Stoichiometry of a Precipitation Reaction

Topics: Stoichiometry, Mole, Reagent Pages: 5 (1470 words) Published: October 24, 2014
Experiment 3: Stoichiometry of a Precipitation Reaction

Abstract:In this experiment, the objective is to use Stoichiometry to predict the amount of product produced in a precipitation reaction. We received working knowledge of how to accurately measure reactants and products of the reaction. We then are able to use the data that we recorded to make assessments of the actual yield opposed to the theoretical yield. When we calculated the percent yield we are able to see how accurate the data we recorded is.

Experiment and Observation: The first thing we needed to do for this experiment was to weigh 1.0 g of CaClᴤ ˣ 2HᴤO and placed it into a 100 mL beaker. Next we measured 25 mL of water and added it to the beaker with the calcium chloride dihydrate and stirred until the granules were dissolved and mixed with the water. The water became slightly cloudy. Next we had to figure out how much NaᴤCoᴣ we needed for this experiment. I know that for this we need mass/molar mass and that equals 1.0/147.015=.006802 mol. Next we use the mol which is .006802 and multiply by the molar mass which equals: .006802 ͯ 105.99= .72 g of CaClᴤ * 2HᴤO. We could then pour this into the beaker with the calcium chloride dihydrate solution. Instantly, the solution turned milky white. I then folded my filter and placed the filter into the funnel and poured the milky solution into the filter lined funnel. The liquid poured through into the paper cup inside the coffee mug. It filtered slowly and I had to continually stop to allow the liquid to drain once all the milky solution went into the liner I waited until no more water was dripping from the funnel. Finally, I was able to remove the filter and placed it on a folded paper towel until dried completely. While I was waiting for the filter to completely dry I calculated the theoretical yield. To do this we use: 1g CaClᴤ*2HᴤO * (1 mol CaClᴤ*2HᴤO/147.015 g CaClᴤ*2HᴤO) * (1 mol CaCOᴣ/1 mol CaClᴤ*2HᴤO) * (100 g CaCOᴣ/1 mol CaCOᴣ) = .6802. To check this we use: .72 g NaCOᴣ * (1 mol NaCOᴣ/106 g NaCOᴣ) * (1 mol CaCOᴣ/1 mol NaCOᴣ) * (100 g CaCOᴣ/1 mol CaCOᴣ) = .68. After the filter was completely dried I placed the filter on the digital scales and weighed the filter. It weighed 1.5 grams then I minus the filter weight previously weighed at .8 g which equals a net weight of .7 g. Calculations and Error: This experiment required many calculations in order to perform the experiment. We had to calculate the amount of chemical to combine with another chemical in order to produce a chemical reaction. For example: We are told to add 1.0 g of CaClᴤ*2HᴤO then we must calculate the amount of NaᴤCOᴣ we need to perform the experiment. To do this we have to convert the 1.0 g of CaClᴤ*2HᴤO into moles. The equation for this is Mass/Molar mass = 1.0/147.015 = .006802 mol. Once we have done this we can calculate how many grams of NaCOᴣ we will need by the following equation: mol * molar mass = .006802 * 105.99 = .72 g. To calculate the theoretical value we used: 1g CaClᴤ*2HᴤO * (1 mol CaClᴤ*2HᴤO/147.015 g CaClᴤ*2HᴤO) * (1 mol CaCOᴣ/1 mol CaClᴤ*2HᴤO) * (100 g CaCOᴣ/1 mol CaCOᴣ) = .6802. Finally we can check this by: .72 g NaCOᴣ * (1 mol NaCOᴣ/106 g NaCOᴣ) * (1 mol CaCOᴣ/1 mol NaCOᴣ) * (100 g CaCOᴣ/1 mol CaCOᴣ) = .68. These calculations prove that my experimental values are in fact accurate with my calculations very close to the theoretical value. However, there were some sources for error. For example I needed .72 g of NaCOᴣ according to my calculations. The digital scale that was provided with our lab paqs only reads grams too one tenth and not hundredth. So I could measure .7 grams of NaCOᴣ with reasonable certainty but .72 had to be a guess. Also when combining the chemical NaCOᴣ I noticed that there was a little of the chemicals granules that wedge themselves in the crease along the bottom of the paper cup. This too would and could cause a error margin. There was some...
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