The purpose of this lab was to figure out the percentage of water in the compound Copper(2) sulfate pentahydrate. To begin I took a evaporating dish and massed it- it was 78.92g. Then, I filled the evaporating dish with around 2 grams of the compound. The evaporating dish now weighed 80.93g. Now, I needed to get rid of the water in the compound. Since water evaporates, I heated the evaporating dish. The mass after the heating was 80.19g. I then repeated this step and the mass after the second heating was 80.18g. I knew I could stop heating up the dish because the mass change between the first and second heating was between .02g. But how did I figure out the percent water you ask? First, I found the mass of the water by subtracting the evaporating dish with the 2 grams of compound with the evaporating dish after all of the heating. So, 80.93-80.18=.75 mass of water. I then found the total mass of the substance by subtracting the mass of the evaporating dish with 2 grams of the compound with the evaporating dish alone. 80.93-78.92=2.01g. To find the percent, I took the amount of water divided by the amount of total substance multiplied by 100. That would be .75/2.01 times 100=37.3% water. These results are fairly reliable because there wasn’t a lot of percent error. Some of the compound stuck to the stirring stick , but none spilled out of the evaporating dish. In order to find out how much water you would lose if you used 6.0g of hydrate you just follow almost the same formula as before. You set 6.0g times .75g/2.01g=2.24g. Then, set 2.24 divided by 6.0 times 100. You would have 37.33% water. As I previously stated, if any of the solid were to be spilled out of the evaporating dish, there would be a huge percent error. The amount spilled out would be accounted for as the amount of water evaporated (which is false). So, you would have a much bigger percent water than there really was, and your data would be unreliable.
Today in class we added information onto our experiment. We found the atomic weight of copper, sulfur, oxygen and water by looking at the periodic table.
O=16 (times 4) =64 amu
H20=18 (times 5)=90amu
I then added all of these together, excluding the H20. So 63.5 amu+32.1 amu+64 amu=159.6 amu. Then I added the h2o. 159.6 amu + 90 amu =249.6amu. In order to find how much water I lost in the experiment, I took the amount of H20 and divided it by the total mass of the compound. 90g H20/249.6g CuSo4 times 5 H20, times 100=36.05%. From this we know that in the experiment I should have lost 36.05% water, but I actually lost 37.3% water.
Now that we have the ideal ratio, I am able to find out how many grams of water I should have lost. (2.01g CuSO4 times 5 H20) X (90.0g H20)/ (249.6g CuSo4 times 5 H20= .728. I should have lost .725 grams of water, but I actually lost .75 grams of water. This was probably caused by percent error. In order to find percent error, I take the amount I should have gotten and subtract the number I actually gotten. Then I divide by the number I should have gotten. .75g H20-.725g H20/.725g H20 X 100= 3.4% error.