Experiment 5 pre-lab questions PDF

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These are for Lab Professor Graeme or Constantino. Calculations and answers are shown....

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Title: Experiment 5: Percent Water in a Hydrated Salt Objective/ Purpose:  

To determine the percent by mass of water in a hydrated salt To learn to handle laboratory apparatus without touching it

Procedure: 1. Prepare a clean crucible. Check the crucible for stress fractures and fissure and once you’ve ensured there’s none, support the crucible and lid on a clay triangle and heat with an intense flame for 5 minutes. Allow it to cool and record the mass. Handle the crucible and lid with the tongs throughout the experiment, and make sure the tongs are clean and dry. 2. Determine the mass of sample. Add 3 grams of hydrated salt to the crucible. Measure and record the combined mass of the crucible, lid, and hydrated salt. Calculate the mass of the hydrated salt. 3. Adjust the crucible lid. Using the tongs, set the lid just off the lip of the crucible to allow the evolved water molecules to escape. 4. Heat the sample. Heat the sample by maintaining a high temperature for 10 minutes. After, cover the crucible with the lid and allow both to cool to room temperature. Determine the combined mass of the crucible, lid, and anhydrous salt. 5. Have you removed all of the water? Reheat the sample for an additional 2 minutes and avoid decomposing the salt. Cool to room temperature and measure the combined mass again. 6. Repeat with a new sample. Repeat the experiment two more times with original hydrated salt samples. 7. Cleanup. Rinse the crucible with 2-3 millimeters of 1 M HCL and discard in the Waste Acids container. Then rinse several times with tap water and finally deionized water. Discard each water rinse in the sink. Pre- lab questions: 3. A 1.803-g sample of gypsum, a hydrated salt of calcium sulfate, CaSO4, is heated at a temperature greater than 170˚C in a crucible until a constant mass is reached. The mass of the anhydrous CaSO4 salt is 1.426 g. Calculate the percent by mass of water in the hydrated calcium sulfate salt. Work: First you would find the difference between the hydrated mass and the anhydrate mass. Mass of hydrated gypsum: 1.803g Mass of anhydrous gypsum: 1.426g Mass of water lost (1.803g – 1.426g) = 0.377g Then you divide the mass of water lost by the mass of hydrate and multiply by 100 (Mass of water / Mass of hydrated gypsum) x 100% Mass of water: 0.377g

Percent by mass of water (0.377g / 1.803g) x 100% = 20.90959512% Finally, I would round my answer Answer: 20.91% 5. The following data were collected from the gravimetric analysis of a hydrated salt. 1. Mass of fired crucible and lid (g) 2. Mass of fired crucible, lid, and hydrated salt (g) 3. Final mass of fired crucible, lid, and anhydrous salt (g) Given: Trial 1

Trial 2

Trial 3

19.437

20.687

18.431

21.626

25.111

22.167

21.441

24.702

21.762

Calculations: 1. Mass of hydrated salt (g): (Mass of fired crucible, lid, and hydrated salt – Mass of fired crucible and lid) Trial 1: 21.626 – 19.437= 2.189g Trial 2: 25.111– 20.687= 4.424g Trial 3: 22.167 – 18.431= 3.736g 2. Mass of anhydrous salt (g): (Mass of fired crucible, lid, and anhydrous salt – Mass of fired crucible and lid) Trial 1: 21.441 – 19.437= 2.004g Trial 2: 24.702 – 20.687= 4.015g Trial 3: 21.762 – 18.431= 3.331g 3. Mass of water lost (g): (Mass of hydrated salt – Mass of anhydrous salt) Trial 1: 2.189 – 2.004= 0.185g Trial 2: 4.424 – 4.015= 0.409g Trial 3: 3.736 – 3.331= 0.405g 4. Percent by mass of volatile water in hydrated salt (%): (Mass of water lost / Mass of hydrated salt x 100) Trial 1: 0.185 / 2.189 x 100= 8.45% Trial 2: 0.409 / 4.424 x 100= 9.25% Trial 3: 0.405 / 3.736 x 100= 10.84% 5. Average percent H2O in hydrated salt (% H2O): (Percent by mass of volatile water in a hydrated salt (the sum) / 3 (the number of values) (8.45+9.25+10.84) / 3= 9.51% 6. Standard deviation of %H2O: (Average percent of H2O in hydrated salt – percent by mass of volatile water in hydrated salt)  square each value and then find the sum  divide the sum by 2 because (N–1) = (3–1) = 2 Trial 1: 9.51– 8.45 = 1.06  (1.06)2 = 1.1236

Trial 2: 9.51 – 9.25 = 0.26  (0.26)2 = 0.0676 Trial 3: 9.51 – 10.84 = -1.33  (-1.33)2 = 1.7689 (1.1236+0.0676+1.7689) = 2.96 2.96 / 2 = 1.48 Take the square root of 1.48 which is 1.22 Standard deviation of % of H2O = 1.22 7. Relative standard deviation of %H2O in hydrated salt (%RSD): (Standard deviation / Average percent H2O in hydrated salt x 100) 1.22 9.51 x 100 = 12.82%

6. a. What is the percent by mass of water in iron (II) sulfate heptahydrate, FeSO4x7H2O (or what percent of the molar mass of FeSO4x7H2O is due to the waters of crystallization)? Work: (Mass of substance / Total mass of the compound) x 100% Atomic Masses of Fe = 55.85, S = 32.06, O = 16, H = 1.008 FeSO4x7H2O = 55.85+32.06+ (16x4) + 7[(1.008x2)+16] = 278.022 7H2O = 7 [(1.008x2)+16] = 126.112 (126.112 / 278.022) x 100 = 45.36% Answer: 45.36% b. What mass due to waters of crystallization is present in a 3.38-g sample of FeSO4x7H2O? Work: ( Mass of sample x Mass of substance) / Total mass of the compound (3.38x126.112) / 278.022 = 1.53g H2O Answer: 1.53g H2O...