1. Analytical chemistry
Second lecture

2. Measurements in analytical chemistry
Analytical chemistry is a quantitative science. Whether determining the concentration of a species, evaluating an equilibrium constant, measuring a reaction rate, or drawing a correlation between a compound’s structure and its reactivity, In this section we briefly review the use of units and significant figures in analytical chemistry.

3. Units of Measurement
A measurement usually consists of a unit and a number expressing the quantity of that unit. We may express the same physical measurement with different units, which can create confusion. For example, the mass of a sample weighing 1.5 g also may be written as lb or oz. To ensure consistency, and to avoid confusion, scientists use a common set of fundamental units, several of which are listed in the Table below; These units are called SI units after the Système International d’Unités.

5. Mass/Amount of Substance:
Mass kilogram (kg) ⇒ gram (g)
Volume liters (L)⇒ milliliters (mL)
Amount of Substance mole (mol)
1 mole = x 1023 particles (e.g. atoms, molecules, ions)
Atomic Mass = number of grams containing Avogadro’s number (6.022 x 1023) of atoms
Molecular Mass = number of grams containing Avogadro’s number (6.022 x 1023) of molecules; sum of atomic masses of elements in a molecule

6. Example: Atomic mass = the sum of proton and neutron in an atom.
Atomic number = number of protons in an atom.
No. of neutrons = atomic mass – atomic no.
No. of protons = no. of electrons;
No. of neutrons = atomic mass – no. protons.
No. of neutrons = atomic mass – no. of electrons.

7. According to that, Atomic mass can be defined as:
It is the mass of an atom or particle in atomic mass unit (amu).
On that scale,
1 atom C12 weighs 12 amu
1 atom H1 = amu
1 atom O16 = amu

8. as defined earlier, moles are the number of atoms, ions or molecules in a substance. We can calculate the weight of a substance if we know the formula, For any element: Atomic mass = molar mass (grams) 1 mole of C12 atom= grams of C12 . 1mole of H1 = g of H.

9. Molecular mass ( molecular weight) :
the sum of the atomic mass of each constituent atom multiplied by the number of atoms of that element in the molecular formula.
Ex:
The molecular mass of H2O = 18 amu
As atomic mass of H=1 and O=16
Molecular mass = (2x1)+ 16 = 18
For any element:
Molecular mass (amu) = molar mass (grams)
So:
1 mole of H2O = 18 grams of H2O/mole

10. Examples: Calculate the molecular mass of H2SO4? If H=1, S=32 ,O =16 ,
Molecular mass = (2x1)+32+(4x16) = 98
1 mole of H2SO4 = 98 gram/mole
CO2
C= 12, O=16,
Molecular mass = 12+ (2x16) = 44
1 mole of CO2= 44 gram/mole

11. How to calculate number of moles in a compound:
From the equation below:
No. of moles = weight (grams) / molecular weight (gram/mole)
The unit is: mole
No. of millimoles = weight (mg) / molecular weight (mg/mmole)
Unit is :mmole
From the equation above we can caculate the weight in grams or mg:
Weight(gram)= no. of moles x molecular weight (g/mole)
Weight (mg) = no. of mmole x molecular weight (mg/mmole)

12. Ex: Calculate the no. of moles in 5 grams of Fe2O3 (ferric oxide)
Ex: Calculate the no. of moles in 5 grams of Fe2O3 (ferric oxide)? Fe= 55.8 ,O=16 Solution: No. of moles = weight (gram)/ molecular weight Weight = 5 grams Molecular weight = (2x55.8)+ (3x16) = gram /mole No. of moles = 5/ = moles

13. Ex2:
Calculate the number of moles in 500 mg of Na2SO4?
Na=23, S=32, O=16?
No. of mmole= wieght(mg) / molecular weight (mg/mmole)
= 500 / 142 = mmole
No. of moles= / 1000 = mole

14. Concentration:
Concentration: is a general measurement unit stating the amount of solute present in a known amount of solution
Concentration= amount of solute/ amount of solvent
And it can be expressed as :
Molarity
Molality
Normality
Percentages.

