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Do now!

Do now!. Can you continue the questions you were doing yesterday? Mr Porter has another sheet if you are finished!. Moles!. Moles!. Equal masses of different elements will contain different numbers of atoms (as atoms of different elements have different masses). Moles!.

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Do now!

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  1. Do now! Can you continue the questions you were doing yesterday? Mr Porter has another sheet if you are finished!

  2. Moles!

  3. Moles! Equal masses of different elements will contain different numbers of atoms (as atoms of different elements have different masses)

  4. Moles! It is sometimes useful for physicists and chemists (but we don’t care about them) to compare the number of atoms or molecules in an amount of substance. To do this we use the idea of moles. A chemist

  5. Moles! You need to learn this definition. • One mole of a substance contains the same number of molecules/atoms as in 12 grams of carbon-12. • This number (of atoms or molecules) is known as the Avogadro constant (NA) which is equal to 6.02 x 1023

  6. How big is 6 x 1023? Imagine the whole of the United states You are here!

  7. How big is 6 x 1023? Imagine the whole of the United states covered in unpopped popcorn

  8. How big is 6 x 1023? Imagine the whole of the United states covered in unpopped popcorn to a depth of six miles!

  9. How big is 6 x 1023? Imagine the whole of the United states covered in unpopped popcorn to a depth of six miles! Count the grains and that is 6 x 1023! 600000000000000000000000

  10. Moles! For example, Hydrogen (H2) has a relative molecular mass of 2, so 2 grams of hydrogen (one mole) contains the same number of molecules as atoms in 12g of carbon-12 (6.02 x 1023)

  11. Moles! It follows therefore that 7g of lithium (atomic mass 7), 20g neon (atomic mass 20) or 39 g potassium (atomic mass 39) all contain the same number of atoms (1 mole or 6.02 x 1023 atoms)

  12. Moles! • The number of moles of a substance can thus be found by dividing the mass of substance by its relative atomic or molecular mass n = mass/RAM

  13. Example How many moles of sulphur atoms are there in 80g of sulphur? How many grams of carbon would have the same number of atoms?

  14. Example How many moles of sulphur atoms are there in 80g of sulphur? How many grams of carbon would have the same number of atoms? N = mass/RAM = 80/32 = 2.5 moles

  15. Example How many moles of sulphur atoms are there in 80g of sulphur? How many grams of carbon would have the same number of atoms? N = mass/RAM = 80/32 = 2.5 moles Mass of carbon = RAM x n = 12 x 2.5 = 30 g

  16. Relative formula mass We can use the idea of moles and apply it to molecules using relative formula mass. C2H5OH RFM = (2 x 12) + (6 x 1) + (1 x 16) = 46 46g of ethanol = I mole of ethanol molecules

  17. Let’s try some questions! Page 81 Questions 1 to 8

  18. Can you read through the “Moles and gases” section?

  19. Equal volumes Luckily, equal volumes of gas contain the same number of particles (at the same temperature and pressure)

  20. Equal volumes For example, 1 litre of nitrogen contains the same number of molecules as there are atoms in one litre of Argon. A litre is 1000 cm3 or 1 dm3

  21. Mole of gas One mole of any gas occupies 24 dm3 (24000 cm3) at standard temperature and pressure (20°C and 1 atmosphere) You will be given this in a question

  22. Moles of gas = volume (cm3)/24000 = volume (dm3)/24 Learn this!

  23. More questions! (pages 342 and 343)

  24. Moles in solution

  25. Concentration/molarity 1M or 1 mol/dm3 means there is one mole of a substance dissolved in 1 dm3 (or 1000 cm3) of solution

  26. # of moles = concentration x volume (cm3)/1000 You need to know this too!

  27. Let’s do some more reading and then try some more questions! (pages 344 to 347)

  28. Working out the formula

  29. Working out the formula • 1.4 g of Nitrogen reacts with 0.3 g of hydrogen to form a compound. What is the formula of the compound?

  30. Working out the formula • 1.4 g of Nitrogen reacts with 0.3 g of hydrogen to form a compound. What is the formula of the compound? • First work out the number of moles • 1.4g Nitrogen = mass/RAM = 1.4/14 = 0.1 • 0.3g hydrogen = mass/RAM = 0.3/1 = 0.3

  31. Working out the formula • 1.4 g of Nitrogen reacts with 0.3 g of hydrogen to form a compound. What is the formula of the compound? • Work out the ratio of the number of moles of each element to the lowest whole numbers N : H 0.1 : 0.3 1 : 3

  32. Working out the formula • 1.4 g of Nitrogen reacts with 0.3 g of hydrogen to form a compound. What is the formula of the compound? • Therefore there are 3 times as many hydrogen atoms as N atoms in the compound. The formula must be NH3 (ammonia)

  33. Working out the formula • A student adds 4.8 g of magnesium to excess dilute hydrochloric acid. What mass of magnesium chloride would be made?

  34. Working out the formula • A student adds 4.8 g of magnesium to excess dilute hydrochloric acid. What mass of magnesium chloride would be made? • First write the balanced equation Mg + 2HCl MgCl2 + H2

  35. Working out the formula • A student adds 4.8 g of magnesium to excess dilute hydrochloric acid. What mass of magnesium chloride would be made? • Find out how many moles of magnesium are being used # moles = mass/RAM = 4.8/24 = 0.2 moles

  36. Working out the formula • A student adds 4.8 g of magnesium to excess dilute hydrochloric acid. What mass of magnesium chloride would be made? • From the equation we can see the number of magnesium atoms is the same as the number of magnesium chloride “molecules”. So if 0.2 moles of Mg are used we should produce 0.2 moles of MgCl2 Mg + 2HCl MgCl2 + H2

  37. Working out the formula • A student adds 4.8 g of magnesium to excess dilute hydrochloric acid. What mass of magnesium chloride would be made? • Work out the mass of 0.2 moles of MgCl2 RFM (RMM) = 24 + (2 x 35.5) = 95 Mass of 0.2 moles of MgCl = 0.2 x 95 = 19g

  38. Let’s try some more questions! Read through the sheet first!

  39. I feel a test coming on moles! Thursday 19th June

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