P1. stoichiometric Relationship SL and HL.pptx

The power point consists of stoichiometric problems in chemical equations. It also contains numerical problems on empirical and molecular formula. Some titration problems such as back titration is also discussed.

1. Stoichiometric Relationship SL and HL Stoichiometry problems involve the calculation of amounts of materials in a chemical reaction from known quantities in the same reaction

2. MATTER : States of matter

3. Matter : Elements , Compounds and Mixtures PURE

4. Changes : Physical and Chemical

5. Write chemical formulae:

6. Balance the following chemical Reactions

7. State the ionic equation for the following reactions. Write the ionic equation between sodium carbonate and sulphuric acid i . BaCl2( aq ) + 2AgNO3( aq )  Ba(NO3)2( aq ) + 2AgCl(s) ii. Na2CO3( aq ) + 2HCl( aq )  2NaCl( aq ) + CO2(g) + H2O(l)

8. Type of reactions

9. Quick Question ( do in class notebook)

10. Atom Economy

11. HOW BIG IS IT ? 602200000000000000000000 (Approximately)... THAT’S BIG !!! It is a lot easier to write it as... 6.02 x 10 23 It is also known as... AVOGADRO’S NUMBER THE MOLE WHAT IS A MOLE ? it is the standard unit of amount of a substance – DOZEN for 12 SCORE for 20 GROSS for 144

12. Mole Chemical reactions involve atoms and molecules. The ratios with which elements combine depend on the number of atoms not on their mass. The masses of atoms or molecules depend on the substance. Individual atoms and molecules are extremely small. Hence a larger unit is appropriate for measuring quantities of matter. A mole is equal to exactly the number of atoms in exactly 12.0000 grams of carbon 12. This number is known as Avogadro’s number. 1 mole is equal to 6.02 x 10 23 particles.

13. Mole and Mass 1 mole of a substance has a mass equal to the formula mass in grams. Examples 1 mole H 2 O is the number of molecules in 18.02 g H 2 O 1 mole H 2 is the number of molecules in 2.02 g H 2 . 1 mole of atoms has a mass equal to the atomic weight in grams. 1 mole of particles = 6.02 x 10 23 particles for any substance ! The Molar mass is the mass of one mole of a substance Avogadro's number is the number of particles (molecules) in one mole for any substance

14. Formula mass /Molecular mass The formula mass is the sum of atomic masses in a formula. If the formula is a molecular formula, then the formula mass may also be called a molecular mass. If the formula mass is expressed in grams it is called a gram formula mass. The gram formula mass is also known as the Molar Mass. The molar mass is the number of grams necessary to make 1 mole of a substance. The units for Molar Mass are g mol -1 .

15. The atomic mass of Carbon 12 is exactly 12.00000. 1 atomic mass unit = 1/12 of the atomic mass of carbon 12. The periodic table gives the average atomic mass for an element relative to Carbon 12. 1 mole of a substance is 6.022 x 10 23 particles. The mole of atomic mass units is equal to 1.000 gram. The formula mass is the sum of the atomic masses in a formula. A gram formula mass is the same number expressed in grams. It is also equal to Avogadro’s Number of particles Example: H 2 O From the Periodic Table - Atomic Masses: H =1.00797, O = 15.999 The formula mass = 2(1.00797)+15.999 = 18.015 Adding the unit “grams” to the formula mass transforms it into a gram formula mass or mole.

16. The mole is connects the macro world that we can measure with the micro world of atoms and molecules. A Mole is also equal to 1 gram formula mass. 22.4 dm 3 of any gas measured at 0 o C and 1.0 atmosphere of pressure.

17. Example 1: Calculating the Molar Mass of a Compound Calculate the gram formula mass or Molar Mass of Na 3 PO 4 . Atom # Atomic Mass Total Na 3 X 22.99 = 68.97 P 1 X 30.97 = 30.97 O 4 X 16.00 = 64.00 Total = 163.94 Therefore the molar mass is 163.94 g mol -1 Calculate Molar Mass Ca(OH) 2 Copper (II) sulphate penta hydrate

18. CALCULATING THE NUMBER OF MOLES OF A SINGLE SUBSTANCE moles = mass / molar mass mass = moles x molar mass molar mass = mass / moles UNITS mass g molar mass g mol -1 Calculate the following: 1) What is the mass of 0.25 mol of Na 2 CO 3 ? 2) Calculate the number of moles of oxygen molecules in 4.00 g oxygen molecules have the formula O 2 MOLE and Mass Relation MOLES = MASS MOLAR MASS MASS MOLES x MOLAR MASS COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED

19. Percentage Composition According to the law of definite proportions, compounds, contain definite proportions of each element by mass. The sum of all of the atomic masses of elements in a formula is called the formula mass . If it is expressed in grams, then it is called a gram formula mass or molar mass. If it represents the sum of all of the masses of all of the elements in a molecule then it is called a molecular mass. To find the percentage of each element in a compound it is necessary to compare the total mass of each element with the formula mass. 19

