ca
40.08
S
32.06
11 O
11 x 16.00
14 H
14 x 1.00
FWt
Σ= 262.14
The mole concept
The mole is the major word we will use throughout the
course. The mole is defined as gram molecular weight
which means that:
Mole
Grams
1 mole H2
2.00 g
1 mole O2
32.00 g
1 mole O
16.00 g
1 mole NaCl
58.5 g
1 mole Na2CO3
106.00 g
Assuming approximate atomic weight of 1.00 , 16.00 , 23.00 , 35.5 and
12.00 atomic mass units of hydrogen, oxygen, sodium, chlorine atom,
and carbon respectively.
The number of moles contained in a specific mass of substance can be
calculated as :
Mol = g substance / FW substance
The unit of formula weight is gm/mol
In the same manner, the number of mmol of a substance contained in a
specific weight of the substance can be calculated as
Mmol = mol/1000
Or
Mmol = mg substance / FW substance
Examples
The number of mmol of Na2WO4 (FW = 293.8 mg/mmol)
present in 500 mg of Na2WO4 can be calculated as
?? Mmol of Na2WO4 = 500 mg / 293.8 (mg/mmol ) = 1.70 mmol
The number of mg contained in 0.25 mmol of Fe2O3 ( FW=
159.7 mg/mmol) can be calculated as
?? Mg Fe2O3 = 0.25 mmol Fe2O3 X 159.7 (mg/mmol) = 39.9 mg
Calculation involving solution:
Molarity
Molarity of a solution can be defined as the number of moles of solute
dissolved in 1L of a solution.
This means that 1 mole of solute will be dissolved in some amount of
water and the volume will be adjusted to 1 L.
The amount of water may be less than 1 L as the final volume of water
and solute is exactly 1 L
Calculation involving molarity
Molarity = mol/L = mmol/ml
This can be further formulated as
Number of moles = molarity X volume in L
or
Mole = M(mol/L) X VL
Number of mmole = molarity X volume in mL
or
Mmole = M(mmol/mL) X VmL
Examples:
Find the molarity of a solution resulting from dissolving 1.26g of
AgNO3 (FW = 169.9 g/mol) in a total volume of 250 ml solution.
First find mmol AgNO3 = 1.26x10³ mg AgNO3/ 169.9
mmol
Molarity = mmol/ml
mg/mmol = 7.42
M = 7.42 mmol/250ml = 0.0297 mmol/ml
Let us find the number of mg of NaCl per ml of a 0.25 M NaCl
solution
First we should be able to recognize the molarity as 0.25 mol/L or 0.25
mmol/ml. of course, the second term offers what we need directly.
?? mg NaCl in 1 ml = 0.25 mmol NaCl/ml x (58.5 mg NaCl/mmol NaCl) = 14.6
mg NaCl/ml
Examples
Find the number of grams of Na2SO4 required to prepare 500 ml of
0.1 M solution.
First, we find mmoles needed from the relation
Mmol = M (mmol/ mL) x VmL
Mmol Na2SO4 = 0.1 mmol/mL x 500 ml = 50 mmol
Mmol = mg substance / FW substance
?? Mg Na2SO4 = 50 mmol x 142 mg/mmol = 7100 mg or 7.1 gm.
Calculation involving solution:
Normality
The second expression used to describe concentration
of a solution is the normality.
Normality can be defined as the number of equivalents
of solute dissolved in 1L of solution. Therefore, it is
important for us to define what do we mean by the
number of equivalents, as well as the equivalents
weight of a substance as a parallel term to formula
weight.
N= eq/L
or
N= meq/ml
Number of equivalent= normality x VL = (eq/L) x L
Number of meq= normality x VmL
= (meq/mL) x mL
Also, Number of equivalents= wt(g)/equivalent weight
Number of meq= wt(mg)/eqw
Equivalent weight= FW/n
Meq= mg/eqw substitute for eqw= FW/n gives:
Meq= mg/(FW/n), but mmol= mg/FW, therefore:
Meq = n x mmol
Dividing both sides by volume in mL, we get,
Where n is the number of reacting units (protons, hydroxide or
electrons), and if you are forming factors always remember that a
mole contains n equivalents.
The factor becomes ( 1 mol/n x eq)
or
(n x eq/1mol)
An equivalent is defined as the weight of substance
giving an Avogadro's number of the
Reacting units (n)
protons or hydroxides
Electrons
(in acid base reaction)
(in oxidation reduction reaction)
For example, HCl has one reacting unit H+ when
reacting with base like NaOH,
Therefore, we say that the equivalent weight of
HCl is equal to its formula weight
In the reaction where Mn (Vll), in KMnO4.
is reduced to Mn(ll), so five electrons are
involved and the equivalent weight of
KMno4 is equal to its formula weight
divided by 5.
sulfuric acid has two reacting units (two protons)
when reacting completely with a base..
equivalent weight of H2SO4 is equal to one half
its formula weight,
Example
Find the equivalent weights of NH3 (FW=17.03),
H2C2O4 (FW= 90.04) in the reaction with excess NaOH
and KMnO4(FW= 158.04)when Mn(Vll) is reduced to
Mn(ll).
Solution:

