CONCENTRATION
EXPRESSION
1
grams
Moles =
formula weight (g / mol)
Where formula weight represents the atomic or
molecular weight of the substance.
Thus,
g
g
Moles Na2 SO 4 =
=
f wt 142.04 g / mol
g
g
+
Moles Ag =
=
f wt 107.870 g / mol
2
Calculate the number of grams in one mole of CaSO4 7H2O
Solution
One mole is the formula weight expressed in grams. The formula
weight is
Ca
40.08
S
32.06
11 O
176.00
14 H
14.11
262.25 g/mol
3
milligrams
Mi lim oles =
formula weight (mg / mmol)
g Na2SO4 = moles X f wt = moles X 142.04 g/mol
g Ag = moles X f wt = moles X 107.870 g/mol
We usually work with millimole quantities, so
Miligrams = millimoles X formula weight (mg/mmol)
Note that g/mol is the same as mg/mmol, g/L the same as
mg/ml, and mol/L the same as mmol/mL.
4
Calculate the number of moles in 500 mg Na2WO4
(sodium tungstate).
Solution
500 mg
293.8 mg/mmol
x 0.001 mol/mmol = 0.00170 mol
How many milligrams are in 0.250 mmol Fe2O3 (ferric
oxide)?
Solution
0.250 mmol x 159.7 mg/mmol = 39.9 mg
5
Millimoles = molarity X milliliters
(or mmol = M X mL)
A solution is prepared by dissolving 1.26 g AgNO3 in a 250 – mL
volumetric flask and diluting to volume. Calculate the molarity
of the silver nitrate solution. How many millimoles AgNO3 were
dissolved?
Solution
M=
1.26g/169.9g/mol
0.250L
= 0.0297mol/L (or 0.0297 mmol/mL)
Then,
Millimoles = (0.0297 mmol/mL)(250 mL) = 7.42 mmol
6
How many grams per milliliter of NaCI are contained
In a 0.250 M solution?
Solution
0.250 mol/L = 0.250 mmol/mL
0.250 mmol/mL X 58.4 mg/mmol X 0.001 g/mg = 0.0146 g/ml
How many grams Na2SO4 should be weighed out to prepare 500
mL of a 0.100 M solution?
Solution
500 mL X 0.100 mmol/mL = 50.0 mmol
50.0 mmol X 142 mg/mmol X 0.001 g/mg = 7.10 g
7
Expressing concentrations of
solution
Molarity = moles of solute (mol)
volume of solution (L)
= mmoles of solute (mmol)
volume of solution (mL)
Normality (N) = no. of equivalents (eq)
volume of solution (L)
= no. of equivalents (meq)
volume of solution (mL)
8
No. equivalents (eq) =
mass (g)
equivalent mass (g/eq)
= normality (eq/L) x vol (L)
No. equivalents (meq) =
mass (mg)
equivalent mass (mg/meq)
= normality (meq/mL) x vol (mL)
Equivalent mass = Relative Molec Mass (g/mol)
No. of reacting units (eq/mol)
Reacting units: H+ (acid/base) or e- (redox)
9
Calculate the normality of a solution prepared by
dissolving 2.3543 g K2Cr2O7 in 1 L water, which
is used to oxidise FeCl2 in the unbalanced
equation. RMM K2Cr2O7 is 294.19
K2Cr2O7 + FeCl2 + HCl
→
CrCl3 + FeCl3 + KCl + H2O
N=
mass K2Cr2O7 (g)
equiv. mass K2Cr2O7 (g/equiv) x mL
Cr2O72- + 14 H+ + 6e → 2Cr3+ + 7H2O
Equiv. mass = RMM K2Cr2O7 = 294.19 (mg/mmol)
no. e
6 (meq/mmol)
= 49.030 mg/meq
= 0.04800 N
N=
2354.3 mg
49.030 mg/meq x 1000 mL
10
Calculate the volume of a conc solution
required to prepare 1 L 0.100 M HCl
solution that was taken from a bottle of
conc. HCl. The label on the bottle
indicates 37.0% HCl, density of 1.18 g/mL.
