Analytical chemistrySecond lecture

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.

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 0.0033 lb or 0.053 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 Systeme International d’Unites.

Mass/Amount of Substance:

• Mass kilogram (kg) gram (g) ⇒• Volume liters (L) milliliters (mL)⇒• Amount of Substance mole (mol)

• 1 mole = 6.022 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

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.

• 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 =1.00794 amu1 atom O16 = 15.9994 amu

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= 12.00 grams of C12 .

1mole of H1 = 1.00794 g of H.

• 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=16Molecular mass = (2x1)+ 16 = 18For any element:Molecular mass (amu) = molar mass (grams)So:1 mole of H2O = 18 grams of H2O/mole

Examples:• Calculate the molecular mass of H2SO4?

If H=1, S=32 ,O =16 ,

Molecular mass = (2x1)+32+(4x16) = 981 mole of H2SO4 = 98 gram/mole

• CO2

C= 12, O=16,Molecular mass = 12+ (2x16) = 441 mole of CO2= 44 gram/mole

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 :mmoleFrom 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)

Ex:Calculate the no. of moles in 5 grams of Fe2O3 (ferric oxide)?

Fe= 55.8 ,O=16Solution:

No. of moles = weight (gram)/ molecular weight

Weight = 5 gramsMolecular weight = (2x55.8)+ (3x16) = 159.6 gram /mole

No. of moles = 5/ 159.6 = 0.0312 moles

• 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 = 3.521 mmole

No. of moles= 3.521 / 1000 = .003521 mole

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 solventAnd it can be expressed as :

1.Molarity2.Molality3.Normality4.Percentages.

1. 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 0.735 mol of it in water where the final volume is 650 ml?

Molarity = no. of moles/ volume (L)No. of moles = 0.735 moleVolume=650 ml /1000 ml = 0.65 LMolarity= 0.735/0.65 = 1.13 molar

2. 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 374.00g of water at 250C.Molality= no. moles/kg solvent

• No. of moles= weight(g)/molar mass of Ba(NO3)2 = 75.0 g/261.32 g/mole = 0.28700 mole

molality = 0.28700 mole/ 0.37400 kg = 0.767 m

3. 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.

• 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

• 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

3. percentages:

• They are a set of concentrations expressions based on the% representation and are of three types:

i. 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)

ii. 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)

iii. 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 = 52.50 mL

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.

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.

• 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.

• 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)

• 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)

Classification of solutes in aqueous 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.

What are the steps for preparing solution?

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.

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.

• 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 294.2 g K2Cr2O7

Weight (g)= 159 grams159 grams of K2Cr2O7 are needed to prepare the

requested solution

• Explain the process of making 1L of 3.0M KCl.M = n/volume (L)n = M x volume (L)n = 3.0M x 1Ln = 4.0 mol of KCl needed

No. of moles= weight (g)/MWweight (g)= moles x MWweight (g)= 4.0 mol KCl x 36.0g KClweight (g)= 144g KCl

Weigh out 144g of KCl. Put in a 1L flask. Add enough distilled H20 to dissolve KCl. Fill flask to 1L.

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.

• 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 25 mL 0.15 mol 50 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

SolventEx.(water)

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)

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

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 ??

• 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

• 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 = 0.03125 L = 31.25 mL

• 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

Summary:

• Expressing Concentration • Solutions• Preparing stock solutions.• Preparing diluted solutions.

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

• 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