Hydrates: Water and Dish

January 4, 2013
Chemistry Honors

Title: Hydrates

Purpose: To find the mass percentage of water in the hydrate.

Materials: * Watch glass * 2.82 g hydrated crystal (CuSO4*5H2O) * Evaporating dish * Bunsen burner * Electronic balance * Metal tongs * Ring stand

Procedure: 1) Determine the mass of evaporating dish and watch glass. 2) Add between 2 and 3 grams of the hydrated crystal to the evaporating dish. 3) Determine the mass of the dish and crystal. 4) Heat the substance until it turns white. 5) Let the dish cool. 6) Determine the mass of the dish and anhydrous crystal. 7) Clean up the lab.

Data Table:

Mass of hydrated and anhydrous crystal | Mass (g) CuSO4*5H2O | Mass (g) CuSO4 | With dish and glass (75.28g) | 78.1 | 77.33 | Without dish and glass | 2.82 | 2.05 |

Questions: 1) Determine the percentage mass of water in your sample according to your experimental results. * 2.82g CuSO4*5H2O - 2.05g CuSO4 = 0.77gH2O
0.77gH2O2.82gCuSO4*5H2O100= 27.30% H2O (experimental) 2) The formula for the hydrate you used is CuSO4*5H2O. Determine the accepted percentage mass for water in this substance. * Cu 1x64= 64g S 1x32=32g O 9x16=144g H 10x1=10g Total= 250g H 10x1= 10g O 5x16= 80g Total= 90g

100H2Ototal = 10090g H2O250g CuSO4 = 36% H2O 3) Determine your percent error. * theoretical % water-experimental % watertheoretical % water100
(36-27.3036)100
Percent error= 24.2%

4) Determine the moles of water evaporated. * 27.30% evaporated H2O = 27.3gH2O
27.3gH2O1x1molH2O18gH2O=1.517molH2O

5) Determine the moles of ANHYDROUS copper II sulfate. (The white stuff in the dish) * 100gCuSO4gCuSO4*5H20
1002.052.82=72.7%CuSO4 (experimental) * 72.7gCuSO41x1molCuSO4160gCuSO4= 0.454 molCuSO4

6) What is your experimental ratio of copper II sulfate to water? * 0.454 CuSO4 : 1.517 H2O = 1 CuSO4 : 3 H2O
.454