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# The Mole 9.6

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### The Mole 9.6

1. 1. Chapter 9.6 Finding the formula of a Compound
2. 2. How can we work out the formula of a compound? <ul><ul><li>We can conduct experiments to find out the formula of a compound. </li></ul></ul><ul><ul><li>1. First , we find out the mass of the reactants taking part in the reaction. </li></ul></ul><ul><ul><li>2. Next , we work out the relative numbers of moles of the reactants used. </li></ul></ul>
3. 3. Example: Working Out the Formula of Magnesium Oxide <ul><ul><li>To work out the formula of magnesium oxide produced by the combustion of magnesium, the following apparatus is used. </li></ul></ul>Magnesium ribbon Clay triangle Tripod stand lid crucible
4. 4. <ul><ul><li>Procedure </li></ul></ul><ul><ul><li>1. Weigh a crucible together with the lid. Put a coil of magnesium ribbon in it and weigh again. </li></ul></ul><ul><ul><li>2. Put the lid on the crucible and heat the crucible gently. When the magnesium catches fire (you will see a white glow through the crucible), heat it more strongly. </li></ul></ul>Magnesium ribbon Clay triangle Tripod stand lid crucible
5. 5. <ul><li>3. Use a pair of tongs to lift the lid slightly from time to time to allow air in. </li></ul><ul><li>Quickly replace the lid to make sure that magnesium oxide formed does not escape. </li></ul><ul><li>4. When the burning is complete, allow the crucible to cool completely. Weigh the crucible together with the lid and the magnesium oxide in it. </li></ul>Magnesium ribbon Clay triangle Tripod stand lid crucible
6. 6. <ul><ul><li>Sample results </li></ul></ul><ul><ul><li>Mass of crucible + lid = 26.52 g </li></ul></ul><ul><ul><li>Mass of crucible + lid + magnesium = 27.72 g </li></ul></ul><ul><ul><li>Mass of crucible + lid + magnesium oxide = 28.52 g </li></ul></ul><ul><ul><li>Calculations </li></ul></ul><ul><ul><li>Mass of magnesium = 27.72 − 26.52 = 1.20 g </li></ul></ul><ul><ul><li>Mass of magnesium oxide produced = 28.52 − 26.52 = 2.00 g </li></ul></ul><ul><ul><li>Mass of oxygen reacted = 2.00 − 1.20 = 0.80 g </li></ul></ul>
7. 7. Deriving the Formula Step 1: List the mass of the element. Element Mg O Mass (from experiment) 1.20 g 0.80 g Relative atomic mass 24 16 Number of moles Molar ratio (divide by the smallest number from the previous row) 1.20 24 = 0.05 0.05 0.05 = 1 0.05 0.05 = 1 0.80 16 = 0.05
8. 8. Deriving the Formula Step 2: State the A r of the element. Molar ratio (divide by the smallest number from the previous row) Number of moles 16 24 Relative atomic mass 0.80 g 1.20 g Mass (from experiment) O Mg Element 1.20 24 = 0.05 0.05 0.05 = 1 0.05 0.05 = 1 0.80 16 = 0.05
9. 9. Deriving the Formula Step 3: Derive the number of moles by dividing the mass with A r . Molar ratio (divide by the smallest number from the previous row) Number of moles 16 24 Relative atomic mass 0.80 g 1.20 g Mass (from experiment) O Mg Element 1.20 24 = 0.05 0.05 0.05 = 1 0.05 0.05 = 1 0.80 16 = 0.05
10. 10. Deriving the Formula Step 4: Obtain the molar ratio. The empirical formula is MgO . Molar ratio (divide by the smallest number from the previous row) Number of moles 16 24 Relative atomic mass 0.80 g 1.20 g Mass (from experiment) O Mg Element 1.20 24 = 0.05 0.05 0.05 = 1 0.05 0.05 = 1 0.80 16 = 0.05
11. 11. Empirical Formula <ul><li>simplest formula of a compound </li></ul><ul><li>shows the types of element present in the compound </li></ul><ul><li>And </li></ul><ul><li>the simplest ratio of the number of the different type of atoms in it </li></ul>
