Determining the Number of Water Molecules in a Hydrate
Mass of empty beaker
Mass of beaker with hydrated copper (II) sulfate
Mass of hydrated copper (II) sulfate
Mass of beaker with dehydrated copper (II) sulfate (last weighing) 57.22 g
Mass of dehydrated copper (II) sulfate
Mass of dehydrated copper (II) sulfate (2nd weighing)
1. Calculate the value of “n”. Show your work. Include an example of ALL calculations involving solving for “n”.
3.02 g of hydrated copper sulfate is heated to drive off water. The dry sample has a mass of 2.11 g. What is the mole ratio between copper (II) sulfate, CuSO4, and water, H2O?
Mass of Hydrate 3.02 g
Mass of Dry Sample - 2.11g
Mass of Water 0.91 g
MM CuSO4= 63.55 + 32.07 + 4(16.00)= 159.62 g/mol
2.11 g CuSO4 (1 mol CuSO4 / 159.62 g) = 0.0132 mol CuSO4
MM H2O = 2(1.01) + 16.00= 18.02 g/mol
0.91 g H2O (1 mol H2O / 18.02 g) = 0.0505 mol H2O
0.0132 mol CuSO4 : 0.0505 mol H2O → divide through by 0.0132 to get CuSO4 ● 4 H2O n=4
In the formula for copper (II) sulfate there are five molecules of water per molecule of copper (II) sulfate. During the experiment there were some mistakes in the data because of human error. Scientists accidentally spilled the hydrated copper sulfate after the second measurement though it lost 0.07 g when it was only supposed to lose 0.02 g at maximum. This meant there was likely still more water in the sample. As is explained in question two below, when not all of the water is driven off there is a lower value of n. This may explain why the n value solved for in this experiment was four when the actual n value is five.
Each of the following errors affects the value of n. in each case, would the value of n increase or decrease? Explain.
2) Not all of the water is driven off.
If not all the water was driven off then the mass of the dehydrated compound would be more...
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