Stoichiometry of a Double Displacement Reaction The objective of this lab is to find the percent yield of a product of a double displacement reaction. Procedure: Refer to handout entitled “Stoichiometry of a Double Displacement Reaction” Materials: Refer to handout entitled “Stoichiometry of a Double Displacement Reaction” Data & Observations: Data Table Calculated Molar Mass of CuSO4•5H2O 249.677 g Calculated Molar Mass of CuO 79.545 g Starting mass of CuSO4•5H2O 2.050 g Mass of 100-ml beaker and filter paper 52.600 g Mass of 100-ml beaker, filter paper, and CuO precipitate 53.450 g Calculations: a. Theoretical yield of CuO 2.050 g CuSO4•5H2O x (1 mol CuSO4•5H2O ÷ 249.677 g CuSO4•5H2O) x (1 mol CuO ÷ 1 mol CuSO4•5H2O) x (79.545 g CuO …show more content…
The reaction should then occur. Next, we heated all the products to the boiling point and covered it so that none of the resulting products left the beaker. This caused another chemical reaction, which lead the copper(II) hydroxide created in the last reaction to decompose into copper(II) oxide and water. Last, in order to calculate how much copper(II) oxide was created, we needed to separate it from the water which we did by putting all the products through filter paper so the water was drained out. To make sure all the water was separated, we heated the filtered copper(II) oxide in an oven before weighing the precipitate we ended with. Then, with the mass we calculated, we compared it with what it should have been to reach a percent yield. The percent yield we reached was 130% yield which is clearly much more than the perfect 100% yield. Oddly, the yield was greater than 100%, meaning that we ended with more copper(II) oxide than should have been possible considering how much copper(II) sulfate we started the reaction with. We indeed did have an experimental yield of 0.850 g which is almost two hundredths of a gram larger than the theoretical yield of 0.6531 g. This would be considered rather unsuccessful as a result of the almost impossible