Mass Relationships in Chemical Reactions Essay Example for Free

Mass Relationships in Chemical Reactions Essay Aim The aim of this experiment is to show that a reaction doesnt have always 100% yield by reacting NaHCO3 and HCl and determining the amount of the products to calculate actual yield. Introduction A chemical reaction will be quantitative if one of the reactants is completely consumed. In this experiment sodium bicarbonate and hydrochloric acid start a reaction. The formula of this reaction is below. NaHCO3 + HCl NaCl + H2O + CO2 Observations In this experiment, sodium bicarbonate is put in an evaporating dish and some amount of HCl is added in the dish and the reaction started. Bubbles are formed and CO2 gas is produced and the reaction started to make sound. There was also water vapor formed. White NaHCO3 started to turn into a colorless liquid after adding HCl. As the reaction takes place water is started to form. NaCl was dissolved in water, so salty water is heated to obtain NaCl. As the liquid is heated it turned into a yellowish color for a few seconds. Then it started bubbling and water vapor is formed. Raw Data: Trial # Mass of Dish+NaHCO3+Lid +- 0.1 (g) Mass of NaCl+Water+Dish+Lid +- 0.1 (g) Mass of NaCl+Dish+Lid +- 0.1 (g) 1 64.14 g. 72.16 g. 63.28 g. 2 65.14 g. 72.95 g. 63.91g. Mass of Evaporating Dish + Lid: 62.14 +-0.1 g Processed Data: Trial #1 64.14 62.14 = 2 g NaHCO3 72.16 62.14 = 10.02 g NaCl + H2O 63.28 62.14 = 1.14 g NaCl Trial # 2 65.14 62.14 = 3 g NaHCO3 72.95 62.14 = 10.81 g NaCl + H2O 63.91 62.14 = 2.07 g NaCl Trial # Mass of NaHCO3 (g) Mass of NaCl + H2O (g) Mass of NaCl (g) 1 2 g 10.02 g 1.14 g 2 3 g 10.81 g 1.77g Calculations Na: 14.01 g/mol, H: 1.01 g/mol, Cl: 35.45 g/mol, O: 16 g/mol, C: 12.01 g/mol NaCl= 49.46 g/mol H2O= 18.02 g/mol NaHCO3: 75.03 g/mol Mole number of NaHCO3 = mole number of NaCl Trial #1 2 / 73.03 = 0.0274 mol NaHCO3 1.14 / 49.46 = 0.0230 mol NaCl Theoretical Yield: 0.0274 mol NaCl Percent Yield: 0.0230 / 0.0274 = 0.8394 x 100 = 83.94% Trial #2 3 / 73.03 = 0.0411 mol NaHCO3 1.77 / 49.46 = 0.0358 mol NaCl Theoretical Yield: 0.0411 mol NaCl Percent Yield: 0.0358 / 0.0411 = 0.8710 x 100 = 87.10% Conclusion The results are 83.94% for trial #1 and 87.10% for trial #2. Trial #2 is more accurate. The accepted value is 100%. The percentage errors are 16.06% for trial #1 and 12.90% for trial #2. The uncertainties are too small to calculate on the results. Random errors presented in this experiment. All the errors were done by human beings. There werent any errors due to a flaw of a machine or the procedure. Evaluation When salty water is heated on the first trial, the substance started to spill around, because the substance is heated with high amount of heat and faster than it should be. As a result, some of the NaCl which stuck on the lid and spilled around was lost, so the result of the first experiment is not accurate. Other reasons that changed the results may be all NaHCO3 may not be dissolved. Too much HCl may be added on the dish. There may be still water molecules left on the salt after heating. To get more accurate results, the experiment should be done more slowly than this experiment. Especially the heating process should be done slowly, so the evaporation can be observed more carefully.