20. Percentage Composition The percent by mass of each element in a compound is equal to the percentage that its atomic mass is of the formula mass. Example: Calculate the percentage of oxygen in potassium chlorate, KClO 3 Atomic masses: K = 39.09, Cl = 35.45 and O = 16.00. Formula mass = 39.09 + 35.45+ 3(16.00) = 122.54 Percent Oxygen = (3(16.00)/122.54) (100) = 39.17% Calculate the percentage by mass of each element in potassium carbonate, K 2 CO 3 First calculate the formula mass for K 2 CO 3 . Find the atomic mass of each element from the periodic table. Multiply it by the number of times it appears in the formula and add up the total 2 Potassium atoms K 2 x 39.10 = 78.20 1 carbon atom C 1 x 12.01 = 12.01 3 Oxygen atoms O 3 x 16.00 = 48.00 Total = 138.21 To find the percent of each element divide the part of the formula mass that pertains to that element with the total formula mass Percent of Potassium K = 78.20 X 100 =56.58 % 138.21 Percent of Carbon C = 12.01 X 100 = 8.69 % 138.21 Percent of Oxygen O = 48.00 X 100 = 34.73 % 138.21 20

21. Empirical Formula Determination The empirical formula is the simplest ratio of the numbers of atoms of each element that make a compound. To find the empirical formula of a compound: Divide the amount of each element (either in mass or percentage) by its atomic mass. This calculation gives you moles of atoms for each element that appears in the formula Convert the results to small whole number ratios. Often the ratios are obvious. If they are not divide all of the other quotients by the smallest quotient 21

22. Example 1 Analysis of a certain compound showed that 32.356 grams of compound contained 0.883 grams of hydrogen, 10.497 grams of Carbon, and 27.968 grams of Oxygen. Calculate the empirical formula of the compound. First divide the amount by the atomic mass to get the number of moles of each kind of atom in the formula 22

23. Elements Mass Ar n ratio Hydrogen 0.883 1.01 g mol -1 0.874 mol 0.874 /0.874 =1 Carbon 10.497 12.01 g mol -1 0.874 mol 0.874/ 0.874 =1 Oxygen 27.968 16.00 g mol -1 1.748 mol 1.748 / 0.874=2 HCO 2

24. Molecular Formula To calculate the molecular formula from the empirical formula it is necessary to know the molecular (molar) mass. Add up the atomic masses in the empirical formula to get the factor Divide this number into the molecular formula mass. If the number does not divide evenly you probably have a mistake in the empirical formula or its formula mass Multiply each subscript in the empirical formula by the factor to get the molecular formula 24

25. Molecular Formula Example Example: Suppose the molecular mass of the compound in the previous example, HCO 2 is 90.0. Calculate the molecular formula. The empirical formula mass of is 1 H 1.01 x 1 = 1.01 1 C 12.0 x 1 = 12.00 2 O 16.0 x 2 = 32.00 Total 45.01 Note that 45 is exactly half of the molecular mass of 90. So the formula mass of HCO 2 is exactly half of the molecular mass. Hence the molecular formula is double that of the empirical formula or H 2 C 2 O 4 25

26. Write empirical formula for the following:

27. Exercise : Naphthalene, best known as ‘mothballs’, is composed of carbon (93.71%) and hydrogen (6.29%). If the molar mass of the compound is 128g, what is the molecular formula of naphthalene?

28. Example: Exercise : Naphthalene, best known as ‘mothballs’, is composed of carbon (93.71%) and hydrogen (6.29%). If the molar mass of the compound is 128g, what is the molecular formula of naphthalene?     carbon hydrogen mass of element 93.71 6.29 number of moles 93.71/12 = 7.8 6.29/ 1 = 6.29   DO NOT ROUND UP OR DOWN most simple ratio 7.8/ 6.29 = 1.25 6.29/6.29 = 1   AGAIN DO NOT ROUND UP OR DOWN BUT FIND LOWEST WHOLE NUMBER RATIO lowest whole number ratio 5 4 empirical formula: C 5 H 4 ratio molecular formula /empirical formula: 128g/ 64g = 2     molecular formula = 2 x C 5 H 4 = C 10 H 8

29. Q1)

30. Q2) Experimental question: Find the empirical formula of oxide of Magnesium. Suggest the possible errors in the experiment

31. Simulation lab: Water of hydration Visit http://introchem.chem.okstate.edu/DCICLA/Empirical.html Perform the experiment and take “print screen” calculate water of hydration by using the readings on screen.

32. Number of particles How many number of cations are there in 250 g of Aluminium Sulphate

33. Reacting Masses: 1

34. Reacting Masses:2

35. Limiting Reagent and % Yield 1) Find the actual number of moles of reactants. 2) Find the limiting Reagent. 3) Find the theoretical Yield. 4) Find the percentage yield of Ammonia.

