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Stoichiometry Lab Report Mg + O2 Maxide

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The labs purpose is to determine magnesium oxides percent yield. In this lab we will create a chemical reaction when two solutions are mixed. Those two solutions are magnesium and oxygen in the reaction Mg + O2 MgO. After these two solutions are mixed, magnesium oxide is produced. The percent yield of magnesium oxide is calculated and reported in this activity. The percent yield is dependent variable and the product amounts is the independent variable in this experiment. Stoichiometry is a method used to determine the most desired results that can occur within a lab (Khan). This is determined by balancing the chemical equation to calculate the amount of reactant and product there are. Ratios are used from balancing equations in order to find the amount of reactants needed for the reaction. These ratios are called the mole ratio, the stoichiometric ratio, or the stoichiometric factor. The mole ratio can be used as a conversion factor between different quantities. The first step to a stoichiometry problem is to balance the equation. Next, convert the known reactant amount to moles. Then, find the moles of the reactant using the mole ratios. Lastly, convert moles to mass. Doing this will give you your limiting and excess amounts (Khan). …show more content…

When all of the limiting reactant is consumed the reaction will stop. The excess reactant is what remains after all of the limiting reactant has been completely consumed. Excess reactant stops due to the absence of anything left to react with (Stoichiometry…). Limiting reactants are completely used in the reaction leaving nothing behind (Theoretical..). However, excess reactants have leftover product once the reaction is completed

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An impure reactant might decrease the percentage yield because impure substances contain less atoms of an element per mole than a pure substance. Consequently, there are less reactants in the reaction than what is expected. This causes the actual yield to be less than the theoretical yield so, an impure reactant should decrease the percentage yield of a reaction. 4.

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9.72 10-3 mol Limiting Reactants : CuCl2 and Na2CO3 Molar Mass CuCO3 = 63.55 g Cu + 12.01g C + (2 16.00g O) Molar Mass CuCO3 = 123.56g Mass of CuCO3 = Moles of CuCl2 1Mole of CuCO3Moles of CuCl2Molar Mass of CuCO3Mole of CuCO3 Mass of CuCO3 = 7.96 10-3 mol11 mol1 mol123.56g1 mol Mass of Moles and Mass of the Precipitate Produced (CuCO3) Moles of CuCO3 = Moles of CuCl2 1Mole of CuCO3Moles of CuCl2 Moles of CuCO3 = 7.96 10-3 mol 11 mol1 mol Moles of CuCO3 = 7.96 10-3 mol Mass of CuCO3 = Moles of CuCl2 1Mole of CuCO3Moles of CuCl2Molar Mass of CuCO3Moles of CuCO3 Mass

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Molar mass is the quantity in grams of one mole of a substance. The molar mass is found by adding the number of grams of each element to find the total amount of grams per mole. For example CaH2 has the elements of Calcium which has the molar mass of (40.078g/mole) and two Hydrogen which is (2 x 1.0079 g/mol).When the mass of both elements are added the molar mass of CaH2 is 42.0938g/mol. Moles are really important because it allows chemist to be able to have a standardize meausment which gives them the chance to compare compounds.

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That is where stoichiometry comes along, which is the quantitative relationships or ratios between two or more substances undergoing a physical or chemical change in a balanced equation. It identifies the mole ratio between reactants and products. To understand stoichiometry you have to understand the products and reactants, which involves how to solve balanced reactions. This allows chemists to determine how many moles of a product will be produced from a specific number of moles of a recant or how many moles of reactant are needed to produce a specific amount of a product.

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Stoichiometry is a method used in chemistry that involves using relationships between reactants and products in a chemical reaction, to determine a desired quantitative data. The purpose of the lab was to devise a method to determine the percent composition of NaHCO3 in an unknown mixture of compounds NaHCO3 and Na2CO. Heating the mixture of these two compounds will cause a decomposition reaction. Solid NaHCO3 chemically decomposes into gaseous carbon dioxide and water, via the following reaction: 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g). The decomposition reaction was performed in a crucible and heated with a Bunsen burner.

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Post Lab Questions: 2Mg + O2 →2MgO 0.29gMg 1molMg24.312molMgO2molMg100= 4.40molMgO ←theoretical yield 2Mg + O2 → 2MgO →2/24 =

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Monitor the reaction and when when the reaction is near completion let some smoke escape by tilting the lid of the crucible When the reaction ceases, turn off the Bunsen burner and let the crucible cool completely before handling it. Weigh the crucible and record the weight Using a pipette, add a small volume of water to the solids in the crucible Stir the mixture with a glass rod until the mixture forms a paste Return the crucible to the Bunsen burner and heat it for several minutes until all the water has evaporated and the solids have turned light grey → indicating conversion to magnesium oxide. Turn off the Bunsen burner and let the crucible cool completely Weigh the crucible with the lid and record the weight Subtract the initial weight of the crucible and lid from the final weight of the crucible, lid, and magnesium oxide to obtain the mass of magnesium that reacted with oxygen. → this information will be used to calculate the empirical formula for magnesium oxide and prove it. Clean the crucible thoroughly to ensure it's ready for the next use.

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Continue taking data until you have enough data to record and make a thorough graph. 3. Plot your data to create two lines with an intersection on a graph. Repeat any measurements that do not fall near the best-fit line. From the intersection, calculate the mole ratio of the reactants.

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g mass of magnesium oxide=0.25 mass of magnesium oxide - mass of Mg + mass of O 0.25g= 0.122g + mass of O mass of O = 0.25g- 0.122g = 0.128g moles of Mg = 0.122g x 1 mol of Mg divided 24.30g Mg ( mass of magnesium) = 0.122 mol Mg moles of O = 0.25 g O x 1 mol of O divided by 16g of O = 0.25 mol o the ratio = 1:1 there is one mol for mg for one mol of O, therefore there will be 1 atom of mg for 1 mol of oxygen. therefore the empirical formula of magnesium oxide is MgO. Conclusion: the main purpose of this experiment is to determine the reaction and the magnesium oxide’s empirical formula which is when magnesium reacts with oxygen in the air.

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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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The central purpose of this experiment was to determine the experimental empirical formula of an oxide of magnesium by performing a synthesis reaction. It was hypothesized that the formula that was derived from the recorded data would be identical to the theoretical empirical formula. After performing calculations with the data that had been collected within the duration of the experiment, it was deduced that the empirical formula of the product generated by the synthesis reaction was Mg5O6. Since the calculated formula did not match the postulated formula, MgO, the inceptive hypothesis was confuted. There was a large discrepancy in the results of students who replicated this experiment, the variation in data possibly having been evoked by the dissimilarity in

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