UNIT 1, Atoms, Bonds and Groups

UNIT 1, Atoms, Bonds and Groups

UNIT 1, Atoms, Bonds and Groups

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Presentation Transcript

1. Module 1 – Atoms and Reactions

2. The Changing Atom L.O. Understand how the model of the atom has changed over the years. Understand that scientific knowledge is always evolving. Starter – So what is an atom?

3. Atomic Models Prepare presentations on, How Ideas on the Atom Have Evolved You might want to look at the Greek model, John Dalton’s ideas, Joseph John Thomson’s, Ernest Rutherford’s, and Neils Bohr’s theories on the atom as well.

4. Atomic Structure L.O. Know what protons, neutrons and electrons are. Understand what isotopes are. Understand the atomic structures of ions. Starter – What are protons, neutrons and electrons?

5. Particles Nucleus is almost all mass of atom, but tiny volume. If the nucleus was the size of a grape, the electrons would be about 1.5 km away.

6. Isotopes Isotopes are atoms of the same element with different numbers of neutrons. ZAX (X – Element) Z – Atomic (proton) number, is the number of protons in the nucleus of an atom. A – Mass (nucleon) number, is the number of particles (protons and neutrons) in the nucleus.

7. Atomic Structures of Ions An ion is a positively or negatively charged atom or group of atoms. +ve ions have lost electrons. -ve ions have gained electrons.

8. Atomic Masses L.O. Understand Relative Isotopic Mass and Relative Atomic Mass. Calculate Relative Molecular Mass and Relative Formula Mass. Starter – Relative mass is measured against the carbon-12 isotope. Why 12C?

9. Relative Masses Relative isotopic mass is the mass of an atom of an isotope compared with one-twelfth of an atom of carbon-12. Relative atomic mass, Ar, is the weighted mean mass of an atom of an element compared with one-twelfth of the mass of an atom of carbon-12.

10. Relative Masses Relative molecular mass, Mr, is the weighted mean mass of a molecule compared with one-twelfth of an atom of carbon-12. Relative formula mass is the weighted mean mass of a formula unit compared with one-twelfth of an atom of carbon-12.

11. The Mole L.O. Explain the terms amount of substance, mole and Avogadro constant. Define and use molar mass. Carry out calculations using amount of substance in moles. Starter – What is pi?

12. Amount of Substance Amount of substance is the quantity whose unit is the mole. It is used to count atoms. A Moleis amount of any substance containing as many particles as there are carbon atoms in exactly 12 g of the carbon-12 isotope. Avogadro constant, NA, (6.02 x 1023 mol-1) is the number of atoms per mole of the carbon-12 isotope

13. Moles and Mass You can find the mass of one mole of atoms of any element, it is the relative atomic mass in grams. 1 mol C atoms – 12.0 g, 1 mol H atoms – 1.0g 2 mol C atoms – 24 g 0.5 mol C atoms – 6 g

14. Molar Mass Molar Mass, Mr, is the mass per mole of a substance. Units: g mol-1.

15. Converting mass  moles n = so m = n.Mr n = amount of substance, in mol m = mass, in g Mr = molar mass, in g mol-1. mMr

16. Questions Calculate the mass in g of, a) 3.00 mol SiO2. b) 0.500 mol Fe2O3. c) 0.250 mol Na2CO3. Calculate the amount in mol of, a) 8.00 g BeF2. b) 23.7 g KMnO4. c) 10.269 g Al2(SO4)3.

17. Questions Calculate the mass in g of, a) 3.00 mol SiO2. 180.3 g b) 0.500 mol Fe2O3. 79.8 g c) 0.250 mol Na2CO3. 26.5 g Calculate the amount in mol of, a) 8.00 g BeF2. 0.170 mol b) 23.7 g KMnO4. 0.150 mol c) 10.269 g Al2(SO4)3. 0.0300 mol

18. Types of Formula L.O. Explain the terms empirical and molecular formula. Calculate empirical and molecular formula. Starter – What does a formula tell you?

19. Empirical Formula Empirical formula: the simplest whole-number ratio of atoms of each element present in a compound. It is always used for giant structure compounds, Ionic compounds such as NaCl, Covalent compounds such as SiO2.

20. Calculating Empirical Formula if you are given the mass of each element in a compound you can calculate the empirical formula. You divide the mass of each element by its molar mass to get the number of moles of each element. Then by dividing by the lowest value you can get the molar ratio.

21. Molecular Formula A molecule is a small group of atoms held together by covalent bonds. Molecular formula: the actual number of atoms of each element in a molecule. It is usually adequate but in organic chemistry the molecular formula gives no information as to the order the molecule is bonded.

22. Determining Molecular Formula Find the empirical formula. Find the empirical formula mass. Divide the molecular mass of the sample by the empirical formula mass to find the number of empirical units. Multiply the number of units by the empirical formula to get molecular formula.

23. Moles and Gas Volumes L.O. Calculate the amount of substance in moles, using gas volumes. Starter –

24. Avogadro Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Molar volume is the volume per mole of gas. Units: dm3 mol-1. Molar volume = 24.0 dm3 mol-1 at RTP RTP: room temp & pressure

25. Calculating Moles from Gas Volumes At room temperature and pressure, If volume is in dm3, n = V / 24.0 If volume is in cm3, n = V / 24000

26. Yes, More Questions… What amount in mol, of gas molecules are in the following at RTP? 36 dm3 1080 dm3 4.0 dm3. What is the volume of the following? 6 mol SO2(g), 0.25 mol O2(g), 20.7 g NO2(g)

27. Yes, More Questions… What amount in mol, of gas molecules are in the following at RTP? 36 dm31.5 mol 1080 dm3 45 mol 4.0 dm3 0.167 mol What is the volume of the following? 6 mol SO2(g), 144 dm3 0.25 mol O2(g), 6 dm3 20.7 g NO2(g) 10.8 dm3

28. You Guessed It……More Questions! What is the mass of the following at RTP: 0.6 dm3 N2, 1920 cm3 C3H8(g), 84 cm3 N2O(g) What is the volume of the following at RTP: 1.282 g SO2, 1.485 g HCN, 1.26 g C3H6. Will there be more????

