Mole concept ok1294991357


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Mole concept ok1294991357

  1. 1. Essential Understandings <ul><li>Define atomic, molecular and formula mass </li></ul><ul><li>Calculate the molecular/formula mass of a compound </li></ul><ul><li>Define Avogadro’s constant </li></ul><ul><li>Define the mole and relate it to mass and molecular mass </li></ul><ul><li>Calculating empirical formula </li></ul><ul><li>Calculating molecular formulae </li></ul><ul><li>Finding the percentage composition </li></ul>
  2. 2. Pre-test <ul><li>For questions 1 to 5 select from the following: </li></ul><ul><li>Cl Cl 2 Cl - HCl CCl 4 H + </li></ul><ul><li>1. A molecule of chlorine …………… </li></ul><ul><li>2. An atom of Chlorine …………… </li></ul><ul><li>3. An ion of Chlorine ………… </li></ul><ul><li>4. A cation …………… </li></ul><ul><li>5. An anion …………… </li></ul><ul><li>6. Name an element (other than chlorine) whose molecules consist of two atoms. ……….. </li></ul><ul><li>7. How many atoms are there in one molecule of sugar C 12 H 22 O 11 ? </li></ul><ul><li>8. How many different atoms are there in one molecule of sugar? </li></ul><ul><li>9. What is an isotope? </li></ul>
  3. 3. Atoms, Atomic Masses and Moles <ul><li>Relative atomic Masses: </li></ul><ul><li>Chemists use a relative atomic mass scale to compare different atoms. In the old days they chose Hydrogen as the standard because it was the smallest atom, but when they were able to find more accurate results and when scientists recognized that elements often consisted of more than one isotope they decided to settle on one isotope. They chose one which was fairly easy to obtain and one which was a solid, so they chose 12 C and they assigned this an atom mass of 12.00000g in this manner none of the previously assigned atomic masses on the old standard of hydrogen had to change too much. They just became more precise. </li></ul>
  4. 4. The Avogadro constant, moles and molar masses <ul><li>Since one atom of carbon is 12 times as heavy as one atom of hydrogen, it follows that 12g of carbon contain the same number of atoms as 1g of hydrogen. In also follows that the relative atom mass in grams of all the elements (we obtain this from the data in the Periodic Table) will contain the same number of atoms. Experiments show that this number is 6.0 X 10 23 or 600,000,000,000,000,000,000,000. This number is called the Avogadro constant </li></ul>
  5. 5. <ul><li>As relative atomic mass in grams of all elements contain 6.0 X 10 23 atoms, chemists call 6.0 X 10 23 atoms of an element 1 mole of atoms. (Symbol mol) </li></ul><ul><li>Strictly speaking, the mole is defined as the amount of substance which contains the same number of particles as there are in 12.0000g of 12 C </li></ul><ul><li>Relative Atomic Mass Ar is the mass of an atom of an element compared to 12 C </li></ul>
  6. 6. <ul><li>When we move from elements to compounds then we use Relative Molar Mass M r </li></ul><ul><li>Now we come to the first of the important mole equations you will need; </li></ul><ul><li>Moles = n = Number of particles </li></ul><ul><li>6.0 X 10 23 </li></ul><ul><li>or </li></ul><ul><li>Moles = n = ____ # _____ </li></ul><ul><li>6.0 X 10 23 </li></ul>
  7. 7. Questions <ul><li>10. How many moles are there in 1 million atoms of Ge (1 X 10 6 ) </li></ul><ul><li>How old are you? (remember IB requires units) </li></ul><ul><li>How many seconds have you been on this Earth? </li></ul><ul><li>How many moles is this? </li></ul><ul><li>If the Earth is 4.5 billion years old, how many seconds is this? </li></ul><ul><li>What fraction of a mole is this? </li></ul>
  8. 8. <ul><li>So, you can see we are dealing in mind boggling numbers so chemists generally just talk about moles or fractions thereof, but we understand that each mole contains 6.0 X 10 23 number of particles. </li></ul><ul><li>The next important mole equation is to relate moles to mass in grams we use: </li></ul><ul><li>Moles = n = Mass in grams </li></ul><ul><li>Relative Molar Mass </li></ul><ul><li>or </li></ul><ul><li>Moles = n = __m__ </li></ul><ul><li>M r </li></ul>
  9. 9. So what have moles to do with the Sahara desert?
