I conducted a lab in which I had to calculate the number of moles of water heated off from copper sulfate. I first weighed copper sulfide in a beaker to in order to get the amount needed to begin the heating of it. If the beaker was wet at the beginning of the experiment for the initial mass of the beaker, how would that affect the experimentally determined number of molecules of hydrate. Increase, decrease, or no affect?
Actually you could get all three under different scenarios. I don't think that is the intent of the question.
To show just one scenario:
1. Water + beaker = xx grams.
2. Add CuSO4.5H2O = 2.0 + xx grams.
Subtracting gives me 2.0 g CuSO4.5H2O as long as none of the water evaporated
If the beaker was wet at the beginning of the experiment and you measured its initial mass as part of calculating the amount of copper sulfate used, it would have an effect on the experimentally determined number of moles of hydrate. Specifically, it would increase the determined number of moles of the hydrate.
The reason for this is that any water present in the beaker, whether from being wet or otherwise, would contribute to the overall mass measured in the experiment. If you were to include this initial wet mass in your calculations, it would erroneously contribute to the total mass used to determine the number of moles of copper sulfate.
To accurately determine the number of moles of the copper sulfate hydrate, it is important to ensure that the beaker is dry before weighing it. This can be achieved by thoroughly drying the beaker prior to the experiment, such as using a clean paper towel or heating it gently to remove any residual moisture. By starting with a completely dry beaker, you can eliminate the effect of any excess water on your calculations and obtain an accurate measurement of the copper sulfate used.