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Chemical Calculations

Chemical Calculations. Relative Atomic Mass. The relative atomic mass of an element is the average mass of one atom of the element when compared with 1/12th of the mass of an atom of carbon-12. A r = average mass of one atom of an element mass of 1/12 th of an atom of carbon-12

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Chemical Calculations

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  1. Chemical Calculations

  2. Relative Atomic Mass • The relative atomic mass of an element is the average mass of one atom of the element when compared with 1/12th of the mass of an atom of carbon-12. • Ar = average mass of one atom of an element mass of 1/12th of an atom of carbon-12 • Ar = symbol for relative atomic mass • Ar has no units

  3. Relative Molecular Mass • The relative molecular mass or Mr of a molecule is the average mass of one molecule of a substance when compared with 1/12th of the mass of one atom of carbon-12. • Mr is derived for any molecule by determining the sum total of the relative atomic masses of all the individual atoms present in the molecule. • Mr has no units.

  4. Questions • Calculate the Mr values for the following substances : • Cu(NO3)2 • CuSO4. 5H2O

  5. Mr of Cu(NO3)2 = 64 + 2(14 + (16 × 3) ) = 188 • Mr of CuSO4. 5H2O = 64 + 32 + (16 × 4) + 5((2 × 1) + 16) = 250

  6. Calculating Percentage Composition of A Compound • The percentage by mass of an element/ ion can be determined by calculating the sum total of the relative masses of the element / ion present in the compound, dividing this value by the Mr of the compound and expressing the value as a percentage.

  7. Questions Calculate the percentages of the elements present in sodium carbonate. Mr of Na2CO3 = (23 × 2) + 12 + (16 × 3) = 106 % of sodium in sodium carbonate = 46 / 106 × 100 = 43.4 %

  8. % of carbon in sodium carbonate = 12 / 106 × 100 = 11.3 % % of oxygen in sodium carbonate = 48 / 106 × 100 = 45.3 % OR % of oxygen in sodium carbonate = 100 – 43.4 – 11.3 = 45.3 %

  9. Calculating the Mass of an Element in a Compound Calculate the mass of nitrogen present in 200 g of ammonium nitrate. Mr of NH4NO3 = 14 + 4 + 14 + (16 × 3) = 80 Mass of nitrogen in 200 g of ammonium nitrate = 28 / 80 × 200 = 70.0 g

  10. Question A sample of copper (II) sulphate has the formula CuSO4. xH2O. The composition of water in the compound is 36.0 %. What is the value of x ? Mr of CuSO4 = 64 + 32 + (16  4) = 160 % of CuSO4 in CuSO4.xH2O = 100 – 36.0 = 64.0 %

  11. Relative mass of water in the salt = 36.0  160 = 90 64.0  x = 90 = 90 = 5 Mr of H2O 18  Formula of salt = CuSO4. 5H2O

  12. The Mole • One mole of any matter (atoms, ions or molecules) contains 6 × 10 23 particles. There are no units for the mole. Symbol = mol • The mass of one mole of any substance is called the molar mass. The value of molar mass is equal to the Ar or Mr of the substance (either element or compound) in grams.

  13. The volume of one mole of any gas at room temperature and pressure is 24 dm3 or 24 000 cm3. This is called the molar volume at r.t.p. d) Avogadro’s Law states that equal volumes of gases contain the same number of molecules under the same conditions of temperature and pressure. e) The volume of a gas is directly proportional to the number of moles of the gas.

  14. Molar mass / mass of one mole of 23 Na = 23 g Mass of one mole of CO2 = 12 + (16 × 2) = 44 g Mass of one mole of H2 = 2 g Mass of one mole of O2 = 32 g

  15. In 23 g of Na, there are 6 × 10 23 atoms of Na. In 44 g of CO2, there are 6 × 10 23 molecules of CO2. In 2 g of H2, there are 6 × 10 23 molecules of H2. In 32 g of O2, there are 6 × 10 23 molecules of O2.

