Erhard a type of chemical reaction where two

Erhard Ruettimann22.01.2018G BlockHow the Volume of a Solution Affects the Percentage YieldIntroduction:Hypothesis:My Hypothesis is that if I increase the volume of the calcium chloride solution, it will also increase the mass of the reactant, this will make my percentage yield become more accurate because the amount of precipitate lost or gained during the procedure due to errors would be less significant when compared to the total mass of the precipitate.Background information:CaCl 2 (aq)+Na2CO3 (aq)?CaCO3(s)+2NaCl(aq) is the balanced chemical reaction that I will use to investigate my hypothesis. Calcium Chloride and Sodium Carbonate are the reactants and Calcium Carbonate and Sodium Chloride are the products, Calcium Carbonate being the precipitate.A double displacement reaction, also known as a double replacement reaction, is a type of chemical reaction where two compounds react, and the positive ions and the negative ions of the two reactants switch places, forming two new compounds or products. ( carbonate is a chemical compound in the formula CaCO?. It is a typical substance found in rocks as the minerals calcite and aragonite and is the fundamental part of pearls and the shells of marine living beings, snails, and eggs. ( concentration is a measure of the concentration of a solute in a solution, or of any chemical species, in terms of amount of substance per unit volume. A commonly used unit for molar concentration in chemistry is the molar which is defined as the number of moles per litre (unit symbol: mol/L or m). A solution with a concentration of 1 mol/L is equivalent to 1 molar (1 m). ( calculated or expected amount of product is called the theoretical yield. The amount of product actually produced is called the actual yield. When you divide actual yield by theoretical yield you get a decimal percentage known as the percent yield of a reaction. ( is a branch of science that studies and measures the amount of matter in chemical reactions. It can be used to predict the number of things that will be made in a chemical reaction. ( waterFilter PaperStirring rodFunnel200ml Beaker50ml Graduated CylinderScaleDrying OvenWeighing BoatsConical flasksCalcium ChloridePetri DishesSodium CarbonateSeparately weigh out 15.9 g of sodium carbonate and 11 g of calcium chloride on two weighing boats.Mix the sodium carbonate with 150 ml of distilled water and the calcium chloride with 100 ml of distilled water to create two 1 M solutions.Mix 20 ml of calcium chloride solution (1M) and 50 ml of sodium carbonate solution (1M) in a conical flask.Place the filter funnel in the neck of another conical flask. Fold the filter paper to fit into a large funnel (in half and half again) and place it inside the funnel which is placed on top of the conical flaskSwirl the reaction mixture gently and pour a little at a time into the filter paper in the funnel. Only pour in enough solution at a time to leave the solution level 1 cm below the rim of the filter paper. Allow filtering through.A clear solution should collect in the flask. If the solution is not clear, and white cloudiness remains in it, you will need to repeat the filtration.Remove the wet filter paper carefully from the funnel and place on a clean dry paper towel. Label with your name and leave in the drying oven, until it has dried completely (a few hours)Repeat the steps one through six, two times increasing the volume of the calcium chloride solution by 10 ml every trial.Individually weigh the dried filter papers with precipitate and record their masses, subtracting the mass of the filter paper. This is the experimental yield.Calculate the theoretical yield using stoichiometry and calculate the percentage yield for each trial by dividing the experimental yield by the theoretical yield.CaCl2 (111.0 g/mol)TrialVolume (ml)Amount of CaCl2 (mol)Mass of CaCl2 (g)Concentration (M)1200.022.21.02300.033.31.03400.044.41.0Table 1: Volume of solution and mass of CaCl2 for individual trialsNa2CO3 (106.0 g/mol)TrialVolume (ml)Amount of Na2CO3 (mol)Mass of Na2CO3(g)Concentration (M)1500.055.31.02500.055.31.03500.055.31.0Table 2: Volume of solution and mass of CaCl2 for individual trialsVariables:VariablesWhat? (with units)How is it measured?Other Info? Range?IndependentVolume of Calcium chloride (CaCl2)Liters (L)Range: 0.2, 0.3, 0.4 LDependentPercentage yield of calcium carbonateMeasured in grams of the precipitate obtained and calculated as experimental yield divided by theoretical yield.The formula for calculating percentage yield:ControlledReactantsSubstances pre-labeledReactants:Calcium Chlorideand Sodium carbonateVolume of sodium carbonate solutionGraduated cylinder0.05 LConcentration of calcium chlorideMolarity (mol/l)Concentration: 1 MConcentration of Sodium carbonateMolarity (mol/l)Concentration: 1 MDrying Oven Same drying oven and equal time on the ovenAll of the samples were put on the same oven for the same duration of one night.SolventThe solvent was taken from the same containerThe solvent is distilled Water Filter paperI used the same type of filter papers in all trialsFilter paper from same brandResults:Raw Data :TrialVolume (ml)YieldMass(g)± (g)± (%)1202.30.14.32303.30.13.03404.20.12.4Table 3: Experimental yield for different volumes of CaCl2 solutionProcessed Data:TrialVolume (ml)Experimental YieldTheoretical YieldPercentage Yield (%)Mass(g)± (g)± (%)Mass (g)1202. 