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Lab Report Balanced Chemical Equation

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1. For the demo experiment, the balanced chemical equation is as follows: (NH4)2Cr2O7(s)=Cr2O3(s)+N2(g)+4H2O(g). After the lightning of Ammonium dichromate, Chromium (III) oxide was formed while the Nitrogen and Water escaped into the atmosphere in a gaseous phase. Ammonium dichromate((NH4)2Cr2O7) gave rise to Chromium (III) oxide (Cr2O3), Nitrogen Gas(N2) and water (H2O) In terms of microscopic level, the ratio between reactants and products is as follows. One mole of Ammonium dichromate will give rise to one mole of 1 mole of Chromium (III) oxide and 1 mole of Nitrogen gas and 4 moles of Water is gaseous phase. To convert these into formula units, 1 mole of any compound will equal 6.022X1023. So based on this, 1 mole of Ammonium dichromate is 6.022X1023 formula units. 1 mole of Chromium (III) oxide is 6.022X1023 …show more content…

Based on the stoichiometric, the theoretical yield, actual yield and percent yield can be determined. In order to calculate the percent yield, the value for theoretical yield is required. Percent yield is determined by diving the actual yield by theoretical yield and multiplying by 100. The theoretical. The theoretical yield for Zinc Sulfide is 0.49 grams but the actual yield is 0.38 grams. So if 0.38 is divided by 0.49 and multiplied by 100 then the percent yield for Zinc Sulfide would be 77.6%. When it comes to Sodium Chloride, the theoretical yield is 0.58 grams and the actual yield is 0.45 grams. So when 0.45 grams is divided by 0.58 grams and multiplied by 100, the percent yield would be 77.5% of Sodium chloride. The actual yield is directly taken from the mass of the products in the experiment while the theoretical yield is determined by using stoichiometric calculations. To determine the theoretical yield, the reactants should be converted from grams to moles based on the coefficients in the chemical equation and the moles should be incorporated into the mass of the reactants. When calculated, this would give rise to the theoretical

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An error that could have been present during the lab includes not letting the zinc react completely with the chloride ions by removing the penny too early from the solution. For instance, the percent error of this lab was 45.6%, which was determined by the subtraction of the theoretical percent of Cu 2.5% and the experimental percent of Cu 3.64% and dividing by the theoretical percent of Cu 2.5%. This experiment showed how reactants react with one another in a solution to drive a chemical reaction and the products that result from the

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The percent yield was calculated to be 129.7%. This high percent yield resulted from the presence of

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As seen in table 1, the theoretical yield was .712 g of C_17 H_19 NO_3. The % yield of this experiment was 7.51 % of C_17 H_19 NO_3. . This low yield can be explained from a poor recrystallization technique combined with potential contamination. Throughout the experiment, the mixture changed color from green, orange, to yellowish lime, and eventually clear.

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In order to begin this experiment, first one must find the balanced chemical equation for the reaction which occurs between the aluminum and copper (II) chloride. This balanced equation being 2Al(s)+3CuCl2 (aq)3Cu(s)+2AlCl3 (aq). After finding this equation, one must use the process of stoichiometry in order to find how many grams of aluminum are needed in order to produce 0.15 grams of copper. In this experiment, the purpose was to produce between 0.1 and 0.2 grams of copper, so one should attempt to produce 0.15 grams of copper seeing as it is the average of those two numbers. The first step in the stoichiometric process which one has to complete is finding how many grams of copper are in one mole of copper.

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(150.22g/mol)(3.5 x 10^-3 mol of nucleophile) = 0.525 g Actual yield = 0.441 g, Percent Yield = (0.441g/0.525g) x 100% = 84% 10. Percent recovery from recrystallization = (0.172g/0.441g) x 100% = 38% 11.

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3. In this experiment, the percent yield was 90%. This number implies that there was little error in this experiment. However, this result could have been caused by certain external factors.

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To test for the presence of the ammonia cation, a scoop of the unknown compound was mixed with NaOH to see if the resulting solution had a noticeable smell that would affirm the presence of ammonia. A moist piece of pH paper was then held near the solution to see if the ammonium would dissolve the water on the pH paper. The next cation test performed was the flame test. A gram of the unknown compound was held by a metal stick above the flame of a Bunsen burner to see what color flame the compound would produce. The chart in the lab manual on page 63 was used to determine what cation correlated with what flame color.

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The percent recovery of the copper was calculated using the equation, percent recovery = (the mass of the copper recovered after all the chemical reactions/the initial mass of the copper) x 100. The amount of copper that was recovered was 0.32 grams and the initial mass of the copper was 0.46 grams. Using the equation, (0.32 grams/0.46 grams) x 100 equaled 69.56%. The amount of copper recovered was slightly over two-thirds of the initial amount.

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Copper Cycle Lab Report Ameerah Alajmi Abstract: A specific amount of Copper will undergo several chemical reactions and then recovered as a solid copper. A and percent recovery will be calculated and sources of loss or gain will be determined. The percent recovery for this experiment was 20.46%.

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The final product weight for percent yield was only the solid E product, which missed one half of the final product produce. If both products were weight, the percent yield would have been larger that it was. Instead of 22.33%, it could have been 44.66%. To prove that both products were obtained, but only one of the two products was analyze, a TLC plate of the DCM layer, that contains both products, and of the final product, was obtain.

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The actual data is the result on our experiment vs theoretical, which is based on the calculations above. I have also learned to pay more attention to draining out all of the product completely before continuing to test the experiment, as any small drop of contaminant can veer our results into a different

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After the reaction is finished, the percentage composition of each element in the product can be found and used to calculate the empirical formula, which is the lowest whole number ratio

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I. Title: Mass and Mole Relationships in a Chemical Reaction II. Background: Percent yield is the ratio of actual yield to theoretical yield. Amount in percent of one product formed in chemical reaction. Actual yield is the information found is experiments or is given.

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To determine the rate of reaction there are many method to be used for example, measuring the mass after the product has been added and measuring the difference in mass on the duration of a digital scale. Another method, which will be used in this experiment is using a gas syringe to measure the volume of the gas which has been produced. The cylinder inside, will be pushed out to show a quantitative presentation of the volume produced by the reaction. Hypothesis

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Properties of Substances Express Lab 1)The purpose of this lab was to compare the physical properties of different types of solids and how the properties of solids are determined by their intermolecular forces and their intramolecular bonds. Then we were to classify each type of solid as either ionic, metallic, non-polar molecular, polar molecular, or network. Paraffin wax classified as a non-polar molecular, Silicon dioxide was classifies as a network, Sodium chloride was classified as ionic, Sucrose was classified as polar molecular and Tin was classified as metallic. (2)The intermolecular forces that are present in Paraffin wax are dispersion forces, because it is non-polar and carries a negative charge. Followed by Sucrose that has

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