Name Post lab Questions -Expe.2-Experimental Analysis of Hydrates In answering t

Question

Name Post lab Questions -Expe.2-Experimental Analysis of Hydrates In answering these questions you may use your textbook, lab maninal, lab You may not confer with any other members of the class nor use your eomic answers. These questions must be completed and urned in to your lab instructor before you leave lab. you may use your textbook, lab manual, lab notebook, or notes from class search for electronic device to 1. (2 pts total (1pt) A student determined the % water of a hydrate in three tria below, Given the list of possibilities in your lab book, which salt was his unknown a. Is. The results were as shown Triall 20.6 % water Trial #2 19.9 % water Trial 3 20.5 % water (l pt) The student claims that the major error that occurred was in not heating the hydrate long enough. Do you agree or disagree with the student's claim? Explain your reasoning b. 2. (2 pts) A Chemistry 131 students obtains 0.480 g of Hydrate #1. Ile then proceeds to heat the sample until all moisture is removed. The final miass of the anhydrous salt was 0.320 g, what was the % water in the hydrate? Show work. Use correct significant figures and units. 3. (2 pts) Perform the following calculations. Use correct significant figures and units if necessary a. 100.300 g-1.00g b. 1,000.00 x 1.0= Note: Insert Divider Under Copy Sheet Before Wrt

Explanation / Answer

Hydrates

1. a. Since the list of possibilites chart from text book is missing, we cannot predict the salt type here.

However, one can do it (while having chart with you) by comparing the molar mass of salt and water and see If the result is 20% (approx) as obtained from three trial runs. [most likely salt is calcium or barium chloride hydrate]

b. The kost likely error comes from not heating the salt for sufficient time. This assumption of the student is correct as not heating the salt long enough would result in higher mass of salt left and lower mass of water lost. therefore giving lower percentage of water of hydration in the salt.

2. Initial mass of hydrate salt (#1) = 0.480 g

after heating to remove moisture,

final mass of dehydrated salt = 0.320 g

mass of water lost = 0.480 - 0.320 = 0.160 g

moles of water = 0.160 g/18 g/mol = 0.0089 mol

moles of hydrate salt (#1) = 0.480 g/molar mass of salt #1

mole ratio of water to salt = moles of water/moles of salt = number of water of hydrate present

percent water in the salt = moles of water/moles of hydrate salt (#1)

[again put molar mass of #1 salt to get correct answer]

3. Calculations using significant figures

a. 100.300 g - 1.00 g = 99.300 g

b. 1,000.00 x 1.0 = 1,000.00

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