Download Solute - Share ITS

ANALYTICAL CHEMISTRY
INTRODUCTION AND REVIEW
What is analytical chemistry?
The science of inventing and applying the concepts,
principles and strategies for measuring the
characteristics of chemical systems and species
often described as the area of chemistry responsible for
characterizing the composition of matter
Qualitative analysis
An analysis in which we
determine the identity of
the constituent species
in a sample
Quantitative analysis
An analysis in which we
determine how much of
a constituent species is
present in a sample.
The analytical problems
characterization analysis
An analysis in which we
evaluate a sample’s chemical
or physical properties
Example, determinations of
-chemical structure,
-equilibrium constants,
-particle size,
-and surface structure
fundamental analysis
Example:
An analysis whose purpose is
to improve an
analytical method’s capabilities
-Extending and improving the
theory
-studying a method’s limitations,
-designing new and modifying
old methods
The analytical process
1.
Identify and define the problem
2.
Design the experimental procedure
3.
Conduct an experiment, and
gather data
4.
Analyze the experimental data
5.
Propose a solution to the problem
The analytical process
Classification of quantitative
method of analysis
Volumetric method
Volume
is
ThemeGallery
is a Design
Digital
Content & Contents
measured
or
mall developed by Guild
used to
Design Inc.
determine
amount of
sample via
concentration
Gravimetric method
Mass is
measured
Instrumental method
Use an
instrumental
technique to
assay the
amount of
sample
Fundamental SI Units
Measurement
Unit
Symbol
kilogram
kg
volume
liter
L
distance
meter
m
temperature
kelvin
K
time
second
s
current
ampere
A
mole
mol
mass
amount of substance
SI Units and non SI Unit
Fundamental SI Units
Exponential
Prefix
Symbol
1012
tera-
T
109
giga-
G
1 mL = 1x10-3 L
106
mega-
M
1 L = 1x103 mL
103
kilo-
k
10-1
deci-
d
1 mL = 1x10-6 L
10-2
centi-
c
1 L = 1x106 mL
10-3
milli-
m
10-6
micro-
μ
10-9
nano-
n
10-12
pico-
p
10-15
femto-
f
10-18
atto-
a
1 attoliter = 1x10-18 L
1 L = 1x1018 attoliter
Review of stoichiometry
A. Empirical vs molecular or structural formulas:
1.
Empirical formulas  give information only about the
simplest ratio between the different elements composing the
molecule.
Example: HO, H2CO
2.
Molecular formulas  give information about the numbers of
atoms of each element found in the molecule.
Example: H2O2, H4C2O2, C2H5OH, C2H4O2, C3H6O3, C6H12O6
3.
Structural formulas  give information about the structure of
the molecule as well as the numbers of atoms of each element
Example: HOOH, (CH3)3COH, CH3CH2OCH2CH3
Solution
Preparation of solution
Solute: a minor species in a solution
Solute
Solvent: a major species in a solution
Solvent
Total
solution
Example: glucose solution
 Glucose is a solute
 Water is a solvent
Molar Concentration
Concentration  a general measurement unit stating
the amount of solute present in a known amount of solution
Molarity:
The number of moles of solute per 1 liter
of solution (M)
number of moles of solute (mole) mol
Cx 

M
volume of solution (L)
L
weight of solute (g)
number of moles of solute (mole) 
MW of solute (g/mol)
Molality and density
Molality number of moles of solute (mole)
C weight 
weight of solvent (kg)
Molality: expresses the mol of substance/solute
per unit mass of solvent
Weight = Volume x density
Percent Concentration
mass of solute
1. Weight percent (w/w) 
100%
mass of solution
volume of solute
2. Volume percent (v/v) 
100%
volume of solution
3. Weight/vo lume percent (w/v)
mass of solute (g)

100%
volume of solution (mL)
Part per hundred (pph), thousand (ppt),
million (ppm) and billion (ppb)
mass of solute
2
C pph (w/w) 
10 pph
mass of solution
mass of solute
3
C ppt (w/w) 
10 ppt
mass of solution
mass of solute
6
C ppm (w/w) 
10 ppm
mass of solution
mass of solute
C ppb (w/w) 
109 ppb
mass of solution
Units for Reporting Concentration
Name
molarity
normality
molality
weight %
volume %
weight-to-volume %
parts per million
parts per billion
EW=equilibrium weight
Units
moles solute
liters solution
number EWs solute
liters solution
moles solute
kg solvent
g solute
100 g solution
mL solute
100 mL solution
g solute
100 mL solution
g solute
106 g solution
g solute
109 g solution
Symbol
M
N
m
% w/w
% v/v
% w/v
ppm
ppb
Example 1
117.0 g of NaCl in 1.00 L of water has
concentration of _____ mol/L
MW of NaCl = 58.5 g/mol
WNaCl
117.0
mole of NaCl 

 2.00 mole
MWNaCl 58.5
C NaCl
mole NaCl 2.00 moles


 2.00 mole/L  2.00M
V (L)
1.00 L
Example 2
294.0 g of H2SO4 in 1.00 L of water has
concentration of _____ mol/L
MW of H2SO4 = 98.0 g/mol
mole of H 2SO 4 
C H 2SO 4 
WH 2SO 4
MWH 2SO 4
mole H 2SO 4
V (L)
294.0 g

