1. Gravimetric analysis Four example problems 1.Determining salt in foodDetermining salt in food 2.Determining sulfate and sulfur content in fertiliserDetermining sulfate and sulfur content in fertiliser 3.Determining the empirical formula of a hydrated moleculeDetermining the empirical formula of a hydrated molecule 4.Determining the empirical formula of an alcoholDetermining the empirical formula of an alcohol

2. Question 1: Determining salt in food A 6.67 g sample of potato chips was crushed, blended with water and filtered. Excess silver nitrate was added to the filtrate. The dried precipitate had a mass of 0.245 g. Calculate the amount of sodium chloride that was in the potato chips (assume all chloride is in the form of the sodium salt). Give your answer in mg of NaCl per 100 g of chips. Where do we start? Write out the relevant chemical reaction: NaCl (aq) + AgNO 3 (aq)  AgCl (s) + NaNO 3 (aq) Identify what else we know: The final mass of AgCl is 0.245 g (3 significant figures) taken from 6.67 g (3 significant figures) of chips Think about where you want to get to: Calculate how many mg of NaCl are present in 100 g of potato chips

3. Calculating the moles of AgCl NaCl (aq) + AgNO 3 (aq)  AgCl (s) + NaNO 3 (aq) m(AgCl)= 0.245 g M(AgCl)= 107.9 g  mol -1 + 35.5 g  mol -1 = 143.4 g  mol -1 n(AgCl)= 0.245 g/143.4 g  mol -1 = 0.001709 mol

4. Calculating the mass of NaCl NaCl (aq) + AgNO 3 (aq)  AgCl (s) + NaNO 3 (aq) n(AgCl)= 0.001709 mol From the chemical equation we know that the moles of NaCl to the moles of AgCl is 1:1, therefore: n(NaCl)= 0.001709 mol M(NaCl)= 23.0 g  mol -1 + 35.5 g  mol -1 = 58.5 g  mol -1 m(NaCl)= 0.001709 mol × 58.5 g  mol -1 = 0.09998 g

5. Calculating the amount of NaCl in mg per 100 g

6. What does this really mean? If you look at the nutrition panel of any food, you will find the mass of sodium reported in mg per 100 g. There are a few questions that you might ask: If there are 1.50 × 10 3 mg of NaCl per 100 g of chips, how mg of Na are there per 100 g of chips? How would the presence of other chloride salts (e.g. KCl) impact calculations of sodium content based on a silver chloride precipitation reaction? Other food additives also contain sodium (e.g. monosodium glutamate or sodium bicarbonate). Would a AgCl precipitation identify other sources of sodium?

7. Question 2: Determining sulfate and sulfur content in fertiliser A student took a 0.998 g sample of fertiliser and added it to 100 mL of water. The solution was mixed well until all soluble fertiliser had dissolved. The solution was then filtered. The filtrate was acidified with dilute hydrochloric acid, then treated with excess barium chloride. The precipitate was filtered, dried to a constant mass and weighed. The weight of the precipitate was 0.387 g. Calculate the percentage of sulfate and the percentage of sulfur in the sample of fertiliser. Where do we start? Write out the relevant chemical reaction: SO 4 2- (aq) + Ba 2+ (aq)  BaSO 4 (s) Identify what is important and what is not for the calculation: Is the volume of water added important? (more about this later) Is the presence of HCl important? (more about this later)

8. Question 2: Determining sulfate and sulfur content in fertiliser A student took a 0.998 g sample of fertiliser and added it to 100 mL of water. The solution was mixed well until all soluble fertiliser had dissolved. The solution was then filtered. The filtrate was acidified with dilute hydrochloric acid, then treated with excess barium chloride. The precipitate was filtered, dried to a constant mass and weighed. The weight of the precipitate was 0.3827 g. Calculate the percentage by mass of sulfate and the percentage by mass of sulfur in the sample of fertiliser. Where do we start (continued)? Identify what we know: The final mass of BaSO 4 is 0.3827 g (4 significant figures) taken from 0.998 g (3 significant figures) of fertiliser Think about where you want to get to: Calculate how many grams of sulfate are present in 100 grams of fertiliser. Do the same for sulfur.

