Download
slide1 n.
Skip this Video
Loading SlideShow in 5 Seconds..
Relative atomic mass The mole Chemical equations PowerPoint Presentation
Download Presentation
Relative atomic mass The mole Chemical equations

Relative atomic mass The mole Chemical equations

148 Views Download Presentation
Download Presentation

Relative atomic mass The mole Chemical equations

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Relative atomic mass • The mole • Chemical equations

  2. Relative Quantities

  3. Turkish coins

  4. Has a RELATIVE value of 10 compared to

  5. Atoms & Elements

  6. The atoms of one element are unique All atoms of a given element are the same size mass

  7. As atoms of different elements have different masses We can compared their RELATIVE masses in exactly the same way we compare the relative value of coins

  8. In the same way that we use the lowest value coin (1 kr) to compare the with the others we use the smallest (and lightest) atom as the mass we compare others to… And the element with the smallest atom is HYDROGEN We say Hydrogen has a relative mass of 1

  9. The mass of an atom is due to the particles that make up that atom, the …… PROTONS NEUTRONS and ELECTRONS Of these the mass of an electron is insignificant, and the mass of a proton and a neutron is (very nearly) the same

  10. Previously we have used the mass of protons and neutrons when we describe the structure of an atom…… The mass number (or atomic mass) 7 Li 3 The proton number

  11. We call these numbers the RELATIVE ATOMIC MASS and use the shorthand symbol MA If you look at a copy of the Periodic Table you will see that the atomic mass of all the elements is given …… SO YOU DO NOT HAVE TO LEARN THESE NUMBERS

  12. Relating Atomic mass to equations

  13. The MA for HYDROGEN is 1 The MA for HELIUM is 4 So 1 atom of HELIUM is 4 times heavier than 1 atom of HYDROGEN

  14. 6 atoms of HELIUM are 4 times heavier than 6 atoms of HYDROGEN As long as we always compare the SAMENUMBER of atoms this will always be true

  15. Oxygen was assigned a relative mass of approximately 16, and was used as the reference element for assigning atomic weights of all the elements, but physics’ found the mass of oxygen as 16,0044. • In realty oxygen has three isotopes (16, 17, 18). There was a conflict, in order to remove that, in 1961 International Union of Pure and Applied Chemistry (IUPAC) accept carbon-12 as a reference.

  16. So, how just how heavy is one atom? The answer is incredibly light, One HYDROGEN atom weighs 0.0000000000000000000000017 grams (or using scientific notation 1.7 x 10-24 grams) These numbers are too small to be easy for everyday use so the mass of any atom can be expressed as the relative atomic weight in units of amu (amu = atomic mass unit, which is 1/12 the mass of one carbon - 12 atom.).

  17. Remember that as long as we always compare the SAMENUMBER of atoms the relative mass will always be the same In 1 gram of HYDROGEN there are 60200000000000000000000000 (or 6,02x 1023) atoms and we call this number Avogadro’s Number. 1g = 6,022 x 1023 amu, 1amu= 1,661 x 10-24g Which means that 4 grams of HELIUM also contains 6,02x 1023 atoms And so on ….

  18. 1 mole of atoms of an element always weighs the MA of that element in grams When the relative atomic weight of an element is given in units of grams, it is known as the gram atomic weight. The amount of a substance that contains Avogadro's number of elementary units is called a mole (the name comes from the latin word for heap)

  19. One atom’s mass of some elements measured by mass spectrometer.

  20. If you divide molar mass by atomic mass you will get some number which is very close to Avogadro’s number.

  21. Mass in chemical reactions

  22. The relative atomic mass of Hydrogen = 1 The relative atomic mass of Carbon = 12 The relative atomic mass of Oxygen = 16 Using Relative Atomic Mass

  23. In exactly the same way that we always compare the SAMENUMBER of atoms when dealing with relative atomic mass the same is true for compounds. 1 mole of Hydrogen atoms weighs 1g 2 moles of Hydrogen atoms weighs 2g 1 mole of Hydrogen molecules contains 2 moles of atoms, it also weighs 2g

  24. One MOLECULE of Hydrogen contains 2 atoms or H2 It has a relative mass of 2 (2x 1 for each H atom) The relative mass of a compound is called either the RELATIVE FORMULAMASS or often the RELATIVE MOLECULAR MASS, in both cases the shorthand symbol is Mw

  25. Example One MOLECULE of Water contains 2 atoms of Hydrogen and one atom of Oxygen or H2O The Mw is made up from 2x 1 (from the hydrogen atoms) and 1x 16 (from the Oxygen atom) = 18

  26. Example One MOLECULE of Carbon Dioxide contains 2 atoms of Oxygen and one atom of Carbon or CO2 The Mw = 12 +16 + 16 = 44

  27. 1 mole of a compound always weighs the Mw of that compound in grams The Mw of a compound can be calculated by adding up the MAvalues for each atom in the formula

  28. Percentage composition You are often asked to calculate the percentage mass of a particular element in a compound e.g. What percentage does Oxygen make in water (H2O) You will either be given a data book or the individual MA values as part of the question

  29. Method 1. Make sure you know the correct formula, e.g. H2O 2. Find the MA value for all the elements in the compound (e.g. H = 1, O = 16) 3. Work out the Mw (e.g. 1 + 1 + 16 = 18) 4. Work out the total MA for the element in the compound (e.g. in water H = 2, O = 16) 5. Calculate the percentage of that element in the total (e.g. for oxygen in water percentage = 16 x100 = 88.9% ) 18

