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All of the elements on the periodic table make up everything in our world!

All of the elements on the periodic table make up everything in our world! The periodic table consisting of most of the known elements was developed by a Russian man called Dimitri Mendelev around 1860. He managed to put all the elements in order of weight ! without sophisticated scientific

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All of the elements on the periodic table make up everything in our world!

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  1. All of the elements on the periodic table make up everything in our world! The periodic table consisting of most of the known elements was developed by a Russian man called Dimitri Mendelev around 1860. He managed to put all the elements in order of weight ! without sophisticated scientific equipment.

  2. Mendeleevs First Periodic Table

  3. Relative atomic Mass (Ar) numbers All atoms were measured against Carbon. The carbon atom was assigned an atomic mass of 12 6 C 12 This tell us that each carbon atom has a mass of 12 It also tells us the number of protons and neutrons We say the relative atomic mass (Ar) of Carbon is 12

  4. Relative atomic Mass (Ar) numbers 1 H hydrogen has a relative mass of 1 compared to carbon 1 We say the relative atomic mass (Ar) of hydrogen is 1

  5. Relative atomic Mass (Ar) numbers 8 O oxygen has a relative mass of 16 compared to carbon 16 We say the relative atomic mass (Ar) of oxygen is 16 All of the relative atomic masses for all the atoms are found on the periodic table

  6. Avogadro's Number Through numerous experiments Avogadro found that one mole (ie 6.02x10 23 atoms ,yes that is a very, very large number) of any element equalled the relative atomic mass (Ar) in grams of each substance This means: 6.02x10 23 oxygen (O) atoms = 16 grams 6.02x10 23 carbon (C) atoms = 12 grams 6.02x10 23 hydrogen (H) atoms = 1 gram 6.02x10 23 sulfur (S) atoms = 32 grams

  7. Atomic Mass (Ar) and Molecular Mass (Mr) We can find the molecular mass (Mr) of a compound by adding the atomic mass (Ar) numbers of the atoms in a compound. Be aware that the terms atomic mass and molecular mass are also known as molar masses

  8. Ar and Mr are very similiar • Ar = the relative atomic mass and is used for single atoms eg Ca, N, O etc in gmol-1 Eg The Ar of Ca = 40 gmol -1 Mr = the relative molecular mass of a substance and is the sum of the Ar values Eg 1 mole of H2 = 1 + 1 = 2 gmol -1 Eg What is the Mr of H2O?(use your PT to find out) (2 x H = 2 ) + (1 x O = 16) = 18gmol -1 What is the Mr of H2SO4 ? (2 x H = 2) + (1 x S = 32) + (4 x O = 64) 2 + 32 + 64 = 98 gmol-1

  9. Measuring in Chemistry A baker uses a dozen (12) of something A chemist uses a mole6.02 x 1023 of something Units for Mr are gmol-1

  10. m n Mr X Moles, Mass and Mr relationship Moles/Mass and Mr are all related by the formula Where: m = mass in grams Mr = molar mass (g mol-1) n = number of moles

  11. Moles,Mass and Mr relationship • Using this relationship we are able to work out: • Moles of a substance given the mass and Mr • Mass of a substance given the number of moles and Mr • Even the Mr of a substance given the mass and number of moles

  12. Questions • Find the number of moles of Ca in 50 grams of the metal. n = 1.25 moles n = 1.25 moles (3sf)

  13. Questions • Find the number of moles in 8 grams of oxygen gas (O2). n = 0.250 moles (3sf)

  14. Questions • Find the number of moles in 12 grams of carbon dioxide gas (CO2). n = 0.273 moles (3sf)

  15. Starter Questions – Using your periodic table Find the molar mass (Mr) for CO2 12 + (2 x 16) = 44gmol-1 (this means that one mole of CO2 has a mass of 44g) 23 gmol-1 Find the atomic mass (Ar) for Na (this means that one mole of Na has a mass of 23g) Find the atomic mass of 0.5 mole of Na metal 11.5 g Find the mass of 0.1 mole of Na metal 2.3 g Toughie? :How many Na atoms are there in 0.1 mol of Na metal? 0.1 x 6.02 x 1023 = 6.02 x 10 22 Na atoms Find the molar mass (Mr) for Na2CO3 (2 x 23) + 12 + (3 x 16) = 106 gmol-1

  16. Moles, Mass and Mr relationship m n Mr X Moles/Mass and Mr are all related by what formula? Where: m = mass in Mr = grams molar mass (gmol-1) n = number of moles

  17. We will use a grid to help in calculations. These become very helpful with complicated questions - just fill what you know then use this to work out your unknown Question- iron oxide (Fe2O3) is used in the production of iron. How many moles of iron oxide would there be in 200 grams? Copy out the grid Fe2O3 (2 x 56) + (3 x 16) = 160 gmol -1 ? = 1.25 moles 200 grams mass