 3.00 moles
98.0 g/mol
3.00 moles

 3.00 mole/L  3.00M
1.00 L
Example 3
What is the w/w % of aqueous ammonia (NH3) solution at
14.3 M with density = 0.9 g/mL (900 g/L)?
Solute: NH3
1) MW of NH3 = 17.0 g/mol
2) Mole of NH3 at 14.3 M in 1.00 L = 14.3 mol/L x 1.00 L
= 14.3 mol
3) Weight of NH3 at 14.3 M in 1.00 L = mole of NH3 x
MW of NH3 = 14.3 mol x 17.0 g/mol = 243 g
4) Weight of 1.00 L solution = volume x density = 1.00 L x
900 g/L = 900 g
mass of solute
243
Weight percent (w/w) 
100% 
100%
mass of solution
900
=27.0%
Example 4
What is the molar concentration of aqueous ammonia (NH3)
solution with density = 0.9 g/mL (900 g/L) and 27.0% (w/w)?
Solute: NH3
MW of NH3 = 17.0 g/mol
Weight of NH3 =mole of NH3 x MW of NH3
= C NH3 x V NH3 x MW of NH3
Weight of 1.00 L solution = volume x density
= 1.00 L x 900 g/L = 900 g
WNH3
mass of solute
Weight percent (w/w) 
100% 
100%  27%
mass of solution
900
W NH3 = C NH3 x V NH3 x MW of NH3 = 243 g
C NH3 = 14.3 mol/L = 14.3 M
Example 5
What is the v/v % of ethanol in a solution prepared by
mixing 5.00 mL of ethanol with enough water to give 1.00 L
of solution?
Solute: ethanol
1) Volume of solute (ethanol) = 5.00 mL = 5.00 x 10-3 L
2) Volume of solution = 1.00 L
3) Volume percent (v/v) =
volume of solute
5 103 L
100% 
100%  0.5%
volume of solution
1.00 L
Stoichiometric Calculation
1.
2.
Stoichiometric calculation are based on the combining
ratios of reactants which result in specific products.
They are expressed in terms of moles
3.
When you are given the mass of reactant or
product, you should first convert the mass to
moles to determine the amount of reactant that
will be consumed or product that will be produced
for a given reaction
4.
If the final answer is to be given in a mass unit, then
the moles must be converted to grams
Volumetric Calculations
Dilution: the number of moles are the same
in dilute and concentrated solution
moles = Cconcentrated Vconcentrated= Cdilute Vdilute
Units:
V=either in L and mL
C=M(mol/L) or mM (mmol/L)
Be sure to match units for both dilute and concentrated solutions
Stoichiometric Calculation (1)
What mass of AgNO3 (MW = 169.9 g/mol) is needed to
convert 2.33 g of Na2CO3 (MW = 106.0 g/mol) to
Ag2CO3 ?
Na 2CO3 (aq)  2 AgNO 3 (aq)  Ag 2CO3 (s)  2 NaNO3 (aq)
1 mol
2 mol
2.33 g
106.0 g/mol
WAgNO3
169.9 g/mol
2.33 g
1 106.0 g/mol

WAgNO3
2
169.9 g/mol
WAgNO3
2.33 g
 2
169.9 g/mol
106.0 g/mol
WAgNO3  7.47 g
Stoichiometric Calculation (2)
What mass of Ag2CO3 (MW = 275.7 g/mol) will be
formed ?
Na 2CO3 (aq)  2 AgNO 3 (aq)  Ag 2CO3 (s)  2 NaNO3 (aq)
1 mol
2.33 g
106.0 g/mol
2.33 g
1 106.0 g/mol

WAg2CO 3
1
275.7 g/mol
1 mol
WAg2CO 3
275.7 g/mol
WAg2CO 3
2.33 g

275.7 g/mol 106.0 g/mol
WAg2CO 3  6.06 g
Stoichiometric Calculation (3)
To prepare a solution with 0.500 M of Cl- from BaCl2.2H2O.
How much of BaCl2.2H2O must be used to prepare 1.00 liter
of solution? Assume BaCl2 completely dissociates:
BaCl 2  Ba 2  2 Cl-
1 mol
x mol
2 mol
0.5 mol/L x 1.00 L
1 mole BaCl 2 mole BaCl 2 mole BaCl 2



2
mole Cl CCl - VCl 0.5  1.00
x mol = mole of BaCl2 (mol) = 0.5/2 = 0.25 mol
MW of BaCl2.2H2O = 244.2 g/mol
W of BaCl2.2H2O = mole of BaCl2 (mol) x MW of BaCl2.2H2O
= 0.25 mol x 244.2 g/mol=61.1 g
Stoichiometric Calculation (4)
How many L of a 50 mol/L solution are required to make 200
mL of a 1 mol L-1 solution?
moles = Cconcentrated Vconcentrated= Cdilute Vdilute
50 mol/L  Vconc.  1 mol/L  0.200 L
1 M  0.2 L
Vconc. (L) 
 0.004 L
50 M