  30. Problems What is the Mw of the following substances NaCl MgCl2 H2SO4 CuSO4 Ba(OH)2

  31. NaCl 23 + 35.5 = 58.5 MgCl2 24 + 35.5 + 35.5 = 95 H2SO4 1+ 1+ 32 + (4x16) = 98 CuSO4 63.5 + 32 + (4x16) = 159.5 Ba(OH)2 137 + (16 + 1) + (16 + 1) = 171 Problems What is the Mw of the following substances

  32. Problems What is the percentage of each element in the the following substances NaCl MgCl2 H2SO4 CuSO4 Ba(OH)2

  33. Problems What is the percentage of each element in the the following substances NaCl Mw = 58.5 % Na = 23/58.5 x 100 =39.3 % % Cl = 35.5/58.5 x 100 =60.7 %

  34. Problems What is the percentage of each element in the the following substances MgCl2 Mw = 95 % Mg = 24/95 x 100 =25.3 % % Cl = (35.5 + 35.5)/95 x 100 =74.7 %

  35. Problems What is the percentage of each element in the the following substances H2SO4 Mw = 98 % H = (1 + 1)/98 x 100 =2.0 % %S = 32/98 x 100 = 32.7% % O = (16 + 16 + 16 + 16)/98 x 100 =65.3 %

  36. Problems What is the percentage of each element in the the following substances CuSO4 Mw = 159.5 % Cu = 63.5/159.5 x 100 =39.8 % %S = 32/159.5 x 100 = 20.1% % O = (16 + 16 + 16 + 16)/159.5 x 100 =40.1 %

  37. Problems What is the percentage of each element in the the following substances Ba(OH)2 Mw = 171 % Ba = 137/171 x 100 =80.1 % %O = (16 + 16)/171 x 100 = 18.7% % H = (1+ 1)/171 x 100 =1.2 %

  38. Mass of reactants = Mass of products In any chemical reaction atoms are not made or destroyed But ……… this only works for the total mass, to work with individual reactants we need to use the idea of moles

  39. BALANCED EQUATIONS To work out the masses of individual reactants/products it is vital that you have a balanced equation, this is usually supplied in an exam question.

  40. Understanding equations Power stations remove the sulphur dioxide produced when fossil fuels are burnt by treating the waste gases with limestone. Limestone + Sulphur Dioxide + Oxygen  Calcium Sulphate + Carbon Dioxide 2CaCO3(s) + 2SO2(g) + O2(g)  2CaSO4(s) + 2CO2(g)

  41. Understanding equations Power stations remove the sulphur dioxide produced when fossil fuels are burnt by treating the waste gases with limestone. Limestone + Sulphur Dioxide + Oxygen  Calcium Sulphate + Carbon Dioxide 2CaCO3(s) + 2SO2(g) + O2(g)  2CaSO4(s) + 2CO2(g) State symbols, s = solid; g = gas; l = liquid; aq = aqueous (dissolved in water)

  42. Understanding equations Power stations remove the sulphur dioxide produced when fossil fuels are burnt by treating the waste gases with limestone. Limestone + Sulphur Dioxide + Oxygen  Calcium Sulphate + Carbon Dioxide 2CaCO3(s) + 2SO2(g) + O2(g)  2CaSO4(s) + 2CO2(g) A subscript, this tells you how many of the atom (or sometimes group of atoms) there are in one formula amount of the substance

  43. Understanding equations Power stations remove the sulphur dioxide produced when fossil fuels are burnt by treating the waste gases with limestone. Limestone + Sulphur Dioxide + Oxygen  Calcium Sulphate + Carbon Dioxide 2CaCO3(s) + 2SO2(g) + O2(g)  2CaSO4(s) + 2CO2(g) This number tells you how many of the formula that follows is needed to allow the reaction to happen, or is formed in the reaction

  44. Understanding equations Power stations remove the sulphur dioxide produced when fossil fuels are burnt by treating the waste gases with limestone. Limestone + Sulphur Dioxide + Oxygen  Calcium Sulphate + Carbon Dioxide 2CaCO3(s) + 2SO2(g) + O2(g)  2CaSO4(s) + 2CO2(g) If you know the relative numbers of particles that react you also know the number of moles that are involved In this example 2 moles of limestone, 2 moles of sulphur dioxide and one mole of oxygen molecules combine to form 2 moles of calcium sulphate and 2 moles of carbon dioxide

  45. Mass of reactants = Mass of products Fe(s) + S(s)  FeS(s) 56g of Iron combines with 32 g of Sulphur, what mass of iron sulphide is formed? This is a simple example , because there is only one product Mass of reactants = 56 + 32 = 88g

  46. 2Fe2O3 + 3C  3CO2 + 4Fe What is the maximum mass of iron that can be obtained from 700g of iron oxide? Each mole of iron oxide forms 2 moles of iron No. of moles of iron oxide = mass / Mw iron oxide Mw = 56 + 56 +16 +16 +16 = 160 No. moles = 700/160 = 4.375 No. moles iron = 2 x 4.375 = 8.75 Mass of iron = moles x MA Iron = 8.75 x 56 = 490g

  47. 1. CaO + 3C  CaC2 + CO; what is the maximum mass of CaC2 that can be formed from 40g of CaO? 2. H2SO4 + 2NH3 (NH4)2SO4; What mass of each of the reactants are needed to form 132g of product 3. 2Al + Cr2O3 2Cr + Al2O3; What masses of reactants are needed to form 180g of Chromium (Cr) 4. 2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(g); If 5.0g of reactant lost 0.31g, what mass of water was formed?