  18. In industrial operations large quantities are used Question - How many moles of iron oxide would there be in 1 tonne (1000kgs) ? Copy out the grid Fe2O3 (2 x 56) + (3 x 16) = 160 gmol -1 ? = 6,250 moles mass 1000kg x 1000 = 1 x 106 grams

  19. How to work out the % of an element in a compound What % of Fe is there in Fe2O3? How much iron would you get from 100kgs of Fe2O3? 0.7 x 100 = 70kg of iron

  20. Reading Equations using Moles 1 Ca + H2SO4 CaSO4 + H2 1 1 1 The above reaction reads: 1 amount of Ca reacts with 1 amount of H2SO4 to give 1 amount of CaSO4 and 1 amount of H2 gas. (we don’t usually put in the ones – but if we did it would look like this) The amount we use is the mole Now the reaction reads: 1 mole of Ca reacts with 1 mole of H2SO4 to give 1 mole of CaSO4 and 1 mole of H2 gas.

  21. Reading Equations using Moles Ca + H2SO4 CaSO4 + H2 • Questions: • If we have 1 mole of Ca how many moles of H2SO4 are needed to fully react the Ca? • If we had 0.5 mole of Ca how many moles of H2SO4 would be required? • If 0.5 mole of Ca reacted with 0.5 mole of H2SO4 how many moles of CaSO4 and how many moles of H2 gas are formed? Ans: 1mole Ans: 0.5 mole Ans : 0.5 mole of H2 gas, 0.5 mole of CaSO4

  22. Reading Equations using Moles Ca + H2SO4 CaSO4 + H2 Questions: If 0.5 mole of Ca reacted with 0.5 mole of H2SO4 how many moles of CaSO4 and how many moles of H2 gas are formed? Ans : 0.5 mole of H2 gas, 0.5 mole of CaSO4 Given the Ar (atomic mass) values H = 1gmol-1 Ca = 40gmol-1 S = 32gmol-1 O =16gmol-1 Find the molar masses (Mr) of H2SO4 , CaSO4 and H2 Then use these and the mole (n) formula to work out the mass of H2 gas and the mass of CaSO4 formed (Hint you will need to rearrange the formula)

  23. Reading Equations using Moles Fe2O3 + CO 2Fe + 2CO2 Questions: If 200kg of Fe2O3 is reacted in the above equation what mass of Fe is produced?

  24. Work out how many moles of HCl is used to completely react 60grams of Mg metal in the following reaction? Draw grid Mg + 2 HClMgCl2 + H2 Mg HCl Mr 24 gmol -1 36.5 gmol -1 mass 60 grams ? = 2.5 moles 1 2 Therefore 2 x moles of Mg = moles of HCl Therefore 2 x 2.5 moles = 5 moles of HCl

  25. What was the mass of HCl used? m = 5 x Mr = 5 mol x 36.5 gmol-1 = 182.5 g = 182 g (3sf)

  26. Sherbet Sherbet balanced reaction NaHCO3 + C4 H6O6 + C6 H7 O7 Na tartarate+H2O + CO2 Tartaric acid Citric acid Sherbet Given 1 gram of NaHCO3 work out how much of each ingredient NaHCO3 Tartaric acid Citric acid Mr 84 gmol -1 150 gmol -1 191 gmol -1 = 0.0119 mol 0.0119 mol 0.0119 mol Molar ratio 1 1 1 mass = n x Mr 0.0119 x 150 = 1.785g 0.0119 x 191 = 2.273g 1g

  27. Work out how many moles of carbon dioxide is created when 100grams of oxygen is reacted in the following reaction? Draw grid CH4 + 2O2 CO2 + 2H2O O2 CO2 Mr 32 gmol -1 44 gmol -1 mass 100 grams ? = 3.125 moles 2 1 Therefore ½ x moles of O2 = moles of CO2 Therefore ½ x 3.125 moles = 1.5625 moles of CO2 Would you give this answer? 1.56 moles of CO2 (3 sig fig)

  28. Empirical Formula This is the simplest ratio of atoms in a compound For example the empirical formula for ethane C2H6 is CH3 the empirical formula for butane C4H10 is C2H5 the empirical formula for methane CH4 is CH4 the empirical formula for hydrogen peroxide H2O2 is HO

  29. Molecular Formula The molecular formula of a compound is the formula giving the actual number of atoms in a compound We will be asked to find the molecular formula given % mass of a compound For example find the molecular formula for a compound which was found to contain 80% carbon and 20% hydrogen

  30. Example: finding the empirical formula for a compound that was found to contain 80% carbon and 20% hydrogen C H Step one : simply convert each % to grams 20g 80g Step two : divide each element by its Ar value Step three : What is the smaller number? Now divide each of these by this smaller number Step four : Now this is the empirical formula for the compound just write out the formula 1 x C 3 x H CH3

