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5.06 percent yield lab report answers free printable version pdf

When complex chemicals are synthesized by many different reactions, one step with a low percent yield can quickly cause a large waste of reactants and unnecessary expense. Place the Magnesium ribbon into the ceramic crucible with the lid on. This is called the theoretical yield, the maximum amount of product that could be formed from the given amounts of reactants. The reason that the product had a higher mass than the reactant is because Mg bonded to O to form MgO, so the product had a higher mass because of the gain of an O atom. The Magnesium ribbon will be heated a fix rate in a ceramic crucible to it form its byproduct: MgO via oxidation/ synthesis. Materials: Percent Yield Virtual Lab Variables: Remember, controlled variables are factors that remain the same throughout the experiment. \[2 \ce{KClO_3} \left( s \right) \rightarrow 2 \ce{KCl} \left( s \right) + 3 \ce{O_2} \left( g \right)onumber \] In a certain experiment, \(40.0 \: \text{g} \: \ce{KClO_3}\) is heated until it completely decomposes. An independent (test) variable changes so that the experimenter can see the effect on other variables. Your theoretical yield calculation serves as your prediction for what you expect the lab to produce, and that will be determined later in the lab. Because this lab is virtual, summarize the steps used to collect your data.In addition, list and explain your controlled variables, independent variable, and dependent variable for this lab. LAB DATA Data Trial #1 Trial #2 Weight of empty crucible + crucible lid (grams) 26.698 g 26.687 g Weight of Mg metal + crucible + crucible lid (grams) 27.060 g 27.046 g Weight of Magnesium oxide + crucible + crucible lid (grams) 27.291 g 27.273 g LAB CALCULATION Mass of magnesium metal. \[\text{Percent Yield} = \frac{14.9 \: \text{g}}{15.7 \: \text{g}} \times 100\% = 94.9\% onumberonumber \] Step 3: Think about your result. See Balanced equation: LABORATORY PROTOCOL: This lab can be performed as an individual or with a partner (Recommended for Grade 10 and above) Obtain the following items: Clamp stand, Bunsen burner/ Hot plate, ceramic tile hot plate, ceramic crucible with its accompanying lid, Tongs, Protective eye ware and gloves.Begin by weighing the ceramic crucible with its accompanying lid; record the value.Obtain approximately 20-25cm of Magnesium ribbon provided to you. You're Reading a Free Preview Page 3 is not shown in this preview. When a chemist synthesizes a desired chemical, he or she is always careful to purify the products of the reaction. Step 1: List the known quantities and plan the problem. Procedure: Access the virtual lab. Since the actual yield is slightly less than the theoretical yield, the percent yield is just under \(100\%\). Of note, some experimental protocols instruct students to reheat the crucible and Magnesium ribbon several times and re-measure the weight each time (continuing until the final mass value is repeatedly constant). For help doing this, please see the video or the example calculation below.] Trial 1: .6 gTrial 2: .596 g Percent yield of magnesium oxide [This is calculated via %yield = (actual yield/theoretical yield) x 100] Trial 1: 98.8%Trial 2: 98.3% Average percent yield of magnesium oxide [This will be done by averaging the % yield of your 2-3 trials] STUDENT LAB DISCUSSION Magnesium (Mg) is weighed in a crucible and then heated to create a reaction with O to produce Magnesium Oxide (MgO). Typically, percent yields are understandably less than \(100\%\) because of the reasons previously indicated. Place the ceramic crucible containing the Magnesium ribbon above the Bunsen burner (see image below). This measurement is called the percent yield. Known Actual yield \(= 14.9 \: \text{g}\) Theoretical yield \(= 15.7 \: \text{g}\) Unknown \[\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% onumberonumber \] Use the percent yield equation above. Now we will use the actual yield and the theoretical yield to calculate the percent yield. The actual yield is the amount of product that is actually formed when the reaction is carried out in the laboratory. Much time and money is spent improving the percent yield for chemical production. What is the percent yield for the reaction? Controlled variables:bunsen burner, ring stand, ball of Mg metal ribbon, steel wool and a crucible Independent Variable: amount of Mg Dependent Variable: the yield Summary of Steps: Begin by weighing the crucible by itself, then with a steel wool, rub a specified length of metal(ex:1in) until it is oxidized and cut it into a ball of metal strips.Then place it on top of a Bunsenburner for approximately 10 minutes and then cool for 5 minutes. Title: Percent Yield Lab Objective(s): I will react Mg with the oxygen in the air by burning metal and determine the percent yield of Mg oxide Hypothesis: No hypothesis needed for this lab. [This will be calculated via direct measurement and subtracting the weight of the empty crucible + lid] Trial 1: .362 gTrial 2: .359 g READ: Decomposition of Hydrogen Peroxide Lab AnswersActual yield of magnesium oxide [This will be