15. Molarity:
It is the number of moles of solutes in a liter of solvent.
Molarity= moles of solute / liter of solvent.
The unit is molar (M).
Example:
calculate the molarity of a solution of Nacl prepared by dissolving mol of it in water where the final volume is 650 ml?
Molarity = no. of moles/ volume (L)
No. of moles = mole
Volume=650 ml /1000 ml = 0.65 L
Molarity= 0.735/0.65 = 1.13 molar

16. molality:
it is the number of moles of solute /kilogram of solvent.
Molality= no. of solute/ kg solvent
The unit is molal (m)
Example:
Determine the molality of a solution prepared by dissolving 75.0g Ba(NO3)2 (s) in to g of water at 250C. Molality= no. moles/kg solvent
No. of moles= weight(g)/molar mass of Ba(NO3)2 =
75.0 g/ g/mole = mole
molality = mole/ kg = m

17. Normality:
Defined as the number of equivalents dissolved in liter of the solvent.
N= no. of eq. / volume(L)
No. of eq. can be calculated as :
No. of eq.= weight (g) / equivalent weight
N = weight (g) / equivalent weight x volume (L)
And the eq. weight can be calculated as:
Molecular weight/ n
Where n= no. of reacting units (التكافؤ)
N varies depending on the reaction.

18. In acid –base reaction :
n = no. of hydrogen if its an acid.
Or the no. of hydroxyl if it’s a base.
Ex: (acids)
HCl ; n = 1
H2SO4 ; n = 2
H3PO4 ; n = 3
Ex: ( base)
NaOH ; n = 1
Ca(OH)2 ; n = 2
Al(OH)3 ; n= 3

19. In redox reactions:
n= no. of electrons (take on or supply).
If we know the no. of equivalents (n), then we can calculate N by:
N = weight (g) x n x 1000 / molecular weight x volume (mL)
or simply:
N = Molarity x n
As molarity = no. of moles / volume (L)
And no. of moles = weight (g) / molecular weight

20. percentages: Weight per volume percentage (w/v )% Ex:
They are a set of concentrations expressions based on the% representation and are of three types:
Weight per volume percentage (w/v )%
Can be defined as the weight of solutes in grams dissolved in mL of the solution and can be expressed as (g/mL) %.
w/v % = weight solute (g)/ volume solution(mL).
Ex:
5 grams of NaOH dissolved in 500 ml of solution, calculate w/v %?
w/v %= 5/ 500 x100 = 1% (g/mL)

21. Weight per weight percentage (w/w)%:
Can be defined as the number of grams of solutes in number of grams of solution.
And its expressed as (g/g)% .
w/w % = weight solute(g) / weight of solution (g) x100
Ex:
Find the weight percentage (w/w)% of HCl, if 20 grams of it was dissolved in 1 kg of solution?
w/w %= 20 (g) / 1000 (g) x 100 = 2 % (g/g)

22. Volume per volume percentage (v/v)% :
can be defined as the volume of solutes (mL) dissolved in (mL) of solution.
And can be expressed as (mL/mL)%
v/v % = volume solute/ volume solution x 100
Ex:
What is the volume of ethyl alcohol dissolved in in water if the v/v% is 15% and volume of solution 350 mL?
15% = volume of alcohol/ 350 mL x100
Volume of alcohol = 15 x 350 /100 = mL

23. Preparing solution:
Now as you know how to calculate M, n, V , what does that mean?
that means you will be able to prepare your own solutions.

24. What is a solution?
Solution: a homogeneous mixture of two or more substances.
Solute: a substance in a solution that is present in the smallest amount.
Solvent: a substance in a solution that is present in the largest amount.
In an aqueous solution, the solute is a liquid or solid and the solvent is always water.
The majority of chemical reactions occur in Aqueous solution.

25. All solutes that dissolve in water fit into one of two categories: electrolyte or non-electrolyte.
Electrolyte: a substance that when dissolved in water conducts electricity.
Non-electrolyte: a substance that when dissolved in water does not conduct electricity.

26. Electrolyte can be divided into; strong or weak electrolyte.
Strong electrolyte: conduct current very efficiently by Completely ionized or dissociate when dissolved in water ( cations (+) and anions (-) ).
Ex: strong acids, strong bases and soluble salts.
NaCl(s) → Na+(aq) + Cl–(aq)
HCl(s) → H+(aq) + Cl–(aq)
Weak electrolyte: conduct only a small current
By Slightly ionized in solution.
Ex: weak acids, weak bases.
CH3COOH(aq) ↔ CH3COO–(aq) + H+(aq)

27. Non electrolyte does not conduct any current as they do not dissociate in the aqueous solution.
No cations (+) and anions (-) in the solution.
Ex:
Sugars( glucose, sucrose), alcohol (ethanol, methanol)
C6H12O6 (s) → C6H12O6 (aq)

28. Classification of solutes in aqueous solution:

29. What are the steps for preparing solution?
Preparing solution of known concentration is perhaps the most common activity in any analytical lab.
Pipets and volumetric flasks are used when a solution’s concentration must be exact; graduated cylinders, beakers when concentrations need only be approximate.
Two methods for preparing solutions are used; preparing stock solution, preparing solution by dilution.

