1. The Mole Concept

2. Understandings The mole is a fixed number of particles and refers to the amount, n, of substance. Masses of atoms are compared on a scale relative to 12 C and are expressed as relative atomic mass (A r ) and relative formula/molecular mass (M r ). Molar mass (M) has the units g mol -1. The empirical formula and molecular formula of a compound give the simplest ratio and the actual number of atoms present in a molecule respectively.

3. Application & Skills Calculation of the molar masses of atoms, ions, molecules and formula units. Solution of problems involving the relationships between the number of particles, the amount of substance in moles and the mass in grams. Interconversion of the percentage composition by mass and the empirical formula. Determination of the molecular formula of a compound from its empirical formula and molar mass. Obtaining and using experimental data for deriving empirical formulas from reactions involving mass changes.

4. SI Units Systeme International de Unites Worldwide Standard set of Units 7 Base Units The International Bureau of Weights and Measures (BIPM) monitors the correct use of SI Units in all applications of science PropertyBase UnitSymbol Masskilogramkg TemperaturekelvinK Timeseconds Amountmolemol Electric current ampereA Luminositycandelacd Lengthmeterm See table 3 p. 13 for prefixes & scales

5. True solutions A solution is a homogenous mixture of two or more substance whose composition can be varied within certain limit. Depending upon no of components present in solution – binary, ternary quaternary solutions If both components are liquid then one in larger quantity is solvent and solute is one in small quantity

6. The Mole Avogadro’s constant (L or N A ) = 6.02 x 10 23 mol -1 Allows us to make comparisons between different chemical species (atoms, molecules, formula units, ions, electrons)

7. Therefore: 6.02 x 10 23 1 mole of 12 C has a mass of 12 g Any

8. Mole Calculations Calculate how many hydrogen atoms are present in 3.0 moles of ethanol, C 2 H 5 OH

9. Mole Calculations Calculate the mass in g of 0.250 mol of magnesium chloride (MgCl 2 )

10. Percentage Composition

11. Applying mole concept Mole ratio to a chemical reaction Stoichiometry, Limiting & Excess Reactants, Theoretical & Percent Yields

12. Suppose you want carry out a reaction that requires combining one atom of iron with one atome of sulfur. How much iron should you use? How much sulfur? Fe + S FeS ? ? You look around in lab there is no device to count number of atoms but there are devices to find relative mass and you have avagadraos rule to find no of atoms One Fe atom has a mass of 55.847 amu and 55.87 g of iron contain 6.02 x 10 23 atoms of Fe Like wise S atom has a mass of 32.06 amu and 32.06 g of iron contain 6.02 x 10 23 atoms of S knowing this you can measure 55.84 g of Fe and 32.06 g of S to make them react react. 1.Mass of substance 2. Amount of substance in moles 3. No of molecules in formula of unit mass Convert using molar mass

13. Atom economy Mole concept allow us to let the chemical reaction happen by know the exact stoichiometry Stoichiometry is the quantitative relationship of reactant and product in a chemical reaction Using a balanced chemical equation to calculate amounts of reactants and products is called stoichiometrystoichiometry Understanding mathematically how in the chemical reaction the participant relate to each other The stoichiometric coefficients/mole ratio are the numbers we use to make sure our equation is balanced. N2 + H2 = NH3 2213 *3*2 626 The coefficient let us understand How reactant and products are related to each other How can it be mathematically be related How this information can be used

14. A limiting reagent is the first reactant that is used up in the chemical reaction When the limiting reactant used up no more product can be formed and the reaction stops Excess reactant is what left over when reaction stops while limiting reactant is all used up

15. Mole ratio

16. Stoichiometry Oxygen gas can be produced by decomposing potassium chlorate using the reaction below. If 138.6 g of KClO 3 is heated and decomposes completely, what mass of oxygen gas is produced? Molar mass of KClO3 - 122.55 g Write equation KClO 3 (s)  KCl (s) + O 2 (g) First Balance the Equation 2 KClO 3 (s)  2 KCl (s) + 3 O 2 (g) What is the mole ratios? 2 : 2 : 3

17. Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products

18. Stoichiometric Calculations From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

20. THEORETICAL YIELD & % YIELD

21. YOUR TURN Aspirin, C 9 H 8 O 4, is made by reacting ethanoic anhydride C 4 H 6 O 3, with 2-hydroxybenzoic acid, C 7 H 6 O 3, according to the equation: 2C 7 H 6 O 3 + C 4 H 6 O 3  2 C 9 H 8 O 4 + H 2 O 13.80 g of 2-hydroxybenzoic acid is reacted with 10.26 g of ethanoic anhydride. a)Determine the limiting reagent. b)The mass obtained in the experiment was 10.90 g. Calculate the theoretical and percentage yield of aspirin.

22. Solution and concentrataion Mole Fraction Molarity Molality Normality

23. Solutions Homogeneous mixture of two ore more pure substances Composed of a solute and a solvent Solute gets dissolved and is distributed evenly throughout the solvent (usually present in smaller amount) Solvent dissolves the solute (usually present in larger amount) When solvent is water the solution is known as an aqueous solution. Weight of solution = weight of solute + weight of solvent

24. Types of Solutions Not all solutions involve a solid dissolved in a liquid. Examples of Solutions: Air - O 2(g) dissolved in mostly N 2(g) Soda water – CO 2(g) dissolved in H 2(l) Vinegar – acetic acid dissolved in water Sugar syrup – sugar dissolved in water Dental filling – Hg (l) dissolved in Ag (s) Brass – Zn (s) dissolved in Cu (s) Steel – C (s) dissolved in Fe (s)

25. Concentrations A solution is said to be saturated when a solvent cannot dissolve any more solute. Concentration is amount of solute to amount of everything Molar Concentration: the amount (mol) of a substance dissolved in 1 dm 3 of solvent.

26. % w/w solution 10 g of sugar is dissolved in water to form a solution of volume 100 ml and density 1 g/ml, calculate its concentration in % w/w Given Wsolute = 10 g W solution = 100 ml D solution = 1 g/ml

27. Try it What would be the concentration in % w/w of a solution obtained by mixing 10 g of Caustic soda (NaOH) in 100 g of water. Given Wsolute =10 g Wsolvent = 100 g Weight of solution = solute + solvent 10 g + 100 g= 110 g The resultant solution will be 9.09% w/w

28. Molarity 1 dm 3 = 1 L What will be molarity of solution of caustic soda when 400g of NaOH is dissolved in water to have a solution of 10litre

29. Determine the molality of 3000. grams of solution containing 37.3 grams of Potassium Chloride KCl. 1. Convert grams KCl to moles KCl 2. Determine the mass of pure solvent from the given grams of solution and solute

30. Determine the molality of 3000. grams of solution containing 37.3 grams of Potassium Chloride KCl. 1. Convert grams KCl to moles KCl 2. Determine the mass of pure solvent from the given grams of solution and solute Total grams = 3000 grams = Mass of solute + Mass of solvent  Mass of pure solvent = (3000 - 37.3) gram = 2962.7 gram

31. . 3. Convert grams of solvent to kilograms 4. Apply the definition for molality

32. Mole Fraction Mole Fraction (  ) A B B B B B A A A A A A A A  A = moles of A sum of moles of all components A B A +  B = moles of B sum of moles of all components B B A + Since A + B make up the entire mixture, their mole fractions will add up to one.

33. Normality It also changes with temperature A normal is one gram equivalent of a solute per liter of solution.

34. Calculate the molarity and normality of a solution containing 0.5 g of NaOH dissolved in 500 ml. Or for monovalent compound like NaOH normality and molarity are same. Solution:

35. A bottle of H2SO4 is marked that it has 98% H2SO4 and density 1.84 g cm-3 at 20 C i) calculate the molarity and normality of the content in bottle ii) also find the volume of the acid required to make 500 cm3 of 0.1 M H2SO4 Solution: 98 g present in 100 g of solution Volume of 100 g of H2SO4 solution= mass/density = 54.3 cm3 Law of dilution to make diluted solution N1V1 = N2V2