Experimental value Accepted value x 100 Percent Error Accepted value In part A,

Question

Experimental value Accepted value x 100 Percent Error Accepted value In part A, you will use pure the first part the experiment to practice technique. By finding the mass of water lost from the sample and using you can how much hydrate must have been present initially You will know how much hydrate you initially weighed out, so comparing the two masses (initial mass from the balance and the experimental mass from stoichiometry) will you to check the accuracy of your method part B, you will be given a mixture of sodium sulfate (an anhydrate) and alum (the hydrate), but the mass percent of the sodium sulfate is unknown. Utilizing the methods employed in part A will allow you to determine th mass your sample, and by difference, you will be able to determine the mass and mass percent of sodium sulfate. You must work very carefully during this experiment because small errors in your technique can cause a relatively large error in your final mass Additionally, careful work will decrease the amount of time necessary to meet the experimental criteria. So be patient, especially with the cooling part of experiment, and use techniques when handling the clean techniques. your glassware. The instructor will illustrate Prelaboratory assignment 1. Answer the following questions and turn them in at the beginning of laboratory class. Write the balanced equation for the thermal decomposition of alum. b. What is the molar mass of alum? c. What is the percent of water in the formula for alum? d. What is the purpose of drying the empty beaker in watch glass before adding the hydrate? e. Draw the set up used for heating the beaker, watch glass, and hydrate What data will allow you to assess accuracy in this experiment? g. What is the equation for relative percent error? h. In part B, what is the mixture being heated in the beaker? In part B, what happens to each material in the mixture upon heating? j. In part B, write the balanced equation for the thermal decomposition of the unknown [Hint: There is only one reactant

Explanation / Answer

Pre lab

a. Equation for thermal decomposition of alum,

KAl(SO4)2.12H2O ---> KAl(SO4)2 + 12H2O

b. molar mass of alum = 474.3884

c. percent water = 12 x 18 x 100/474.3884 = 45.53%

d. We do not want any external water to be added to alum and thus drying of beaker prior to alum addition is done.

f. The data accuracy depends upon correct weight measurement and controling no extra water is added from the outside environment. The heating is done until we get a constant weight.

g. Relative percent error = (experimental value - actual value) x 100/actual value

h. Mixture of alum with water of crystallization is heated.

i. H2O gets evaporated and anydrous alum salt is obtained.

j. with alum being the reactant, thermal decomposition equation is same as in part a).

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