Quantitative Analysis A.S. 2.1 (Chemistry 91161) Year 12 4 Internal Credits

Quantitative Analysis

A.S. 2.1 (Chemistry 91161)

Year 12

4 Internal Credits

The Mole and Molar The Mole and Molar MassMass

6.02 X 6.02 X 10102323

The amount (n) of a substance that contains

particles is called a mole

The MoleThe Mole

• A counting unit• Similar to a dozen, except instead

of 12, it’s 602 billion trillion 602,000,000,000,000,000,000,000

• 1 mole = 6.02 X 1023 particles• Avagadro’s number • Enough soft drink cans to cover

the surface of the earth to a depth of over 300 Km.

The MoleThe Mole

• 1 dozen cars = 12 cars

• 1 mole of cars = 6.02 X 1023 cars

• 1 dozen Al atoms = 12 Al atoms

• 1 mole of Al atoms = 6.02 X 1023 atoms

Note that the NUMBER is always the same, but the MASS is very different!

Mole can be abbreviated, but only to mol!!!

1. Number of atoms in 0.500 mole of Ala) 500 Al atoms

b) 6.02 x 1023 Al atomsc) 3.01 x 1023 Al atoms

2.Number of moles of S in 1.8 x 1024 S atomsa) 1.0 mole S atomsb) 3.0 mole S atomsc) 1.1 x 1048 mole S atoms

Learning CheckLearning Check

• The Mass of 1 mole of a substance (in grams)

• Equal to the atomic mass number

1 mole of C atoms = 12.0 g

1 mole of Mg atoms = 24.3 g

1 mole of Cu atoms = 63.5 g

• Unit gmol-1

• 7gmol-1 means 1 mole of the substance would

weigh 7 g

Molecular Mass (M)Molecular Mass (M)

Mass (in grams) of 1 mole is equal to the sum of the atomic masses e.g.

What is the Molar mass of CaCl2

1 mole Ca x 40.1 g/mol

+ 2 moles Cl x 35.5 g/mol = 111.1 g/mol CaCl2

You try

1 mole of N2O4 =

Molecular Mass of Molecules and Molecular Mass of Molecules and CompoundsCompounds

92.0 g/mol

A. Molar Mass of K2O = 94 Grams/mole

B. Molar Mass of antacid Al(OH)3 = 78 gmol-1

C. Prozac, C17H18F3NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. Find its molar mass.

Learning Check!Learning Check!

?

?

309 gmol-1

Balancing Equations

• 2Ca(s) + O2(g) 2CaO(s) + energy• This means that 2 mols of Ca atoms react with every 1

mol of Oxygen molecules to produce 2 mols of CaO and “n” mols of energy is released.

• If 4 mols of Ca was burnt how many mols of oxygen would be required and how much energy is produced?

• How many mols of oxygen are reacted when 6 mols of CaO are produced?

• If 40g of Ca are reacted what mass of CaO is produced?

• What is the mole ratio of ions formed when MgCl2 dissolves?

Use a balanced equation to calculate quantities

• If 50g of ethane burns what is the mass of water produced?

• 2C2H6 + 7O2 4CO2 + 6H2O• M(C2H6) = 30• n=m/M =50/30 = 1.67 moles ethane• C2H6:3H2O Mole ratio STOICHIOMETRY• 1.67 x 3 = 5 moles water• M(H2O)= 18• m = nxM = 5x18 = 90g water

m

Mn

Converting Moles (n), Mass (g) Converting Moles (n), Mass (g) and Molecular Mass (M)and Molecular Mass (M)

m

Mn

Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al?

m = n x M = 3 x 27 = 81g Al

n = m/M

The artificial sweetener aspartame

(Nutra-Sweet) formula C14H18N2O5 is

used to sweeten diet foods, coffee and

soft drinks. How many moles of

aspartame are present in 225 g of

aspartame?

Learning Check!Learning Check!m

Mn

0.77 moles

Learning Check!Learning Check!How many atoms of K are present in 78.4 g of K?

