Repeat the titration until there are two titres within 0.1cm3 of each other. Record results in a suitable table. Results: Rough Titre: 7.653 First Run: 6.553 Second Run: 6.453 Third Run: 6.553 Calculations: During the titration, iron(II) ions are oxidised to iron(III) ions and manganate(VII) ions are reduced to manganese(II) ions. The equation is as follows: 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) ? 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) The above equation shows that one mole of manganate(VII) ions reacts with 5 moles of iron(II) ions in acid solution.
Calculate the percent error in the molar mass value. Enter both values in the Data Table. Molar Mass Na2CO3 = 105.99 g/mol – this is the closest molar mass to what I calculated, so the unknown M2CO3 must be sodium carbonate. Percent Error: 93.08g/mol – 105.99 g/mol (100) = 12.18% error 105.99 g/mol DISCUSSION Review the procedure and list the possible sources of error that would cause either the molar mass of the unknown to be (a) too high or (b) too low. The goal of this lab was to discover the unknown group 1 metal (M) of the compound M2CO3 by dissolving the compound in water and adding a solution of calcium chloride, CaCl2 to the solution in order to precipitate the carbonate ions to reveal the molar mass of the unknown element, thus determining the identity of the unknown element.
This was done by taking small amounts of the unknown acid. It was then placed into capillary tubes and inserted into the Bibby Sterilin Device. Starting with a high plateau to find a quick melting point and then proceeding to find an actual melting point. The next experiment that aided in finding more characterizations of unknown #2651145-PLF13 was equivalent weight. In order to find the equivalent weight a titration of the unknown acid had to be conducted.
Let’s experiment and observe Whether these ions will have a reaction or indeed soluble in water. Experimental Procedure Let’s begin this experiment by obtaining a 100ml beaker from our equipment tool bar list, then we are going to add 50ml(s) of Potassium Chromate 1M solution to that beaker, The next step we are going to add another ionic compound of 50ml of Lead (11) nitrate 1M into the same beaker as the Potassium Chromate. As we observe the beaker, we notice a reaction has occurred and a precipitate has formed and settled on the bottom of the beaker. The next step is we are going to filter the resulting precipitate into a 250 ml Erlenmeyer Flask with a Buchner funnel. Now that we have filtered the precipitate, we will then place the resulting precipitate into a test tube to measure the weight.
5.51: Which Reagent is limiting and How Much Precipitate is formed? SCH-3UI-03 David Yu Mrs. Hatton Due Date: May 5, 2012 Cut-Off Date: May 12, 2012 Purpose: To experience and use what you have learned in class about gravimetric stoichiometry by predicting and determining the mass of precipitate of two reactants and then comparing what you experience and what you calculated. Background: Avogadro’s constant is 6.02 x 1023 to find the number of entities. A mole is a useable amount of chemicals that is practical to use. The molar mass of a compound or atom is the mass of 1 mole of anything; this is relative to the atomic mass from the periodic table.
In each trial, the initial reading, final reading and the volume of HCl used was recorded down as quantitative results. The average volume of hydrochloric acid was found to be 12.03mL. The amount of sodium carbonate in the 10.00ml of solution was found to b 0.05 mol. The amount of hydrogen chloride that was dissolved in the average volume of acid is 0.365g. Through these calculations, the concentration of hydrochloric acid was found to be 8.3 mol•L-1.
For reaction (II) Pb (NO3)2 + 2KI -> PbI2+2KNO3 Lead nitrate is soluble, so it gets written as ions. The same goes for potassium iodide and potassium nitrate. Complete Ionic equation: 2 K+1 + I-1 + Pb2+ + NO32- -> PbI2 + 2K+1 + NO3-1 Net Ionic Equation: Pb+2 + 2 I-1 -> PbI2 Warm-Up Exercise 2 In this lab you will mix 25 mL of 0.05M lead nitrate with 1.4 mL of 0.025M sodium carbonate. After the reaction occurs, you will filter the solution to remove the precipitate. You will then test the remaining solution for excess lead ion and for excess carbonate ion… Imagine that you mix the two volumes and then freeze frame the reaction so the it does not proceed: 1.
Calculate the surface area of the circle formed (πd2/4): Surface area = .785 cm2 2. Calculate the number of molecules on the top layer. We must convert the surface area in centimeters squared to nanometers squared and then multiple that by the surface area of a sodium stearate molecule. Convert the surface area of the circle formed (#1) to molecules per layer using the matrix below: Answer = 4.76*10^14 molecules/top layer 3. Calculate the concentration of grams of sodium stearate per milliliter of diluted solution.
After 10 seconds, the colorless mixture suddenly turns blue. Concept: Demonstrates a typical clock reaction; shows the effect of the interaction between chemical reactions that have different rates. Materials: • Solution A o 0.6 grams Starch o 30 mLs of Acetic Acid o 4.1 grams of Sodium Acetate o 50 grams of Potassium Iodide o 4.7 grams of Sodium Thiosulfate ▪ Allow mixture to cool and dilute to 1 liter with distilled water o 1 liter flask • Solution B o 500 mLs of 3% Hydrogen Peroxide o 500 mLs of distilled water o 1 liter flask o Safety: Hydrogen Peroxide can be irritating to skin and eyes. Wear safety goggles and gloves. Procedure: Mix the two solutions together.
Graduated cylinder 8. Calculator 9. Wash bottle Procedure: To commence, first I gathered all the material needed to keep from being too messy with the experiment. Next I went to collect my unknown liquid and metal since the material and purpose was to find the mass and volume of them and located their place of origin from the data collected. Once all the equipment has be gather, step one was to clean the flask because a dirty flask might have an effect on the precise mass of it.