Explain How To Make 1 Grams Of Anhydrous Equation Lab Report

799 Words2 Pages

Goal: To make 1 grams of Cu and 2.54 grams of ZnSO4 Anhydrous
Equation: Zn + CuSO4.5H2O ------> ZnSO4+ Cu
Procedure:
First, we measured out 1.03 grams of Zn and 3.93g grams of CuSO4. Next we took the two and mixed them in a flask together. After we put them in the flask together we added enough water to get the reaction started. We swirled in around to make sure all the Zn reacted with the SO4. After the reaction took place, we let the flask sit for awhile to let all the copper separate from the ZnSO4. After the cooper settle to the bottom, we weighted our filter paper and the beaker that we are going to put the zinc sulfate. After we weighed the materials for the next step, we took the filter and fitted it into the beaker. After the paper was in the beaker my group ported the solution in the …show more content…

The first major set back we had is we had to restart twice because the first time we didn’t measure out the amounts correctly and the second time is because we used tap water in the reaction instead of distilled water. Another major problem we had was our copper was a mixture of copper and copper(ll). This was really hard to find the amount if each type of copper in the solution.
Actual Calculations: We were planning on getting 1.00g of Cu and 2.54g of ZnSO4 anhydrous crystal. What we ended up getting were 2.54g of ZnSO4 so exactly the right amount, but for the Cu we didn’t quite get what we were after. The final weight of our Cu was 1.12g we calculated out that if we had 0.0157 moles of Cu we would have ended up with 1.00g of Cu we got 1.12g so what we suspect to have happened is that we got 48% 2CuO which with 0.0157 moles of it we would have had 1.25g of it. By the look of our copper we can assume we got 48% 2CuO which should have resulted in 0.60g of it. Pairing that with a 52% Cu we should have gotten 0.52g of it added together they equal 1.12g that is exactly we got in the end.

Open Document