calculated via direct measurement and subtracting the weight of the empty crucible + lid] Trial 1: .593 gTrial 2: .586 g Theoretical yield of Magnesium oxide [This will done by taking the weight of your magnesium ribbon and converting it to Moles and solving via the Molar ratio. Step 2: Solve. Besides spills and other experimental errors, there are often losses due to an incomplete reaction, undesirable side reactions, etc. However, percent yields greater than \(100\%\) are possible if the measured product of the reaction contains impurities that cause its mass to be greater than it actually would be if the product was pure. By doing this repeated heating and measurement until a constant value is obtained would ensure all of the Mg ribbon had be oxidized to MgO therefore ensuring max percent yield. This relates to the law of conservation of matter which states that matter is not destroyed or created from nothing, however it can convert from one form to another. You will submit your completed report. To compute the percent yield, it is first necessary to determine how much of the product should be formed based on stoichiometry. These are the countries currently available for verification, with more to come! United States Canada Mexico American Samoa Guam Northern Mariana Islands Moldova Guernsey South Africa China Israel Japan South Korea Taiwan Australia New Zealand Argentina Brazil Chile Venezuela Percent Yield Lab Report Instructions: For this investigative phenomenon, you will need to determine the percent yield of magnesium oxide from the given reaction to determine if it is a useful commercial process. The amount of MgO formed will be dependent on the amount of Magnesium ribbon used, along with avoidance of human error (i.e incorrect weighing or loss of experimental product). This is an oxidation reaction because the Magnesium gained Oxygen and oxidized to create MgO. Known Given: Mass of \(\ce{KClO_3} = 40.0 \: \text{g}\) Molar mass \(\ce{KClO_3} = 122.55 \: \text{g/mol}\) Molar mass \(\ce{O_2} = 32.00 \: \text{g/mol}\) Unknown Apply stoichiometry to convert from the mass of a reactant to the mass of a product: \[\text{g} \: \ce{KClO_3} \rightarrow \text{mol} \: \ce{KClO_3} \rightarrow \text{mol} \: \ce{O_2} \rightarrow \text{g} \: \ce{O_2} onumberonumber \] Step 2: Solve. A great deal of research takes place to develop better ways to make drugs faster and more efficiently. \[40.0 \: \text{g} \: \ce{KClO_3} \times \frac{1 \: \text{mol} \: \ce{KClO_3}}{122.55 \: \text{g} \: \ce{KClO_3}} \times \frac{3 \: \text{mol} \: \ce{O_2}}{2 \: \text{mol} \: \ce{KClO_3}} \times \frac{32.00 \: \text{g} \: \ce{O_2}}{1 \: \text{mol} \: \ce{O_2}} = 15.7 \: \text{g} \: \ce{O_2} onumberonumber \] The theoretical yield of \(\ce{O_2}\) is \(15.7 \: \text{g}\). Chemical reactions in the real world do not always go exactly as planned on paper. Mass of empty crucible with lid Mass of Mg metal, crucible, and lid This is a laboratory report which examines the oxidation of Magnesium and its percent yield. In the course of an experiment, many things will contribute to the formation of less product than would be predicted. Chemists need a measurement that indicates how successful a reaction has been. What is the theoretical yield of oxygen gas? The percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage: \[\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%onumber \] Percent yield is very important in the manufacture of products. It is a very common lab report assignment for high school students across North America and the UK. Record your data and calculations in the lab report below. Analysis of how much of a compound is produced in any given reaction is an important part of cost control. Related Posts Weight it and record the value. The report will look to understand the percent yield of Magnesium Oxide (MgO) after heating varying amounts of Magnesium via a conventional Bunsen Burner or Ceramic hot plate. Many drugs have several steps in their synthesis and use costly chemicals. Cover the crucible and heat strongly for approximately 5-10 minutes.Turn off the Bunsen burner and when safely cooled, weight the entire ceramic crucible containing the Magnesium with the lid on; record the value . The mass of oxygen gas must be less than the \(40.0 \: \text{g}\) of potassium chlorate that was decomposed. Step 3: Think about your result. The world of pharmaceutical production is an expensive one. Example \(\PageIndex{1}\) Potassium chlorate decomposes upon slight heating in the presence of a catalyst, according to the reaction below. Hint: Using the same instrument, you should have the same number of digits to the right of the decimal. Don't forget to record measurements with the correct number of significant figures. The experiment is performed, the oxygen gas is collected, and its mass is found to be \(14.9 \: \text{g}\). The dependent (outcome) variable will change in response to the test variable. Solution First, we will calculate the theoretical yield based on the stoichiometry. Repeat the same experiment with a different cut of metal(ex: 2 in) Data: Type the data in the data table below.

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