Data Table 1 Mass of aluminum cup 2.5g Mass of aluminum cup +alum 4.5g Mass of c

Question

Data Table 1 Mass of aluminum cup 2.5g Mass of aluminum cup +alum 4.5g Mass of cup + alum after 1st heating 3.5g Mass of cup + alum after 2nd heating 3.6g Mass of cup + alum after 3rd heating 3.1g Mass of alum 0.6 Mass of water lost 1.4 Please include all units and conversions to receive lifesaver points! A. Determine the moles of anhydrous KAl(SO4)2. B. Determine the moles of H2O lost. C. Calculate the ratio of moles H2O to moles KAl(SO4)2. D. Calculate the percent water: mass of water lost / mass of hydrated salt x 100%. E. Write the empirical formula for hydrated alum, based on your experiment.

Explanation / Answer

Take the mass of the water lost and divide that number by the mass of hydrated calcium sulftate. Then multiply your answer by 100. For example: To find the amount of water, subtract the mass of the anhydrous sample from the hydated one like this: Mass of hydrated salt: 21.626 Mass of anhydrous salt: 21.441 21.626 - 21.441 = 0.185 (amount of water) To find the percentage of water, divide the water mass by the hydrated mass number: .185/21.626 = 0.00855 Now all you need to do is multiply your answer by 100 to get the percentage: 0.00855(100) = .8554% To see this last part algebraically, you can say, "what percent of the hydrated salt (21.626) is the mass of water (0.185)?" This would look like: x%(21.626)=0.185

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