What is the mass (in grams) of 1.20 X 1024 molecules of glucose (C6H12O6)?

How many Na+ in 25g of Na2CO3?

X 2 = 0.472 moles

2.01 x 6.02 x 1023 = 12.1 x 1023

n=1.20 X 1024 / 6.02 x 1023 =2 moles

M= 180g/mole so m=nxM =2x180 = 360g

M(Na2CO3)= 106g/mol

N=m/M = 25/106 = 0.236moles

What is the percent carbon in C5H8NO4 (the

glutamic acid used to make MSG monosodium

glutamate), a compound used to flavour foods and

tenderise

meats?

a) 8.2 %C

b) 27.8 %C

c) 41.1 %C

Percent CompositionPercent CompositionGives the breakdown of mass between different atoms

e.g. Find the percentage of Carbon in Carbon monoxide M(C) =12 M(O) =16

Percentage(C) = M(C) /M(CO) x 100 = 42.86%

Types of FormulasTypes of FormulasEmpirical FormulaEmpirical Formula

The formula of a compound that has the The formula of a compound that has the smallestsmallest whole number ratiowhole number ratio of the atoms of the atoms present. present.

e.g. CHe.g. CH33 means for every C there are 3H means for every C there are 3H

Molecular FormulaMolecular Formula

The formula that gives the The formula that gives the actualactual number of number of each kind of atom found in the molecule.each kind of atom found in the molecule.

e.g. Ce.g. C22HH66 means every molecule has 2C and means every molecule has 2C and

6H atoms6H atoms

Chemical Formulas of Chemical Formulas of CompoundsCompounds

• Formulas give the relative numbers of atoms or Formulas give the relative numbers of atoms or moles of each element in a formula - always a moles of each element in a formula - always a whole number ratio.whole number ratio.

NONO22 2 atoms of O for every 1 atom of N 2 atoms of O for every 1 atom of N

1 mole of NO1 mole of NO22 : 2 moles of O atoms to : 2 moles of O atoms to

every 1 mole of N atomsevery 1 mole of N atoms

• If we know or can determine the relative number If we know or can determine the relative number of moles of each element in a compound, we can of moles of each element in a compound, we can determine a formula for the compound.determine a formula for the compound.

Find Find Empirical Formula Empirical Formula given given the compositionthe composition

1.1. Determine the mass in grams of each Determine the mass in grams of each element present, if needed.element present, if needed.

2.2. Calculate the number of Calculate the number of molesmoles of of each each atom.atom.

3.3. Divide each by the smallest number of Divide each by the smallest number of moles to obtain the moles to obtain the simplest whole simplest whole number ratio.number ratio.

4.4. Write empirical formula and determine its Write empirical formula and determine its molar mass.molar mass.

m

Mn

A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine the empirical formula for this substance.

moles of N = 2.34g of N = 0.167 moles of N

14.01 g/mole

moles of O = 5.34g of O = 0.334 moles of O

16.00 g/mole

Formula:0.167 0.334 2

0.167 0.167

N O NO

Therefore N0.167O0.334


Find Molecular FormulaFind Molecular Formula given given the molar massthe molar mass

1. Identify the empirical formula and calculate its molar mass.

2. Divide the given molar mass by the empirical molar mass.

3. Multiply the empirical formula by the answer from step 2.

m

Mn

CH4

M(CH4) =16

64/16 = 4

C4H16

Calculate the molecular formula of a compound with empirical formula CH4 and a molar mass of 64g/mol

A compound has an empirical formula of NO2. The colourless liquid, used in rocket engines has a mass of 92.0 g. What is the molecular formula of this substance?


Molecular Formula from % Molecular Formula from % CompositionComposition

A substance contains 60.80 % Na ; 28.60 % B ; 10.60 % H by mass. Its molar mass is 114g/mole.What is the molecular formula of the substance?