  16. 44 g of CO2 occupy 24 dm3 at r.t.p. 2 g of H2 occupy 24 dm3 at r.t.p. 32 g of O2 occupy 24 dm3 at r.t.p.  10 cm3 of O2 have the same no. of molecules as 10 cm3 of CO2. 10 cm3 of O2 and 10 cm3 of CO2 consist of the same number of moles.

  17. Calculating number of moles of a substance • Amount of an element in mol = Mass of sample (g) Ar b) Amount of a compound in mol = Mass of sample (g) Mr

  18. c) Amount of an element / compound in mol = Number of particles 6 × 10 23 • Amount of a gas in mol = Volume of gas in dm3 (or cm3) 24 (or 24 000)

  19. Mass of sample (g) = Amount in mol  Ar (or Mr) • Mr of molecule = Mass of sample in grams Amount in mol

  20. Questions • Determine the mass of 3 moles of ammonia. No. of moles of ammonia = mass of sample (g) Mr of NH3  Mass of ammonia = 3  17 = 51.0 g

  21. 0.25 moles of ethene has a mass of 7 g. What is the relative molecular mass of ethene ? Mr of ethene = Mass of sample No. of moles = 7 / 0.25 = 28

  22. Determine the number of moles present in a 100 g sample of calcium oxide. Mr of CaO = 40 + 16 = 56 Amount of CaO in mol = Mass of sample Mr = 100 / 56 = 1.79 mol

  23. Calculate the number of moles of nitrogen gas present in 1500 cm3 sample of the gas. Amount of nitrogen gas in mol = 1500 / 24 000 = 0.0625 mol.

  24. Determine the volume occupied by 0.2 moles of sulphur dioxide gas at r.t.p. Volume of SO2 gas at r.t.p. = Amount of SO2 in mol  Molar Volume = 0.2  24 = 4.80 dm3

  25. What is the number of moles of hydrogen atoms in 54 g of water ? Amount of water in sample in mol = 54 / 18 = 3 mol Ratio of moles of H2O : H : O = 1 : 2 : 1  Amount of hydrogen atoms in mol = 3  2 = 6.00 mol.

  26. Find the mass of 60 dm3 of oxygen gas at r.t.p. Amount of oxygen gas in mol = Volume (dm3) = 60 = 2.5 mol 24 24 Mass of oxygen sample = 2.5  (16  2) = 80.0 g

  27. How many atoms are there in 36 g of carbon ? Amount of carbon atoms in mol = Mass in g = 36 = 3 mol Ar 12 No. of oxygen atoms in sample = 3  6  10 23 = 1.8  10 24 atoms

  28. What is the mass in grams of 5.4  10 24 molecules of chlorine gas ? Amount of chlorine gas in mol = No. of particles = 5.4  10 24 Avogadro’s No. 6 10 23 = 9 mol Mass of chlorine gas = No. of moles  Mr = 9  (35.5  2) = 639 g

  29. What is the volume at r.t.p of 66 g of carbon dioxide ? Amount of carbon dioxide in mol = Mass in g = 66 = 1.5 mol Mr 44 Volume of CO2 gas at r.t.p. = Amount of CO2 in mol  Molar Volume = 1.5  24 = 36.0 dm3

  30. Empirical Formulae • It is the simplest formula of the compound. • It shows the ratio of atoms of each element present in each molecule. • It does not indicate the number of atoms present per molecule. e.g. Ethene has a molecular formula of C2H4. Empirical formula = CH2

  31. Calculating Empirical Formulae • To calculate the empirical (simple) formula of a substance, you must know the masses or percentages of the individual elements / ions present in the substance. • Two steps are involved : • The mass / % of the element / ion divided by the Ar (or Mr). (Amount in mol is determined) • Each value obtained is divided by the smallest of the values obtained. (Ratio of amounts in mol is determined)

  32. A hydrocarbon P of mass 28 g consists of 24 g of carbon. Determine the empirical formula of P. The empirical formula is CH2.

