This investigation should aid in the understanding of the composition of hydrates and simple decomposition reactions. The purpose is to determine the percent of water in a hydrate.
Equipment/Materials • Evaporating Dish • Wire gauze • Laboratory balance • Copper (II) Sulfate hydrate (Approximately 2.00g) • Iron Ring • Microspatula • Safety glasses • Crucible tongs • Bunsen burner • Ring stand
Procedure 1. Prepare the setup shown in figure 1 [pic]
2. Heat the dish with the hottest part of the flames for 3 minutes 3. Using the crucible tons, remove the evaporation dish from the apparatus. Place it on an insulated pad and allow it to cool for 3 minutes. 4. Find the mass of the evaporation dish to +/-. 01g. Record the mass in the data selection 5. With the evaporation dish on the balance, measure into it approximately 2.00g Copper (II) sulfate hydrate. Record the mass of the evaporating dish and the hydrate in the data selection 6. Place the evaporating dish with the hydrate on the wire gauze. Gently heat the dish by moving the burner back and forth around the base of the evaporating dish. Avoid any popping and spattering caused by keeping the burner stationary. 7. Heat strongly for 5 minutes or until the blue color has disappeared. During heating, a microspatula solid appear to be turning brown, remove the heat momentarily and resume heating at a gentle rate. 8. Allow the evaporation dish to cool for about 5mminutes. Immediately find the mass of the dish with the anhydrous salt and record it in the data table
Observations and Data:
Mass of evaporating dish: 22.20g
Mass of evaporating dish+ hydrate: 24.70g Mass of evaporating dish+ anhydrous salt: 23.81g
Observations: As the hydrate we being heated in the evaporating dish the hydrate color was removed and became a pale grey dust looking color turning into anhydrous salt.
Analysis/Calculations: Show work on a separate sheet of paper
1. The color of the hydrate before being heated was a bright neon blue. 2. Mass of hydrate= Mass of evaporating dish+ hydrate subtract the mass of evaporating dish +anhydrous salt = 2.50g 3. The color of the anhydrous salt was a pale grey. 4. Mass of anhydrous salt= Mass of evaporating dish + anhydrous salt subtract the mass of the evaporating dish= 1.61g. 5. The mass of the anhydrous salt and the hydrate was different because the heat evaporated the water in the hydrate, which therefore made the anhydrous salt a lighter color. 6. The molar mass for one copper is 63.55g then add one mole of sulfur which is 32.06g next add 4 moles of oxygen which will equal 64g when added up you get which is the molar mass for 1 mol of CuSo4 becomes 159.5g. Then to find the number of moles take the mass of the anhydrous salt and multiple by 1 mol over the molar mass of CuS04, which is 159.56g. Which equals 0.01 moles in the anhydrous salt. 7. The mass of the water lost is 0.89g. The evaporating dish + hydrate (24.70g)- the mass of the evaporating dish (2220g) = 2.50g.Then the mass of the evaporating dish + the anhydrous salt (23.81g) – the mass of the evaporating dish (22.20g) = 1.61g. Then the mass of the hydrate used (2.50g) – the mass of the anhydrous salt (1.61g)= 0.89g which is the mass of the water lost. 8. The mass of 2 Hydrogen is 2.02g and the mass of one Oxygen is 16.00g so those two added together = 18.02g, then to figure out the number of moles divide 18.02g by .89g=