Consider a sample size of 100 gramsDetermine the number of moles of each atomDetermine the simplest whole number ratioWrite empirical formulaCalculate molar mass of empiricalDivide molar mass by empirical molar massMultiply answer by empirical formula

Na3B3H12

HELP!

Concentration of Solution

• n = amount of solute (moles)

• c = concentration of solution (moles/litre)

• V = volume of solution (litres)

nc V

• c=n/V• =0.025/0.050• =0.5 molL-1

• n = cV• = 0.3 x 0.021• = 0.0063 moles

nc V

Problem 1. Calculate concentration if Problem 1. Calculate concentration if 0.025 moles of HCl are present in 0.025 moles of HCl are present in 50ml of solution.50ml of solution.

Problem 2. Calculate moles of NaProblem 2. Calculate moles of Na22COCO33

present in 21mL of 0.3molLpresent in 21mL of 0.3molL-1-1 solution solution

• n = m/M

• = 2.4/106

• = 0.023 moles

• c = n/V

• = 0.023/0.250

• = 0.092molL-1

2.4g of sodium carbonate is dissolved in water to make 250ml of solution. Calculate the concentration. M(Na2CO3) = 106gmol-1

nc V

mn M

Volumetric Analysis

• Acid Base Titrations– Used to determine an

unknown concentration

ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations

ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations

HH22CC22OO44(aq) + 2 NaOH(aq) ---> Na(aq) + 2 NaOH(aq) ---> Na22CC22OO44(aq) + 2 H(aq) + 2 H22O(liq)O(liq)

acidacid basebase

Carry out this Carry out this neutralisationneutralisation reaction using a reaction using a TITRATION.TITRATION.

Oxalic acid,Oxalic acid,

HH22CC22OO44

Setup for titrating an acid with a baseSetup for titrating an acid with a base

TitrationTitrationTitrationTitration1.1. Use Use pipettepipette to measure your (unknown to measure your (unknown

concentration) solution into the flask.concentration) solution into the flask.2.2. Add known solution from the Add known solution from the buretburet to to

find the average titre.find the average titre.3.3. Indicator shows when exact Indicator shows when exact

stoichiometric reaction has occurred. stoichiometric reaction has occurred. (Acid = Base)(Acid = Base)

4.4. Calculate moles (n) of known solution.Calculate moles (n) of known solution.5.5. Use balanced equation to find the Use balanced equation to find the

unknown amount of moles.unknown amount of moles.6.6. Calculate the concentration of Calculate the concentration of

unknown.unknown.

PROBLEM #1: Standardise a solution of PROBLEM #1: Standardise a solution of NaOH — i.e., accurately determine its NaOH — i.e., accurately determine its concentration.concentration.


Titre 1 2 3 4 Av.

Vol. ml

25.0 25.4 29.4 25.2 25.2

35 mL of NaOH is 35 mL of NaOH is pipettedpipetted

into a flask and neutralised into a flask and neutralised

(by titration) with (by titration) with ?? mL of mL of

0.0998 M HCl. What is the 0.0998 M HCl. What is the

concentration of the NaOH? concentration of the NaOH?

25.2

CONCORDANCE

3 consistent results

2. Find average titre of HCl (L)

NaOH + HCl NaCl + H20

1:1 ratio, therefore 0.025 moles NaOH reacts with 0.025 moles HCl



4. Calculate moles (n) of HCl

nc V

n = cV

= 0.0998 x 0.0252

= 0.025 moles HCl used

5. Use balanced equation to find the unknown amount of moles NaOH

• c=n/V• = 0.025/0.035

• = 0.71molL-1



6. Calculate the concentration of NaOH nc V

Your turnYour turn

20 mL of Al(OH)20 mL of Al(OH)33 is neutralised (by titration) with 18.6mL is neutralised (by titration) with 18.6mL

of 0.2 M HCl. What is the concentration of the Al(OH)of 0.2 M HCl. What is the concentration of the Al(OH)33??