9. Calculating the moles of BaSO 4 SO 4 2- (aq) + Ba 2+ (aq)  BaSO 4 (s) m(BaSO 4 )= 0.3827 g M(BaSO 4 )= (137.3 + 32.1 + 4 × 16.0) g  mol -1 = 233.4 g  mol -1 n(BaSO 4 )= 0.3827 g/233.4 g  mol -1 = 0.001640 mol

10. Calculating the mass of SO 4 2- SO 4 2- (aq) + Ba 2+ (aq)  BaSO 4 (s) n(BaSO 4 )= 0. 001640 mol From the chemical equation we know that the moles of SO 4 2- to the moles of BaSO 4 is 1:1, therefore: n(SO 4 2- )= 0. 001640 mol M(SO 4 2- )= (32.1 + 4 × 16.0) g  mol -1 = 96.1 g  mol -1 m(SO 4 2- )= 0. 001640 mol × 96.1 g  mol -1 = 0.1576 g

11. Calculating the percentage by mass of SO 4 2- in the fertiliser

12. Calculating the percentage by mass of sulfur in the fertiliser

13. Are there other considerations? Is knowing the exact volume of water used to dissolve the sulfate containing compounds important? To answer this, consider the purpose of the water and whether it ever came into play in any of the calculations. What is the purpose of the hydrochloric acid? To answer this, you need a bit more information. Fertiliser also contains carbonate containing compounds. Will carbonate precipitate in the presence of barium ions? What happens to carbonate in the presence of acid?

14. Question 3: Determining the empirical formula of a hydrated molecule Magnesium sulfate readily takes up water and is usually found in a hydrated form with a set number of water molecules per molecule of magnesium sulfate. A student conducted an experiment to calculate the empirical formula for hydrated magnesium sulfate. Mass of empty crucible:45.39 g Mass of crucible, cover and magnesium sulfate before heating:50.14 g Mass of crucible, cover and magnesium sulfate after 1 st heating:47.70 g Mass of crucible, cover and magnesium sulfate after 2 nd heating:47.63 g Mass of crucible, cover and magnesium sulfate after 3 rd heating:47.62 g Where do we start? Think about where you want to get to  mole ratio of water to magnesium sulfate

15. Calculate the number of moles of water m(H 2 O)= starting mass – dried mass = 50.14 g – 47.62 g = 2.52 g M(H 2 O)= (2 × 1.0 + 16.0) g  mol -1 = 18 g  mol -1 n(H 2 O)= 2.52 g/18 g  mol -1 = 0.1400 mol

16. Calculate the number of moles of anhydrous magnesium sulfate m(MgSO 4 )= dried mass – crucible = 47.62 g – 45.18 g = 2.44 g M(MgSO 4 )= (24.3 + 32.1 + 4 × 16.0) g  mol -1 = 120.4 g  mol -1 n(MgSO 4 )= 2.44 g/120.4 g  mol -1 = 0.0203 mol

17. Calculate the mole ratio and determine the empirical formula

18. Question 4: Determining the empirical formula of an alcohol When 12 g of an unknown alcohol was fully combusted, it was found to produce 26.4 g of carbon dioxide and 14.4 g of water. What is the empirical formula for the alcohol? Where do we start? Write out the relevant chemical reaction: C x H y O z + O 2  CO 2 + H 2 O We know that alcohols contain carbon, hydrogen and oxygen, but we don’t know the numbers of each, so we will just call them x, y and z. Think about what you know and how you can use it to get started: Knowing the mass of the products will allow the calculation of the number of moles of the products, which will give the number of moles of carbon and hydrogen. Oxygen will be a bit more tricky.

19. Determining the number of moles of carbon and hydrogen C x H y O z + O 2  CO 2 + H 2 O Mass (g):26.414.4 Molar mass (g  mol -1 ):44.0 18.0 Moles (g/g  mol -1 ):0.600.80 Moles (C or H)0.601.60 (note: the number of moles of carbon equals the number of moles of CO 2 ; the number of moles of hydrogen is twice the number of moles of H 2 O)

20. Determining the number of moles of oxygen C x H y O z + O 2  CO 2 + H 2 O Knowing the number of moles of carbon and hydrogen, we can work out the mass of carbon and hydrogen in the alcohol and from that derive the mass of oxygen in the alcohol.

21. Determining the number of moles of oxygen C x H y O z + O 2  CO 2 + H 2 O Moles of C and H(mol)0.61.6 Molar mass (g  mol -1 )12.01.0 Mass (mol × g  mol -1 )7.21.6 Mass of oxygen = 12 g – 7.2 g – 1.6 g = 3.2 g Moles of oxygen in the alcohol= 3.2 g/16.0 g  mol -1 = 0.2 mol

22. Determining the empirical formula

23. C 3 H 8 O – propyl alcohol (IUPAC name: propan-1-ol)