NH3 react with one proton only
Equivalent weights of NH3= FW/1 = 17.03 g/eq

Two protons of oxalic acid react with the base
Equivalent weight of H2C2O4= FW/2 = 90.04/2=45.02 g/eq

H2C2O4+ 2 NaOH
Na2C2O4 + 2 H2O
5 electrons are involved in the reduction of Mn(Vll) to Mn(ll)
Equivalent weight of KMnO4 = FW/5= 158.04/5=31.608 g/eq
Example
Find the normality of the solution containing 5.300 g/L of
Na2CO3 (FW=105.99), carbonate reacts with two protons.
Solution:
Normality is the number of equivalents per liter, therefore we
first find the number of equivalents.
Eq wt= FW/2= 105.99/2= 53.00
Eq= Wt/eq wt= 5.300/53.00= 0.100
N= eq/L= 0.100 eq/1L= 0.100 N
Example
Find the normality of the solution containing 5.267g/L
K2Cr2O7 (FW=294.19) if Cr +6 is reduced to Cr+3.
Solution:
N= eq/L, therefore we should find the number of eq
Eq= wt/eq wt, therefore we should find the number of eq wt
Eq wt= FW/n, here: each contributes 3 electrons and since the
dichromate contains 2 Cr atoms we have 6 reacting units.
Eq wt= (294.19g/mol)/(6 eq/mol)
Eq= 5.267g/(294.19g/mol)/(6eq/mol)
N= eq/L=5.267g/(294.19g/mol)/(6eq/mol)/L=0.1074eq/L
Example
How many mg of I2 (FW=254 mg/mmol) should you weigh to
prepare 250 ml of 0.100N solution in the reaction
I2 + 2e
2 I⁻
meq= N x Vml
meq= 0.1 x 250 = 25meq
mg I2= meq x eq wt= meq x FW/2= 25 X 354/2= 3180mg
Example
What volume of concentrated HCL (FW= 36.5g/mol,
concentration=32%, density=1.1g/ml) are required to prepare 500ml
0f 2.0 M solution.

Always start with the density and find how many grams of solute in each ml of
solution
Density= g solution/ml
Remember that only a 32% of the solution is solute.
Mg HCl/ml = 1.1 x 0.32 x 1000 mg HCL/ml

The problem is now simple as it require conversion of mg HCL to mmol since:
molarity= mmol HCL/ml
Molarity= mmol HCL/ml = {1.1 x 0.32 x 1000mg HCL}/ 36.5 mg/mmol =9.64 M