4: Percentage yield for different volumes of CaCl2 solutionGraph 1: A graph to show the relationship between the volume of a solution and the percentage yield.Calculations:Here are my example calculations for the amount of sodium carbonate solute to add to the solvent:Sodium Carbonate:Concentration: 1 MVolume:0.15 L or 150 MlMolar Mass:106.0 g/mol Na2CO30.15 L * 1 M =0.15 mol0.15 mol * 106.0 g/mol =15.9 gMass of sodium carbonate required: 15.9 g Example calculation of theoretical yield:20 ml of calcium chloride solution (1M)50 ml of sodium carbonate solution (1M)CaCl2 +Na2CO3 = CaCO3 + 2NaCl(0.02)        (0.05)            (0.02)Molar mass of CaCO3: 100.1 g/molTheoretical yield : 100.1g/mol *0.02 mol = 2.0 gExample calculation of percentage yield for trial 1:Percentage yield = actual yield theoretical *100Percentage yield =2.3 g/2.0 g * 100 = 115%Conclusion and Evaluation:Conclusion:My hypothesis was that as the volume of the calcium chloride solution increases, then the percentage yield will become more accurate because there would be a smaller amount of precipitate that could be lost or gained during the procedure due to errors. My data supports my hypothesis. I conducted three trials, each with a different volume of calcium chloride solution, and measured the mass of the created precipitate, as seen in table 2. Then, using stoichiometry, it is possible to calculate the theoretical yield of each reaction. Using this information, I then calculated the percentage yield by dividing the experimental yield by the theoretical yield, as seen in table 3. Percentage yield is an indicator of how efficient your reaction and procedure were in achieving the theoretical yield. As seen in table 3, the percentage yield for 20 ml of solution was 115%, whereas my percentage yield for 40 ml was 105%. This clearly supports that with the increase in the volume of solution, the accuracy of the percentage yield increases significantly, in this case, by 10%. This trend of increasing accuracy with increasing volume is also displayed in graph 1. All of my percentage yields were above 100%, which is impossible, as this would contradict conservation of matter, but it still shows a relevant trend of increasing accuracy.One possible reason why my percentage yield became more accurate with increasing volume of solution could’ve been due to the mass of precipitate gained or lost through errors during the procedure became less relevant with a greater total mass of precipitate. In the first trial, when we using only 20 ml of the solution, this would have impacted the data significantly, whereas in trail 3, when we used 40 ml of the solution, the error of a few millilitres would have had a minor impact on the total mass of calcium carbonate. Another thing similar can be seen in table 2, where the percentage error on the mass of calcium carbonate due to the scale decreases with an increase of volume, even though the absolute error remains constant. I assume that this is the most significant reason for this trend. We also always added an excessive amount of sodium carbonate solution to avoid a limiting factor and to make sure all of the calcium chloride reacted. All in all, my data supports my hypothesis of increasing percent yield accuracy with the increased volume of solution. My evaluation:Benefit:Strength:This altogether spared time and ensured the molarity of the arrangements stayed consistent amid all trials.I made one bigger batch of each solution at the beginning of the experiment rather than during each trial.This made data analysis easier and made comparing percentage yield between different trials more valuable. It also made it easier to recognize certain trends.Throughout my trials, the different volumes used each differed by 20 ml.This allowed me to measure solely how the change in volume of calcium chloride solution affected the percentage yield without having to worry about the other reactant possibly limiting the experimental yield.In all of my trials, I always used an excess amount of sodium carbonate solution to ensure that it wasn’t a limiting factor. Weakness/LimitationImpact on dataModifications suggestedSome of the liquid being filtered passed through the funnel without being filtered. This occurred due to the filter paper overflowing and the unfiltered liquid flowing into the conical flask.This would change the mass of the precipitate as more than only calcium carbonate would be filtered out. This would produce a less accurate experimental yield and therefore also a less accurate percentage yield.Conduct the experiment over a longer period of time and poor in the liquid more gradually to avoid overflowing.Only one trial was completed for each volume of calcium chloride solution.Due to there only being one trial for each volume, it is not possible for us to know whether or not that trail was faulty or an outlier. Conducting multiple trials reduces the impact or errors and produces a more accurate trend and percentage yield.It would have made my experiment more accurate if we completed at least 3 trials per volume.There might have been little leftover precipitate left in the glassware and equipment. Less of the precipitate was weighed and therefore results may have turned out lower. Which changed the accuracy of my percentage yield.Rinse the glassware and equipment with distilled water to remove all remaining precipitate