36. Quick Question

37. n = V stp (dm 3 ) / 22.7 (dm 3 / mol ) Standard Temperature and Pressure STP = 273 K (0 °C) and 100 kPa (1atm) The molar volume of a gas at STP is 22.7 dm 3 /mol. Room Temperature and Pressure RTP = 298 K (25°C) and 101.3 kPa (1atm) The molar volume of a gas at RTP is 24 dm 3 /mol.

38. Molar volume

39. Avogadro’s Law: Directly relates gas volumes to moles

40. Quick questions

41. Solutions

42. CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION UNITS concentration mol dm -3 volume dm 3 BUT IF... concentration mol dm -3 volume cm 3 THE MOLE MOLES CONC x VOLUME COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED MOLES = CONCENTRATION x VOLUME MOLES = CONCENTRATION (mol dm -3 ) x VOLUME (dm 3 ) MOLES = CONCENTRATION (mol dm -3 ) x VOLUME (cm 3 ) 1000

43. Units of concentration of solution

44. Use n=CV or m/M = CV

45. The original solution has a concentration of 0.100 mol dm -3 This means that there are 0.100 mols of solute in every 1 dm 3 (1000 cm 3 ) of solution Take out 25.00 cm 3 and you will take a fraction 25/1000 or 1/40 of the number of moles moles in 1dm 3 (1000cm 3 ) = 0.100 moles in 1cm 3 = 0.100/1000 moles in 25cm 3 = 25 x 0.100/1000 = 2.5 x 10 -3 mol 250cm 3 25cm 3 250cm 3 CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask = 0.100 mol dm -3 volume pipetted out into the conical flask = 25.00 cm 3 Moles in part of the solution

46. Standard solution : Solution whose concentration is known accurately ( Activity on preparation of standard solution)

47. Dilution of a solution C 1 V 1 = C 2 V 2

48. Titration: A  titration  is a technique where a solution of known concentration is used to determine the concentration of an unknown solution.  Acid -base Titrations. Redox Titrations . Precipitation Titrations. Complexometric Titrations.

49. Q2) Back titration

50. Challenge Q1 and Q2

51. Gas Laws No definite shape No definite volume Very easily compressed High rate of diffusion Gas particles exert pressure on their surroundings. Kinetic theory of gases 1. The volume of the gas particles is assumed to be zero. 2. The gas particles are in constant motion. 3. The collisions of the gas particles with the sides of the container cause pressure. 4. The particles exert no forces on each other. 5. The average kinetic energy is directly proportional to the Kelvin temperature of the gas.

52. Celcius to Kelvin It is important to remember that you ALWAYS use Kelvin temperatures when working gas law problems. K = o C + 273 .15 Zero K (-273 o C ) is called absolute zero and this is the temperature when all motion theoretically ceases to exist.

53. BOYLE’S LAW Robert Boyle, an Irish chemist, discovered that the volume of a gas was inversely proportional to the pressure applied at constant temperature. In other words, as pressure is increased, volume decreases. P 1 V 1 = P 2 V 2

54. CHARLES’S LAW Jacques Charles, a French physicist, discovered the relationship between temperature and volume at constant pressure. He was the first person to make a solo balloon flight and it was on his flight that he discovered that the volume of a gas was dependent upon the temperature. He discovered that the volume of a gas is directly proportional to the Kelvin temperature. In other words, if you increase the temperature, the volume increases. T 1 V 2 = T 2 V 1

55. GAY LUSSAC’S LAW Joseph Gay Lussac , a French physicist and chemist, another avid balloonist discovered the relationship between temperature and pressure. He discovered that at constant volume, if you increase the temperature, the pressure increases. He also invented many types of chemical glassware that is still in use today. T 1 P 2 = P 1 T 2

56. Combined Gas Laws You can combine the first three gas laws into one equation. Remember to always use Kelvin temperature. In this equation, you will be given 5 knowns and solve for 1 unknown. = Or P 1 V 1 T 2 = P 2 V 2 T 1  

57. Question1

58. Question 2

59. IDEAL GAS LAW Use the ideal gas law when given moles or mass in a gas law problem or if you are asked to solve for moles or mass. PV = nRT Where P is pressure, V is volume, n is moles, T is Kelvin temp, and R is the universal gas law constant. R = 8.31 J K -1 mol -1 or 8.31 m 3 Pa / K mol (A J is a N-m and a pascal is a Nm -2 ) You must use the correct units. P has to be in Pa and V must be in m 3 . OR You must use the correct units. P has to be in kPa and V must be in dm 3 .

60. REAL GASES An ideal gas exactly obeys the gas laws. Real gases actually have attractive forces between them. Real gases actually take up some volume. A gas most behaves like an ideal gas at high temperatures and low pressures.

61. Q1)

62. Q2)

63. Q3)

64. Q4 Data based question

65. Calculate % Error and %Uncertainty from the data below

66. List the different Errors you might have come a crossed and how will you eliminate them?