Concentration of Al(OH)3 = 0.062molL-1

PROBLEM 3: You have 50.0 mL of PROBLEM 3: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M 3.0 M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?

PROBLEM 3: You have 50.0 mL of PROBLEM 3: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M 3.0 M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?

Add water to the 3.0 M solution to lower its Add water to the 3.0 M solution to lower its

concentration to 0.50 Mconcentration to 0.50 M i.e.i.e. Dilute it!Dilute it!

3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

But how But how much much water water do we do we add?add?

PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?

PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?

c • V = Amount (n) of NaOH in original solution

(3.0 mol/L)(0.050 L) = 0.15 mol NaOH

Amount of NaOH in final solution must also = 0.15 mol NaOH

n/C = Volume of final solution

(0.15 mol NaOH)/(0.50 mol/L) = 0.30 L

or 300 mL

n

c V

PROBLEM: You have 50.0 mL of PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M 3.0 M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?

PROBLEM: You have 50.0 mL of PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M 3.0 M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?

Conclusion:Conclusion:add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of

0.50 M NaOH.

3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

Diluting Solutions

Finding the new concentrationYour volume Total volume

e.g. 20ml of 0.5molL-1 is diluted to 100ml. What is the final concentration?

0.02/0.1 X 0.5 =0.1molL-1

Find the original concentration of HCl when 20ml is pipeted into a flask and made up to 250ml giving a final concentration of 4x10-3molL-1

X concentration = diluted

concentration

0.05molL-1

No need to convert ml to L if both units are the same

Test Yourself• You have 50.0 mL of 3.0 M NaOH and you want 0.50

M NaOH. How much water do you need to add?

250ml• What is the concentration of Cu2+ if 20ml of 0.64molL-1

of CuSO4 is diluted to 100ml?

0.128molL-1

• 25ml CuSO4 solution was diluted to 200ml. The new concentration is 7.36x10-4 molL-1. What was the original concentration?

5.89x10-3 molL-1

Density

• Convert L g and g L• E.g. Calculate the volume of 6

moles of ethanol given the density of ethanol (CH3CH2OH) is 0.789gml-1

• M = 46gmol-1

• m = Mn = 46 x 6 = 276g• 276g/0.789gml-1 = 349.8ml or

0.350L

Apple Juice• 20ml apple juice was titrated against

0.1molL-1 NaOH and the average was 10.36ml. Assuming all apple juice is citric acid and that 1mole reacts with 3 moles of NaOH. Calculate the concentration of citric acid in gL-1. Apple juice must contain between 0.3 and 0.8g per 100ml. Is this considered apple juice or apple drink?

• M(citric acid)=192gmol-1

Gravimetric analysis

Decomposing of solids

• Hydrated magnesium sulphate (MgSO4.XH2O) is heated in a crucible (total mass 37.95g) so the water is lost from the crystals. When it is reweighed it has a constant mass of 36.82g. The crucible weighs 35.75g what is the mass of the water and anhydrous MgSO4?

MgSO4 = 1.07g H20 = 1.13g


• Calculate X the water of crystallisation in MgSO4.XH2O

• M(MgSO4) = 120gmol-1 M(H2O) = 18gmol-1

• Calculate the number of moles of each and the mole ratio.

• n(MgSO4) = 1.07/120 = 0.00892moles• n(H2O) = 1.13/18 = 0.0628moles• 0.0089:0.0628 = 1:7• X = 7

m

Mn


• AgNO3 is added to 25ml of water containing Cl- so that the ions are precipitated out.

• Ag+(aq) + Cl-(aq) AgCl(s)

• They are found to weigh 0.32g. Calculate the concentration of Cl- in solution.

• n= m/M = 0.32/143.5 = 0.00223 moles AgCl

• Cl- : AgCl so 0.00223 moles Cl-

• c = n/V = 0.00223/0.025 = 0.089molL-1 m

Mn

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Quantitative Analysis A.S. 2.1 (Chemistry 91161) Year 12 4 Internal Credits