  10. 10. <ul><li>16. Make a guess of how many moles of sand grains there are in the Sahara desert.1,000,000 100,000 10,000 1,000 100 10 1 0.1 </li></ul><ul><li>17. If the Sahara desert is 4,500km from East to West, 1500km north to South and the sand is 10m deep and 1cm 3 of sand contains 800 grains. How many grains of sand are there in the Sahara desert? </li></ul><ul><li>18. How many moles is this? </li></ul>
  11. 11. <ul><ul><ul><li>What is the number of moles in 10.0g of 12 C? </li></ul></ul></ul><ul><ul><ul><li>What is the mass in grams of 2.5 moles of 24Mg? </li></ul></ul></ul><ul><ul><ul><li>What is the mass of 0.00125 moles of 31 P ? </li></ul></ul></ul><ul><ul><ul><li>So dealing with elements is relatively easy. What about diatomic molecules such as Chlorine Cl 2 ? </li></ul></ul></ul><ul><ul><ul><li>Here we use 2 X 35.5 for M r and plug into the previous equation. </li></ul></ul></ul><ul><ul><ul><li>22. How many moles are there in 2.45g of chlorine molecules? </li></ul></ul></ul><ul><ul><ul><li>23. What is the mass of 0.55 moles of Bromine liquid (Br 2 ) if A r of Bromine is 80g mol -1 </li></ul></ul></ul><ul><ul><ul><li>24. How many moles of oxygen molecules are there in 1.00g of O 2 ? </li></ul></ul></ul><ul><ul><ul><li>25. Notice this is different from how many atoms of oxygen are there in 1.00g of O 2 because there are …….. atoms of Oxygen in one …………. of oxygen. </li></ul></ul></ul>
  12. 12. <ul><li>Now let us go one step further and move to compounds. The M r for compounds is found by adding up the A r of all the elements in the compound. For example CuO is 64 for copper + 16 for oxygen making a total of 80g mol -1 </li></ul><ul><li>25. What is the M r of NaOH? </li></ul><ul><li>26. How many moles are there in 10g of NaOH? </li></ul><ul><li>27. What is the M r of CuCO 3 ? (Cu + C + O + O +O) </li></ul><ul><li>28. What is the mass of 0.245mol of CuCO 3 ? </li></ul>
  13. 13. <ul><li>29. The next step is to try more complicated formulae. For example what is the Mr of FeSO 4 .7H 2 O ? This will be </li></ul><ul><li>Fe + S + O + O + O + O + 7( H + H + O) </li></ul><ul><li>30. What is M r of CuSO 4 .5H 2 0 ? </li></ul><ul><li>31. How much hydrated copper(II)sulphate (the stuff in above question) would you need to weigh out if you needed 0.20 mole ? </li></ul><ul><li>32. What is M r K 2 Cr 2 O 7 ? </li></ul><ul><li>33. How many moles of potassium dichromate are there in 1.00 g of K 2 Cr 2 O 7 ? </li></ul>
  14. 14. <ul><li>Sometimes the chemical formulae are displayed showing the bonds but you can easily add up all the atoms to find M r . For example Methyl orange </li></ul><ul><li>H 3 C O </li></ul><ul><li> II </li></ul><ul><li> N----C 6 H 4 ----N=N----C 6 H 4 ----S----O----H </li></ul><ul><li> II </li></ul><ul><li>H 3 C O </li></ul><ul><li>Can be reduced to N 3 C 14 H 15 SO 3 </li></ul>
  15. 15. <ul><li>Percentage composition </li></ul><ul><li>Sometimes it is important to know a percentage composition. For example a farmer needs a nitrogen based fertilizer and he wants to get as much nitrogen as possible for his money. </li></ul><ul><li>Let’s say we have a choice of 3 nitrogen based fertilizers, ammonium sulphate, (NH 4 ) 2 SO 4 , ammonium nitrate, NH 4 NO 3 and Urea , CO(NH 2 ) 2 and each costs 150 bombles(the Universal currency) a kilo. Which should he buy? </li></ul><ul><li>One way of looking at it is to consider the amount of nitrogen as a fraction of the whole and then multiply by 100 to obtain a % </li></ul>
  16. 16. <ul><li>For ammonium sulphate there are 2 Nitrogen atoms </li></ul><ul><li>N 2 X 100 = 28 X 100 = 21.2% </li></ul><ul><li>(NH 4 ) 2 SO 4 132 </li></ul><ul><li>For ammonium nitrate there are also 2 Nitrogen atoms in the formula </li></ul><ul><li>N 2 X 100 = 28 X 100 = 35% </li></ul><ul><li>NH 4 NO 3 80 </li></ul><ul><li>33. Please complete the calculation for Urea and state which fertilizer the farmer should buy. </li></ul>
