1. Unit 2 Stoichiometry – Gravimetric analysis

2. Gravimetric Analysis The mass of an element or compound present in a substance is determined by changing that substance into another chemical substance of known chemical composition that can be easily isolated, purified and weighed. (e.g. precipitation followed by filtration)

3. The product of the reaction must be: insoluble so that all the product is precipitated have a particle size large enough for it to be able to be filtered stable at 100 o C to allow drying before weighing if required. Analysis in this way can be difficult due to errors in precipitation, filtering and weighing.

4. Another method of analysis is to use heat to change one substance into another. (e.g. dehydration of a hydrated salt to calculate the number of moles of water) The product must be continually heated and weighed until a constant mass is achieved to ensure that the reaction is complete. Successive weighings, at room temperature, must be within 0.01g of each other.

5. A compound containing only carbon, hydrogen and oxygen was analysed by burning a sample and collecting the products. 1.365g of the compound burned to produce 1.98g of carbon dioxide and 0.945g of water. Calculate: (a) The masses of carbon, hydrogen and oxygen present in the compound. (b) The number of moles of carbon, hydrogen and oxygen present. (c) The empirical formula of the compound.

6. (a) The mass of carbon = RAM of C x mass of CO 2 gfm of CO 2 = 12 x 1.98 = 0.54g 44 The mass of hydrogen = gfm of H 2 x mass of H 2 O gfm of H 2 O = 2 x 0.945 = 0.105g 18 The mass of oxygen = 1.365 – (0.54 + 0.105) = 0.72g

7. (b) Number of moles of C = m = 0.54 = 0.045mol gfm 12 Number of moles of H = m = 0.105 = 0.105mol gfm 1 Number of moles of O = m = 0.72 = 0.045mol gfm 16

8. (c) ElementCHO Number of moles0.0450.1050.045 Mole ratio0.045 = 1 0.045 0.105 = 2.33 = 2 1/3 0.045 0.045 = 1 0.045 Whole number ratio 1 x 3 = 32 1/3 x 3 = 71 x 3 = 3 The empirical formula = C 3 H 7 O 3

9. Example 2 A 9.36g sample of hydrated copper(II) sulphate, CuSO4.xH2O, was heated and weighed until no more loss of mass was detected. The remaining anhydrous copper(II) sulphate, CuSO4, weighed 5.985g Calculate the value of x in the formula CuSO4.xH2O.

10. The mass of water in the hydrated copper(II) sulphate = 9.36 – 5.985 = 3.375g So the value of x = 5 and the formula of the hydrated copper(II) sulphate is CuSO4.5H2O CompoundCuSO4H2O Mass5.985g3.375g GFM159.6g18.0g Number of moles0.03750.1875 Mole ratio0.0375 = 1 0.0375 0.1875 = 5 0.0375 Whole number ratio15

11. Example 3 14.91g of an impure sample of aluminium sulphate was dissolved in water and filtered to remove any insoluble material. Excess sodium hydroxide was then added to the filtrate forming a precipitate of aluminium hydroxide. Al2(SO4)3(aq) + 6NaOH(aq) → 2Al(OH)3(s)+3Na2SO4(aq) The precipitate was filtered, washed to remove any soluble substances, dried and weighed. The mass of the precipitate was 5.69g.

12. Calculate: (a) the number of moles of aluminium hydroxide in the precipitate. (b) the number of moles of aluminium sulphate in the original sample (c) the mass of aluminium sulphate in the original sample (d) the percentage purity by mass of aluminium sulphate in the original sample.

13. (a) n Al(OH)3 = m  gfm = 5.69  78 = 0.073mol

14. (b) From the balanced equation: 1 mol Al2(SO4) 3 → 2 mol Al(OH)3 ∴ 0.073mol Al(OH)3 will be formed from 0.0365 mol Al2(SO4) 3

15. (c) m Al(SO4)3 = n x gfm = 0.0365 x 342.3 = 12.49g

16. (d) % purity = 12.49 x 100 = 83.77% 14.91

17. Exercise AH Chemistry Calculations textbook P31 – 34