RMM for HCl is 36.5
11
37.0% HCl =
37.0 g HCl
100 g conc. solution
Mass HCl = 1.18 g conc soln x 37.0 g HCl
Vol HCl
mL conc soln
100 g conc.
= 0.437 g/mL
Mass HCl required =
1000 mL x 0.100 mmol x 0.0365 g
mL
mmol
= 3.65 g
= 8.36 mL
Vol HCl required = 3.65 g
0.437 g/mL
12
Solid Samples
• Express as weight % (% w/w)
% = mass solute (g)
x 100
mass of sample (g)
For trace concentrations, expressed as ppt, ppm,
ppb
Ppt = mass solute (g)
x 103
mass of sample (g)
Ppm = mass solute (g)
x 106
mass of sample (g)
Ppb = mass solute (g)
x 109
mass of sample (g)
13
• Mass Units
mg 10-3 g
µg 10-6 g
ng 10-9 g
Volume Units
L
mL 10-3 L
µL 10-6 L
14
A sample weighing 1.3535 g contains 0.4701 g .
Calculate the % Fe in the sample. What is the Fe
content in ppt and ppm?
% Fe = 0.4701 g x 100 = 34.73%
1.3535 g
Ppt Fe = 0.4701 g x 103 = 347. 3 ppt
1.3535 g
Ppm Fe = 0.4701 g x 106 = 3.473 105 ppm
1.3535 g
15
Liquid Samples
Normally expressed as % mass/volume (% w/v)
% w/v = mass solute (g) x 102
vol of sample (mL)
Ppt = mass solute (g) x 103
vol of sample (mL)
Ppm = mass solute (g) x 106
vol of sample (mL)
Ppb = mass solute (g) x 109
16
vol of sample (mL)
% volume/volume (% v/v)
% v/v = volume of solute (mL) x 102
vol of sample (mL)
Ppt = volume of solute (mL) x 103
vol of sample (mL)
Ppm = volume of solute (mL) x 106
vol of sample (mL)
Ppb = volume of solute (mL) x 109
vol of sample (mL)
17
Common Units For Expressing Trace Concentrations
Unit
Parts per million
Abbreviation
wt/wt
wt/vol
vol/vol
Ppm
mg/kg
mg/L
µL/L
µg/g
µg/mL
nL/mL
µg/kg
µg//L
nL/L
ng/g
ng/mL
pL/mLa
(1ppm = 10-4%)
Parts per billion
Ppb
(1ppb = 10-7% =
10-3ppm)
Milligram percent
A
mg%
mg/100g mg/100mL
pL = picoliter = 10-12 L
18
DILUTIONS – PREPARING THE RIGHT CONCENTRATION
We often must prepare dilute solutions from
more concentrated stock solutions.
The millimoles taken for dilution will be the
same as the millimoles in the diluted solution ,
i.e.,
Mstock X mLstock = MDiluted X mLDiluted
19
You wish to prepare a calibration curve for the
spectrophotometric determination of permanganate.You have
a stock 0.100 M solution of KMnO4 and a series of 100 –mL
volumetric flasks. What volumes of the stock solution wll you have
to pipet into flasks to prepare standards of 1.00, 2.00, 5.00, and
10.00 X 10-3 M KMnO4 solutions?
Solutions
A 100 mL solution of 1.00 X 10-3 M KMnO4 will contain
100 mL X 1.00 X 10-3 mmol/mL = 0.100 mmol KMnO4
We must pipet this amount from the stock solution
0.100 mmol / mL X χ mL = 0.100 mmol
χ = 1.00 mL stock solution
For other solutions, will need 2.00, 5.00 and 10.0 mL of stock
solution.
20
A 2.6 – g sample of plant tissue was analyzed and found to
contain 3.6 µg zinc. What is the concentration of zinc in the
Plant in ppm ? In ppb?
Solution
3.6 µg
= 1.4 µg / g 1
ß.4 ppm
2.6 g
3.6 X10 3 ng
= 1.4 X10 3 ng / g 1400
ß
ppb
2.6g
21
(a) Calculate the molar concentrations of 1.00 ppm solution
each of Li+ and Pb2+
Solution
(a) Li concentration = 1.00 ppm = 1.00 mg/L
Pb concentration = 1.00 ppm = 1.00 mg/L
M
1.00 mg Li / L X10 3 g / mg
4 mol / L Li
=
1
.