12. 12. Worked Example 1 A sample of an oxide of copper contains 8g of copper combined with 1g of oxygen. Find the empirical formula of the compound . Empirical formula of the compound is Cu 2 O 1 or usually written as this Cu 2 O. Element Cu O Step 1 Mass of element 8 1 Step 2: A r 64 16 Step 3: No. of Moles (Mass/ A r ) 8 = 0.125 64 1 = 0.0625 16 Step 4: Molar ratio (Divide by smallest number) 0.125 = 2 0.0625 0.0625 = 1 0.0625
13. 13. Worked Example 2 <ul><li>A compound has the following percentage composition: Sodium 32.4 %, Sulphur 22.6%, Oxygen 45.0 % </li></ul>Empirical formula: Na 2 SO 4 Element Na S O Step 1 % composition by mass 32.4 22.6 45.0 Step 2 A r 23 32 16 Step 3 No. of Moles 32.4 = 1.4 23 22.6 = 0.7 32 45.0 = 2.8 16 Step 4 Molar ratio ( divide by smallest number) 1.4 = 2 0.7 0.7 =1 0.7 2.8 = 4 0.7
14. 14. Molecular Formula
15. 15. Molecular Formula <ul><ul><li>Its actual formula is P 4 O 10 . We call this the molecular formula . </li></ul></ul><ul><ul><li>The empirical formula of phosphorus(V) oxide as determined by experiment is P 2 O 5 . </li></ul></ul>
16. 16. Molecular Formula <ul><li>shows the exact number of atoms of each element in a molecule </li></ul><ul><li>Is sometimes the same as the empirical formula , i.e. in actual the compound exists in the simplest ratio </li></ul><ul><li>Example: water (H 2 O) and ammonia (NH 3 ) </li></ul>
17. 17. Molecular Formula <ul><li>However most of the compounds do not have a molecular formula that is similar to the empirical formula. </li></ul><ul><li>If the compound’s molecular formula and empirical formula are different, the molecular formula is just a multiple of the empirical formula </li></ul><ul><li>For example, the molecular formula and empirical formula of phosphorus(V) oxide are P 4 O 10 and P 2 O 5 respectively. </li></ul><ul><li>The multiple is 2. </li></ul>
18. 18. Molecular Formula <ul><li>possible for different compound s to have same empirical formula . </li></ul><ul><li>For example: ethane has a molecular formula of C 2 H 4 </li></ul><ul><li>propane has a molecular formula of C 3 H 6 , </li></ul><ul><li>Therefore, they have the same empirical formula which is CH 2 . </li></ul>
19. 19. Molecular Formula <ul><li>To find the molecular formula of a compound we use this method, n is the multiple to the empirical formula </li></ul><ul><li>n = relative molecular mass of the compound </li></ul><ul><li>M r of the empirical formula </li></ul>
20. 20. Example 3 Propane has the empirical formula CH 2 . The relative molecular mass of propane is 42. Find the molecular formula of propane n = relative molecular mass of the compound Mr of the empirical formula n = 42 (12 X 1) + 2 = 3 Molecular formular = (Empirical formula ) n = (CH 2 ) 3 Hence the molecular formula for propane is C 3 H 6 = C 3 H 6
21. 21. Example 4 Compound X contains 40.0% carbon, 6.6% hydrogen and 53.3% oxygen. Its relative molecular mass is 180. What is the molecular formula of X? Molar ratio Number of moles 16 1 12 Relative atomic mass 53.3 6.6 40.0 Percentage in compound O H C Element = 3.3 12 40.0 = 3.3 16 53.3 = 1 16 3.52 = 2 16 3.52 = 6.6 1 6.6 = 1 3.3 3.3
22. 22. Example 2 (continued) Compound X contains 40.0% carbon, 6.6% hydrogen and 53.3% oxygen. Its relative molecular mass is 180. What is the molecular formula of X? The empirical formula of X is CH 2 O . = M r from empirical formula Relative molecular mass 180 30 = 6 Hence, the molecular formula of X = (CH 2 O) 6 = C 6 H 12 O 6 M r of CH 2 O = 12 + (2 x 1) + 16 = 30 n =
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