  17. 17. <ul><li>Empirical formula </li></ul><ul><li>This is the simplest whole number ratio of atoms in a chemical formula. So if the real formula (called the molecular formula if it is a molecule) of glucose is C 6 H 12 O 6 then the empirical formula of glucose, which is a carbohydrate is CH 2 O since it can be simplified by dividing through by 6. CH 2 O is the empirical formula for a whole range of carbohydrates. </li></ul><ul><li>Empirical formula, like most chemical formulae, are found by experiment. </li></ul><ul><li>Suppose an experiment on ethene, a hydrocarbon (containing C and H only) found that it contained 85.72% carbon and 14.28% hydrogen, then here is how we find the empirical formula. </li></ul><ul><li>Remember that all formula are in whole number mole ratios, so the first thing to do is to convert each mass into moles. A second good idea is to remember this template for all empirical formula calculations. </li></ul>
  18. 18. Remember this template C H Mass in grams 85.72 14.28 Moles 85.72/12 14.28/1 Ratio 7.14 14.28 Simple ratio 1 2 Empirical formula CH 2
  19. 19. What to do if you have horrible numbers <ul><li>If you have ‘horrible’ numbers in the ratio line just divide both sides by the smallest ‘horrible’ number then you will have 1: something. </li></ul><ul><li>If the ratio is 1 : something which is not near a whole number, then you have to keep multiplying each side by 2, then 3 then 4 etc until both sides become whole numbers, but all columns must be X same number </li></ul>
  20. 20. <ul><li>Often the experimental results are not given as a percentage and may look something like this: </li></ul><ul><li>A 1.31g sample of sulphur was allowed to react with an excess of chlorine to produce 4.22g of a product that contains only sulphur and chlorine. What is it’s empirical formula? </li></ul>S Cl Mass/g 1.31 4.22-1.31 Moles 1.31/32 2.91/35.5 Ratio 0.04094 0.08197 Simple ratio 1 2 Empirical formula SCl 2
  21. 21. <ul><li>Besides there being ‘horrible’ ratios there are often more than two elements in a compound. </li></ul><ul><li>Vanillin, has the following composition by mass carbon, 63.2%; hydrogen 5.26% and the remainder is oxygen. What is the empirical formula of vanillin? </li></ul>C H O Mass 63.2 5.26 100 -(63.2+5.26) Moles 63.2/12 5.26/1 31.54/16 Ratio 5.2666 5.26 1.9713 Ratio 2.67 2.67 1 Simple Ratio 8 8 3 Empirical formula C 8 H 8 O 3
  22. 22. <ul><li>34. A propellant used in an aerosol can was analysed and found to contain 0.423g of carbon, 2.50g of chlorine, and 1.34g of fluorine. What is the empirical formula of this compound? </li></ul>C Cl F Mass/g 0.423 2.50 1.34 Moles Ratio Ratio Simple Ratio Empirical
  23. 23. <ul><li>Going from an empirical formula to a Molecular formula </li></ul><ul><li>Not only does the molecular formula give us the ratio of atoms in the compound, it also tells us the actual number of atoms of each kind in a molecule of the substance. </li></ul><ul><li>In our previous example with glucose, the empirical formula is CH 2 O so how do we get to the molecular formula of C 6 H 12 O 6 ? We know there is a whole range of carbohydrates so how can we distinguish which carbohydrate? One important way in which these substances differ is that they have different molecular masses. For instance the molecular mass of glucose is 180g mol -1 and this must be a whole number multiple of the mass of CH 2 O In other words we find out how many units of CH 2 O add up to make a mass of 180, the mass of 1 mole in the following manner: </li></ul>
  24. 24. <ul><li>x(CH 2 O) = 180 </li></ul><ul><li>x(12+2+16) = 180 </li></ul><ul><li>30x = 180 </li></ul><ul><li>x = 6 </li></ul><ul><li>(There are 6 units of CH2O in glucose) </li></ul><ul><li>The molecular formula is written as C 6 H 12 O 6 </li></ul>
  25. 25. <ul><li>35. If the empirical formula of a compound was found to be NaS 2 O 3 and it’s Formula Mass ( the name we use for ionic compounds instead of molecular mass for covalent molecules ) is 270.4. What is the correct formula? </li></ul>
  26. 26. This looks pretty Complicated so let us just Consider Room Temperature and Pressure, then 1mole Of any gas occupies 24 liters =24,000cm 3 You can thank Mr Avogadro for this
  27. 27. Let us now consider chemical equations