44
X
10
Li =
6.94 g Li / mol
M
1.00 mg Pb / L X10 3 g / mg
6 mol / L Pb
=
4
.
83
X
10
Pb =
207 g Pb / mol
22
What weight of Pb(NO3)2 will have to be dissolved
in 1 liter of water to prepare a 100 ppm Pb2+
solution?
(b) 100 ppm Pb2+ = 100 mg/L = 0.100 g/L
0.100 g Pb
= 4.83 X 10 4mol Pb
207g / mol
Therefore, we need 4.83 X 10-4 mol Pb (NO3)2
4.83 X 10-4 mol X 283.3g Pb (NO3)2/mol = 0.137 g Pb (NO3)2
23
• KAT 141nm2
24
25
• A selective reaction or test is one that can occur with
other substances but exhibits a degree of preference for
the substance of interest.
• A specific reaction or test is one that occurs only with
the substance of interest.
Unfortunately, few reactions are specific but many
exhibit selectivity. Selectivity may be achieved by a
number of strategies. Some examples are:
Sample preparation (e.g., extractions, precipitation)
Instrumentations (selective detectors)
Target analyte derivatization (e.g. derivatize specific
functional groups with detecting reagents)
Chromatography, which provides powerful separation
26
27
Material
Max. Working
Temperature
(C)
Borosilicate
glass
200
Sensitivity to
thermal
Shock
Chemical
Inertness
Notes
1500C
change OK
Attacked
somewhat
by alkali
solutions on
heating
Trademarks :
Pyrex, Kimax,
Soft glass
Poor
Attacked by
alkali
solutions
Boron – free.
Trademark :
Corning
Alkaliresistant
glass
More
sensitive
Than
Borosilicate
Resistant to
most acids,
halogens
Quartz
crucibles
used for
fusions
Fused
quartz
1050
Excellent
28
Material
High – silica
glass
Porcelain
Platinum
Max. Working
Temperature (C)
1000
1100 (glazed)
1400 (unglazed)
ca.1500
Sensitivity to
thermal
Shock
Chemical
Inertness
Excellent
More resistant
to alkalis than
borosilicate
Good
Excellent
Resistant to
most acids,
molten salts.
Attacks by
aqua regia,
fused nitrates,
cyanides,
chlorides at >
10000C. Alloys
with gold,
silver, and
other metals
Notes
Similar to
fused quartz
Trademark :
Vycor
(Corning)
Usually alloyed
with iridium or
rhodium to
increase
hardness.
Platinum
crucibles for
fusions and
treatment with
HF.
29
Material
Max.
Working
Temperature
(C)
Sensitivity
to thermal
Shock
Nickel and
iron
Stainless steel
400 – 500
Polyethylene
115
Excellent
Chemical Inertness
Notes
Fused samples
contaminated with
the metal
Ni and Fe
crucibles
used for
peroxide
fusions
Not attacked by
alkalis and acids
expect conc. HCI,
dil H2SO4 and
boiling conc. HNO3
Not attacked by
alkali solutions or
HF. Attacked by
many organic
solvents (acetone,
ethanol OK)
Flexible
plastic
30
Sensitivity to
thermal
Shock
Material
Max. Working
Temperature
(C)
Chemical
Inertness
Notes
Polystyrene
70
Not
Somewhat
attacked by brittle
HF. Attacked
by many
organic
solvents
Teflon
250
Inert to most
chemicals
Useful for
storage of
solutions and
reagents for
trace metal
analysis
31
Some Common Drying Agents
Capacity
Deliquescenta
High
Yes
CaSO4
Moderate
No
CaO
Moderate
No
MgCIO4
(anhydrous)
High
Yes
Silica gel
Low
No
AI2O3
Low
No
P2O5
Low
Yes
Agent
CaCI2
(anhydrous)
Q
Trade Name
Drierite (W.A. Hammond
Drierite Co).
Anhydrone (J.T. Baker
Chemical Co.);
Becomes liquid by absorbing moisture
32