CHEMISTRY. Unit 3, Area of Study 1: Chemical Analysis. Gravimetric Analysis

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1 Watch this lesson online: CHEMISTRY Unit 3, Area of Study 1: Chemical Analysis Gravimetric Analysis 1

2 Outcome 1 On completion of this unit the student should be able to evaluate the suitability of techniques and instruments used in chemical analyses. Key Knowledge Gravimetric analysis the application of chemical equations to (volumetric and) gravimetric analyses 2

3 Gravimetric analysis This form of quantitative analysis the mass of a part of the sample and relating it back to the sample, often as a percentage by mass The part of the sample we are interested in may be the water in the product, or perhaps we might be focussing on a particular ion that is present. 3

4 Measuring water content Many solids have a high proportion of water. Dehydration removes the water and decreases the mass This may be heating at just above 100 C Mass loss = mass of water originally present % water content = m(water) 100 m(sample) 4

5 Weigh the sample Heat the sample in an oven just above 100 C Allow the sample to cool in a desiccator This cycle is repeated until the sample assumes a constant mass Reweigh the sample 5

6 Finding the composition of a mixture 1. Dissolve the mixture in water 2. Add an ionic solution which will form a precipitate with the ion that is required (eg SO 2-4 ). 3. This new solid is out, dried and weighed. 6

7 The most common precipitates Element you wish to analyse Substance that is added to create precipitate Formula of precipitate Chlorine Silver nitrate AgCl Bromine Silver nitrate AgBr Sulfur Barium chloride BaSO 4 7

8 Application of gravimetric analysis Gravimetric analysis is used in the food industry to determine the amount of salt in products. When an AgNO 3 solution is added to the dissolved food, chloride ions (present as NaCl) are precipitated out as AgCl. 8

9 Calculations Find n(agcl) using: n(agcl) = Then: n(agcl) = n(cl - ) = n(nacl) Convert n(nacl) to m(nacl) using m(nacl) = n(nacl) M(NaCl) m(agcl) M(AgCl) %(NaCl) = m(nacl) 100 m(food) 1 9

10 Experiment Analysis of Lawn food Lawn food contains many nutrients required by plants such as sulfate ions, SO 4 2-, nitrate ions, NO 3 - and phosphate ions, PO Using gravimetric analysis to ;ind the amount of sulfate ions in a sample of lawn food 1. Dissolve lawn food in water 2. Add barium chloride solution 3. Sulfate ions precipitate out as barium sulfate BaCl 2 (aq) + SO 4 2- (aq) BaSO 4 (s) + 2Cl - (aq) 10

11 4. Filter, wash and dry the BaSO 4 to obtain a pure sample of precipitate. 6. During washing is tested for presence of left over chloride ions by adding silver nitrate. AgNO 3 (aq)+ Cl - AgCl(s) + NO 3- (aq) If ions are in then they have probably also stuck to the solid and its mass will be too high. 11

12 6. Barium sulfate is dried, then weighed. 8. Using M(BaSO 4 ), n(baso 4 ) is calculated so n(so 4 2- ) can be deduced. n(so 4 2- ) = n(baso 4 ) 8. m(so 4 2- ) = n(so 4 2- ) M(SO 4 2- ) 9. m(s) = n(so 4 2- ) M(S) 10. % S = m(s) 100 m(sample) 1 12

13 Sample exam question An 8.64g sample of rock was thought to be made up of CaCO 3 which reacted with HCl according to: 2HCl(aq) + CaCO 3 (s) CaCl 2 (aq) + H 2 O(l) + CO 2 (g) and 1.55g of insoluble SiO 2. (a) Calculate the expected percentage of CaCO 3 in the rock sample (b) The dissolved calcium ions were precipitated out as CaC 2 O 4.H 2 O which was collected, washed and dried. It was converted to CaO (M = 56.1gmol- 1) and 3.87g was obtained. Calculate the % CaCO 3 in the rock sample. (VCAA June 2007 Q3) 13

14 Answers (a) m(caco 3 ) = = 7.09g %(CaCO 3 ) = = 82.1% b) n(cao) = mol 56.1 = n(ca 2+ ) = mol = n(caco 3 ) m(caco 3 ) = = 6.905g %(CaCO 3 ) = = (3 79.9% 14

15 QUESTION 1 Xylose is a compound that has five carbon atoms in each molecule and contains 40% carbon by mass. What is the molar mass of xylose? A. 30 B. 67 C. 150 D. It cannot be determined without further information. (VCAA Q3 June 2008) 15

16 QUESTION 1 (ANSWER) Xylose is a compound that has five carbon atoms in each molecule and contains 40% carbon by mass. What is the molar mass of xylose? A. 30 B. 67 C. 150 D. It cannot be determined without further information. Alternative C is the correct answer M(C) = 12.0 gmol -1 1 mol xylose molecules contains 5 mol C atoms (5 C atoms per molecule, so 5 mol of C atoms per 1 mol of molecules) m(5 mol C atoms) = 60.0 g 40% of xylose is carbon, so 40% of M(xylose) = 60.0 This can also be written as 0.40 M(xylose) = M(xylose) = 0.40 = gmol -1 (VCAA Q3 June 2008) 16

17 QUESTION 2 The amount of calcium carbonate (:molar mass = g ) in the ore dolomite can be determined by gravimetric analysis. The dolomite sample is dissolved in acid and the calcium ions () present are precipitated as calcium oxalate ( :molar mass = g ). The calcium oxalate is foltered, dried and strongly heated to form calcium oxide (CaO); molar mass = 56.1 g ). In one analysis the mass of dolomite used was 3.72 g. The mass of calcium oxide formed was found to be 1.24 g. The percentage of calcium carbonate in the dolomite sample is closest to A B C D (VCAA Q8 June 2005) 17

18 QUESTION 2 (ANSWER) The amount of calcium carbonate (:molar mass = g ) in the ore dolomite can be determined by gravimetric analysis. The dolomite sample is dissolved in acid and the calcium ions () present are precipitated as calcium oxalate ( :molar mass = g ). The calcium oxalate is foltered, dried and strongly heated to form calcium oxide (CaO); molar mass = 56.1 g ). In one analysis the mass of dolomite used was 3.72 g. The mass of calcium oxide formed was found to be 1.24 g. The percentage of calcium carbonate in the dolomite sample is closest to A B C D

19 Alternative D is the correct answer n(cao) = = 59.5% n(ca 2+ ) = n(cao) = n(caco 3 ) = m(caco 3 ) in dolomite = = 2.21g %(CaCO 3 ) = = 59.5% (VCAA Q8 June 2005) 19

20 QUESTION 3 When 2.54 g of solid iodine reacts with excess chlorine and the unreacted chlorine is evaporated, 4.67 g of a yellow product remains. The empirical formula of the product is A. ICl 2 B. ICl 3 C. ICl 4 D. ICl 5 (VCAA Q4 June 2007) 20

21 QUESTION 3 (ANSWER) When 2.54 g of solid iodine reacts with excess chlorine and the unreacted chlorine is evaporated, 4.67 g of a yellow product remains. The empirical formula of the product is A. ICl 2 B. ICl 3 C. ICl 4 D. ICl 5 The empirical formula is ICl 3 so the correct answer is B The yellow product is only made up of chlorine and iodine In 4.67g of the yellow product, 2.54g is iodine and ( ) 2.13g is chlorine Empirical formula: I : Cl 2.54g : 2.13g : (VCAA Q4 June 2007) : : 3 21