30. Preparing stock solution:
A Stock Solution is a concentrated solution that will be diluted to some lower concentrated for actual use. Stock solutions are used to save preparation time, conserve materials, reduce storage space, and improve the accuracy with which working lower concentration solutions are prepared.
A stock solution is prepared by weighing out an appropriate portion of a pure solid or by measuring out an appropriate volume of a pure liquid and diluting to a known volume.

31. For example, to prepare a solution with a desired molarity you weigh out an appropriate mass of the reagent, dissolve it in a portion of solvent, and bring to the desired volume.

32. How many grams of Potassium Dichromate, K2Cr2O7, are required to prepare a 250mL solution with a concentration of 2.16M?
250mL x 1L/ 1000mL = 0.250L
M= n/v
n= M x v
n= 2.16M x .250L
n= 0.54 mol But in the lab we weigh grams not moles, so …
No.moles = weight (g)/MW
Weight (g) = no.moles x MW
Weight (g)= 0.54 mol K2Cr2O7 x g K2Cr2O7
Weight (g)= 159 grams
159 grams of K2Cr2O7 are needed to prepare the requested solution

33. Explain the process of making 1L of 3.0M KCl.
M = n/volume (L)
n = M x volume (L)
n = 3.0M x 1L
n = 4.0 mol of KCl needed
No. of moles= weight (g)/MW
weight (g)= moles x MW
weight (g)= 4.0 mol KCl x 36.0g KCl
weight (g)= 144g KCl
Weigh out 144g of KCl. Put in a 1L flask. Add enough distilled H20 to dissolve KCl. Fill flask to 1L.

34. Preparing solution by dilution:
Solutions are often prepared by diluting a more concentrated stock solution. A known volume of the stock solution is transferred to a new container and brought to a new volume.
So we can define Dilution as the procedure for preparing a less concentrated solution from a more concentrated one.

35. e.g.: 25 mL of 6.0 M HCl contains 0.15 mol HCl
A given volume of a stock solution contains a specific number of moles of solute.
e.g.: 25 mL of 6.0 M HCl contains 0.15 mol HCl
(How do you know this???) =
25 mL mL
0.15 mol mL
0.15 mol
If 25 mL of 6.0 M HCl is diluted with 25 mL of water, the number of moles of HCl present does not change.
Still contains 0.15 mol HCl
Solvent
Ex.(water)

36. no. moles solute = no. moles solute
before dilution after dilution
Although the number of moles of solute does not change, the volume of solution does change.
The concentration of the solution will change since:
Molarity = no .of moles / volume (L)

37. Dilution calculations:
When a solution is diluted, the concentration of the new solution can be found using the formula below:
Mc x Vc = Md x Vd
Where,
Mc= initial concentration (mol/L)= more concentrated
Vc = initial volume of more conc. solution
Md = final concentration (mol/L) in dilution
Vd = final volume of diluted solution

38. Practice:
What is the concentration of a solution prepared by diluting 25.0 mL of 6.00 M HCl to a total volume of 50.0 mL?
Given: Vc = 25.0 mL
Mc = 6.00 M
Vd = 50.0 mL
Find: Md ??

39. Mc x Vc = Md x Vd Md = Mc x Vc / Vd Md = 6.00 x 0.025/ 0.05 = 3 mol/L
By using the formula:
Mc x Vc = Md x Vd
Md = Mc x Vc / Vd
Md = 6.00 x 0.025/ 0.05 = 3 mol/L

40. How many mL of a 14 M stock solution must be used to make 250 mL of a 1.75 M solution?
Mc x Vc = Md x Vd
Mc = 14 M,
Vc = ?,
Md = 1.75 M,
Vd = 250 mL
Vc = Md Vd / Mc = (1.75 M) x (0.250 L) / (14 M)
Vc = L = mL

41. 100 mL of 6.0 M CuSO4 must be diluted to what final volume so that the resulting solution is 1.5 M?
Mc = 6 M,
Vc = 100 mL,
Md = 1.5 M,
Vd = ?
Vd = McVc / Md = (6 M) x (0.100 L) / (1.5 M)
Vd = 0.4 L or 400 mL

42. Summary: Expressing Concentration Solutions Preparing stock solutions.
Preparing diluted solutions.

43. Exercises:
How many grams of nitric acid are present in 1.0 L of a 1.0 M HNO3 solution?
63 g
Give the molarity of a solution containing 10 g of H2SO4 solute in 2.5 L of solution?
0.041 mol/L

44. Water is added to 4 L of 6 M antifreeze until it is 1. 5 M
Water is added to 4 L of 6 M antifreeze until it is 1.5 M. What is the total volume of the new solution?
Vd = 16 L