  33. Find the empirical formula of a compound with composition 48.6 % carbon, 43.2 % oxygen and 8.1 % hydrogen by mass. Empirical formula = C3H6O2

  34. Molecular Formula • It is the exact formula of the compound. • It shows the exact number of atoms present per molecule.

  35. Determining Molecular Formulae To determine the molecular formula : • the empirical formula must be determined first • the relative molecular mass / molar mass must be known Molecular formula = n  Empirical formula  n = relative molecular mass (Mr) relative mass of empirical formula

  36. Question Butene has an empirical formula of CH2. The relative molecular mass of butene is 56. Determine the molecular formula of butene. Relative mass of CH2 = 12 + 2 = 14  n = relative molecular mass (Mr) relative mass of empirical formula = 56 / 14 = 4  Molecular formula = 4  CH2 = C4H8

  37. Calculations from Equations Equations tell us : a) what the reactants and products of a reaction are. • the ratio of amount in mol of the reactants to the products. • the ratio of volumes of reactants to products in reactions involving gases.

  38. Questions • Copper (II) oxide reacts with ammonia gas as follows : 3CuO + 2NH3 3Cu + 3H2O + N2 If 4.0 g of CuO was reacted completely, calculate : i) the amount of CuO in mol that reacted

  39. Amount of CuO in mol = Mass in g Mr = 4.0 = 0.05 mol 64 + 16 • The amount of NH3 in mol that reacted with the CuO

  40. Ratio of mol of CuO : NH3 = 3 : 2 Amount of NH3 in mol = 0.05  2 3 = 0.03333 mol • The volume of NH3 that reacted = Amount in mol  Molar volume = 0.03333  24 = 0.800 dm3 (3sf)

  41. the volume of nitrogen gas produced. Ratio of mol of CuO : N2 = 3 : 1 Amount of N2 in mol = 0.05  1 3 = 0.01666 moles Volume of N2 = No. of moles  Molar Vol. = 0.01666  24 = 0.400 dm3

  42. the mass of copper produced Ratio of mol of CuO : Cu = 1 : 1 Amount of Cu in mol = 0.05 mol Mass of Cu = Amount in mol  Ar = 0.05  64 = 3.20 g

  43. Alternative method : 3CuO + 2NH3 3Cu + 3H2O + N2 3(64 + 16)g CuO 3(64) g Cu 4.0 g CuO ? Mass of copper formed = 4.0  3(64) 3(80) = 3.20 g

  44. Nitrogen combines with hydrogen to form ammonia as shown below : N2 + 3H2 2NH3 30 cm3 of nitrogen gas is reacted with 60 cm3 of hydrogen gas at r.t.p. • What is the volume of ammonia gas formed ? • What is the total volume of gaseous products formed ?

  45. Ratio of mol of N2 : H2 = 1 : 3 Therefore, 20 cm3 of nitrogen reacts with 60 cm3 of hydrogen gas. The hydrogen gas is the limiting reagent & the nitrogen gas is in excess. Ratio of mol of H2 : NH3 = 3 : 2 Volume of NH3 gas formed = 2  60 3 = 40.0 cm3

  46. Volume of excess nitrogen = 30 – 20 = 10.0 cm3 Total volume of gaseous products = 10 + 40 = 50.0 cm3

  47. Calculating Percentage Yield in a Reaction • During experiments, the amount of chemicals produced is always less than the maximum amount you would expect. This is expressed as the percentage yield. • Percentage yield of substance = Experimental mass of substance  100 % Maximum possible mass of product

  48. Questions • Zinc reacts with sulphur according to the equation : Zn + S ZnS 6.5 g of zinc was reacted with sulphur to make zinc sulphide. 9.0 g of zinc sulphide was obtained. Calculate the % yield.

  49. Amount of Zn in mol = Mass = 6.5 Ar 65 = 0.1 mol Amount of ZnS in mol = 0.1 mol Expected mass of ZnS = Amount in mol  Mr = 0.1  (65 + 32) = 9.70 g

  50. % yield of ZnS = 9.0  100 % = 92.8 % 9.7

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