29. You Guessed It….. More Questions! What is the mass of the following at RTP: 0.6 dm3 N2, 1920 cm3 C3H8(g), 84 cm3 N2O(g) 0.7 g 3.52 g 0.154 g What is the volume of the following at RTP: 1.282 g SO2, 1.485 g HCN, 1.26 g C3H6. 0.48 dm3 1.32 dm3 0.72 dm3.

30. Moles and Solutions L.O. Calculate the amount of substance in mol, using solution volume and concentration. Describe a solutions concentration using the terms concentrated and dilute. Starter – What is concentration?

31. Concentration The concentration of a solution is the amount of solute (mol) dissolved per 1 dm3 (1000 cm3) of solution. Units: mol dm-3 If you know the concentration in mol dm-3 you can find the amount in moles in any volume of a solution.

32. Concentrated or Dilute? The terms concentrated and dilute refer to the amount in moles of dissolved solute in a solution Concentrated – a large amount of solute. Dilute – a small amount of solute per dm3. [you may see ‘M’ on some bottles. This is an old nomenclature 1M = 1 mol dm-3]

33. Standard Solutions Standard solution: a solution of known concentration. Standard solutions are normally used in titrations to determine unknown information about another substance.

34. Calculating Moles mMr From mass, n = n = No. of moles (amount) m = mass of species Mr = relative Molecular mass

35. Calculating Moles From solutions, n = c x V n = No. of moles (mol) c = concentration of solution (mol dm-3) V = volume of solution, in dm3. To convert cm3 to dm3 divide by 1000.

36. Questions Find the amount in moles of solute dissolved in the solutions. 4 dm3 of a 2 mol dm-3 solution. 25 cm3 of a 0.15 mol dm-3 solution. 24.35 cm3 of a 0.125 mol dm-3 solution.

37. Questions Find the amount in moles of solute dissolved in the solutions. 4 dm3 of a 2 mol dm-3 solution. 8 mol 25 cm3 of a 0.15 mol dm-3. 3.75x10-3 mol 24.35 cm3 of a 0.125 mol dm-3. 3.04x10-3 mol

38. More Questions Find the concentration in mol dm-3 for the following, 6 moles dissolved in 2 dm3 of solution. 0.500 moles dissolved in 250 cm3 of solution. 8.75x10-3 moles dissolved in 50.0 cm3 of solution.

39. More Questions Find the concentration in mol dm-3 for the following, 6 moles dissolved in 2 dm3 of solution. 3 mol dm-3 0.500 moles dissolved in 250 cm3 of solution. 2.00 mol dm-3 8.75x10-3 moles dissolved in 50.0 cm3 of solution. 0.175 mol dm-3

40. Even More Questions Find the mass concentration, in g dm-3 for the following solutions, 0.042 moles of HNO3 dissolved in 250 cm3. 0.500 moles of HCl dissolved in 4 dm3. 3.56x10-3 moles of H2SO4 dissolved in 25 cm3.

41. Even More Questions Find the mass concentration, in g dm-3 for the following solutions, 0.042 moles of HNO3 dissolved in 250 cm3. 10.584 g dm-3 0.500 moles of HCl dissolved in 4 dm3. 4.5625 g dm-3 3.56x10-3 moles of H2SO4 dissolved in 25 cm3. 13.969 g dm-3

42. Chemical Equations L.O. Construct balanced chemical equations for known reactions and for others given the reaction. Starter – What is a Chemical Equation?

43. Reactants and Products A chemical reaction does not create or destroy atoms or change them into different atoms. It simply rearranges the atoms into different combinations. Reactants  Products

44. Equations You can show molecules by their molecular formula. Giant structures are shown using their empirical formula. We can show the state of the species using state symbols: (s) solid, (l) liquid, (g) gas, and (aq) aqueous.

45. Balancing Equations Equations will always balance as we do not create or destroy atoms. We can balance the equation by making sure that the number of atoms are the same on both sides of the equation.

46. Balance With State Symbols Li + O2 Al + Cl2  Al + H2SO4(aq)  C3H8(g) + O2  Zn + HNO3  Zn(NO3)2(aq) + H2O + NO2 Cu + HNO3  Cu(NO3)2(aq) + H2O + NO

47. Balance With State Symbols 4Li(s) + O2(g)  2Li2O(s) 2Al(s) + 3Cl2(g)  Al2Cl6(s) 2Al(s) + 3H2SO4(aq)  Al2(SO4)3(aq) + 3H2(g) C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) 4HNO3(aq) + Zn(s)  Zn(NO3)2(aq) + 2H2O(l) + 2NO2(g) 8HNO3(aq) + 3Cu(s)  3Cu(NO3)2(aq) + 4H2O(l) + 2NO(g)

48. Moles and Reactions L.O. Deduce the quantities of reactants and products from balanced equations. Starter – What is stoichiometry?

49. Stoichiometry Stoichiometry is the molar relationship between the relative quantities of substances taking part in a reaction. You use a balanced chemical equation: H2SO4 + 2NaOH  Na2SO4 + 2H2O 1 : 2 : 1 : 2