  31. Example : Now find the molecular formula for the compound that was found to contain 80% carbon and 20% hydrogen. Given that the compound was found to have a molar mass of 30 gmol-1 Step five: use the empirical formula you just found ie CH3 and follow the method below Step six : divide your given molar mass by the molar mass of CH3 Step seven: now multiply your empirical formula by 2 to get the molecular formula - 2 x CH3 = C2H6

  32. A compound was found to have a % composition of 2.0% Hydrogen, 32.7% Sulfur, 65.3% Oxygen find the empirical formula

  33. A compound was found to have a % composition of 2.0% Hydrogen, 32.7% Sulfur, 65.3% Oxygen find the empirical formula H S O Step one : simply convert each % to grams 2g 65.3g 32.7g Step two : divide each element by its Ar value Step three : What is the smaller number? Now divide each of these by this smaller number Step four : 2 1 4 Now round and just write out the empirical formula H2 S O4

  34. A sample with composition 40% C, 6.7% H and 53.3% O had a molar mass 60 g mol-1 find the Empirical formula and also the molecular formula

  35. A sample with composition 40% C, 6.7% H and 53.3% O had a molar mass 60 g mol-1 find the Empirical formula and also the molecular formula C H O Step one : simply convert each % to grams 40g 6.7g 53.3g Step two : divide each element by its Ar value Step three : What is the smaller number? Now divide each of these by this smaller number Step four : 1 2 1 Now round and just write out the empirical formula C H2 O

  36. Step five: use the empirical formula you just found ie CH2O and follow the method below Step six : divide your given molar mass (60 gmol-1) by the molar mass of CH2O (30 gmol-1) Step seven: now multiply your empirical formula by 2 to get the molecular formula - 2 x CH2O = C2H4O2

  37. Water of Crystallisation • Some hydrated salts have H2O molecules in their structure eg CuSO4 5H2O (hydrated means with water) • This means that each particle of CuSO4 is surrounded by 5H2O molecules • These H2O molecules are called the water of crystallisation • The water in the structure can be removed by heating to form the anhydrous (without water) salt ie CuSO4

  38. Finding the Empirical Formula for a Hydrated salt by Gravimetric Analysis • To find the mass of H2O in a hydrated (with water) salt • Firstly find the mass of the hydrated (with water) salt • Heat strongly for 10 minutes cool then reweigh • Then heat again and reweigh to check the salt is at constant mass

  39. Then subtract the mass of the anhydrous(dry) salt from the hydrated(wet) salt • This will give you the water mass • Then use your anhydrous salt mass and water mass to workout the empirical formula of the hydrated salt using grid

  40. Find the empirical formula for a hydrated salt Na2CO3XH2O Given: Mass of hydrated Na2CO3 salt = 2.58g Mass of salt after heating to constant weight = 0.95g Therefore: 2.58g – 0.95g = 1.63g (mass of water)

  41. Calculation for finding the empirical formula for a hydrated salt Na2CO3XH2O Na2CO3 H2O Step one : 106gmol-1 18gmol-1 Find the Mr ‘s Step two : Given masses find moles of each Step three : What is the smaller number? Now divide each of these by this smaller number Step four : 1 x Na2CO3 10H2O Make small adjustments to give whole number To give empirical formula Na2CO310H2O

  42. Concentration Particles per volume Can be in grams per litre but chemists usually express concentration in moles per litre This are related by the formula: Where: n = moles V = volume in litres c = concentration in molL-1

  43. Write the formula for: 1. Moles using mass and Molar mass 2. Concentration using Volume and Moles

  44. Concentration Example 15 grams of NaCl is dissolved in 1 litre of water, what is the concentration of the solution in moles per litre? (M NaCl = 58.5 gmol-1) Find moles of NaCl Divide moles by volume in litres

  45. Concentration Example 20 grams of NaOH is dissolved in 500mls of water, what is the concentration of the solution in moles per litre? (M NaOH = 40gmol-1) Find moles of NaOH Divide moles by volume in litres

  46. If solution A (25mls of 0.23 molL-1 AgNO3) is mixed with solution B (82 mls of O.15molL-1 of AgNO3 ) What is the final concentration of the solution when they are mixed Find total moles by adding moles of A to moles of B Divide total moles by total volume in litres

  47. How many grams of water vapour H2O(g) will be produced in the complete combustion of 500g methane, CH4(g)? CH4 + 2O2 CO2 + 2H2O CH4 H2O Mr 16 gmol -1 18 gmol -1 62.5x18 =1,125g = 1.13kg H2O mass 500 grams ? = 31.25 moles 1 2 Therefore 2 x moles of CH4 = moles of H2O Therefore 2 x 31.25 moles = 62.5 moles of H2O

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