Now, we can calculate the volume required from the relation
M1V1 = M2V2
9.64 x Vml = 2.0 x 500 ml
Vml= 10.4ml of the concentrated HCL should be added to dis.H2O and the volume
adjusted to 500ml
An easy short cut:
M= Density x percentage of solute x 1000
Formula weight
The percentage is a fraction : ( i.e : is written as
0.35)
Stoichiometric Calculations:
Volumetric Analysis
In this section we look at calculations involved in titration processes as well as
general quantitative reactions. In a volumetric titration, an analyte of
unknown concentration is titrated with a standard in presence of a
suitable indicator.
Volumetric Analysis
Titration type
Redox
Precipitation
Complexometric
Volhard's method
Back titration
Mohr's method
In-direct method
Direct method
Titration method
Acid base
For a reaction to be used in titration the
following characteristics should be
satisfied:
stoichiometry of the reaction should be exactly known. This means that we
should know the number of moles of A reacting with 1 mole of B.
reaction should be rapid and reaction between A and B should occur
immediately and instantly after addition of each drop of titrant
(the solution in the burette).
There should be no side reactions. A reacts with B only.
There should exist a suitable indicator which has distinct color change.
The reaction should be quantitative. A reacts completely with B.
Standard solution:
Standard solution:
Readily absorb CO2, from the atmosphere
NaOH, KOH
Readily absorb H2O, from the atmosphere
Non Primary
Standards solution
HCl, and H2SO4,
The % written on the reagent bottle is a
claimed value and should not be taken as
guaranteed.
HNO3
It always contains a little HNO2, which
has a destructive action on many acidbase indicators.
Standard solution:
anhydrous sodium carbonate,
Na2CO3, weak base
Prepare a dilute solution of HCL,H2SO4
with an approximate concentration, and then
determine the concentration accurately by
titration with a primary standard solution of
Na2CO3
Primary Standards
solution
potassium hydrogen phthalate,
KH(C8H4O4), a weak acid
you can prepare a NaOH solution with an
approximate concentration. This solution
can then be titrated with an acidic primary
standard solution, such as KH(C8H4O4).
Molarity volumetric analysis
What you really need is to use the stoichiometry of the
reaction to find how many mmol of A as compared to the
number of mmoles of B
Moles of
Compound A
Molar ratio
Moles of
Compound B
Example
A 0.4671 g sample containing NaHCO3 (FW = 84.01 mg/mmol) was dissolved
and titrated with 0.1067 M HCl requiring 40.72 mL. Find the percentage of
bicarbonate in the sample.
We should write the equation in order to identify the stoichiometry
NaHCO3 + HCl
NaCl + H2CO3
Now it is clear that the number of mmol of bicarbonate is equal to the number of mmol HCl
Mmol NaHCO3 = mmol HCl
Mmol = M x VmL
Mmol NaHCO3 = ( 0.1067 mmol/ml ) x 40.72 mL = 4.345 mmol
Now get mg bicarbonate by multiplying mmol times FW
Mg NaHCO3 = 4.345 mmol x (84.01 mg/mmol) = 365.0
% NaHCO3 = (365.0 x 10-3 g/0.4671 g) x 100 = 78.14%
Example
A 0.4671 g sample containing Na2CO3 (FW = 106mg/mmol) was
dissolved and titrated with 0.1067 M HCl requiring 40.72 mL.
Find the percentage of carbonate in the sample.
The equation should be the first thing to formulate
Na2CO3 +2 HCl = 2NaCl + H2CO3
Mmol Na2CO3 = ½ mmol HCl
Now get the number of mmol Na2CO3 = ½ x ( MHCl x VmL (HCl) )
Now get the number of mmol Na2CO3 = ½ x 0.1067 x 40.72 = 2.172 mmol
Now get mg Na2CO3 = mmol x FW = 2.172 x 106 = 230 mg
% Na2CO3 = (230 x 10-3 g/0.4671 g ) x 100 = 49.3 %
Example
How many mL of 0.25 M NaOH will react with 5.0 mL of 0.10 M H2SO4.
H2SO4 + 2 NaOH = Na2SO4 + 2 H2O
Mmol NaOH = 2 mmol H2SO4
Mmol NaOH = 2 {M (H2SO4) x VmL (H2SO4)}
Mmol NaOH = 2 x 0.10 x 5.0 = 1.0 mmol
mmol NaOH = MNaOH x VmL (NaOH)
0.1mmol NaOH = 0.25 x VmL
VmL = 4.0 mL
Example
Find the volume of 0.1 M KMnO4 that will react with
50 ml of 0.2 M H2O2 according to the following
equation:
5H2O2 + 2KMnO4 + 6H+
2Mn+2 +5O2 + 8H2O
The following step is to formulate the relationship between
the number of moles of the two reactants.
We always start with the one we want to calculate, that is
Mmol KMnO4 = 2/5 H2O2
Mmol = molarity x Vml
0.1 x Vml = (2/5) x 0.2 x 50
Vml = 40 ml KMnO4
Example
Find the volume of 0.1 M KMnO4 that will react with
50 ml of 0.2 M MnSO4 according to the following
equation:
3MnSO4 + 2KMnO4 + 4 OHmmol KMnO4 = 2/3 mmol MnSO4
0.1 M x Vml = 2/3 x 0.2 M X 50
Vml = 66.7 ml KMnO4
5MnO2 + 2H2O + 2K+ + 3SO4 -2
Normality volumetric calculation
We have seen previously that solving volumetric problems
required setting up a relation between the number of
mmoles or reacting species. In case of normality,
calculation is easier as we always have the number of meq
of substance A is equal to meq of substance B, regardless
of the stoichiometry in the chemical equation. Of course
this is accounted for in the calculation of meqs.
Therefore, the first step in a calculation using normalities
is to write down the reaction
meqA = meq B
Example
A 0.4671 g sample containing sodium bicarbonate NaHCO3 was titrated
with HCl requiring 40.72 ml. the HCL was standardized by titrating
0.1876 g of sodium carbonate Na2CO3 (FW= 106 mg/mmol) requiring
37.86 ml of the acid. Find the percentage of NaHCO3 (FW= 84.00
mg/mmol) in the sample.
Eq wt Na2CO3 = FW/2= 53.0
Now we can find meq of NaHCO3
Eq wt NaHCO3 = FW/1=84.0
meq NaHCO3 = meq HCl
meq HCl = meq Na2CO3
Mg NaHCO3/ eq wt = N X Vml
Normality x volume (ml) = wt (mg) / eq wt
Mg NaHCO3/ 84 = 0.0935 x 40.72
N x 37.86 = 187.6 mg / (53 mg/meq)
Mg NaHCO3 = 84 x 0.0935 x 40.72
N HCL = 0.0935 eq/L
% Mg NaHCO3= (0.67) x 100 = 67.2%
Example
Use normality to calculate how many ml of a 0.1 M H2SO4 will
react with 20 ml of 0.25 M NaOH
We can first convert molarities to normalities:
N= n x M
N (H2SO4)= 2 x 0.1= 0.20
N (NaOH)= 1 x 0.25= 0.25
Meq H2SO4 = meq NaOH
Substitution for meq as usual (either N xVml or mg/eq wt)
0.2 x Vml = 0.25 x 20
Vml= 25 ml