based on the data below determined the complete formula for a barium chloride hy
ID: 1085108 • Letter: B
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based on the data below determined the complete formula for a barium chloride hydrate (#2 in picture) W. Post-Laboratory Assignment egin a new page in your laboratory notebook and answer the following questions. Explain your easoning and write legibly. It is not necessary to copy the questions into your notebook, but early identify each question by number. Evaluate the accuracy of your hydrated formula based on the percent error in the mass of anhydrous compound produced. Be sure to discuss how the "x" in your formula is 1. affected: is it too high or too low? 2. Based on the data below determine the complete formula for a barium chloride hydrate. Mass of empty crucible Mass of crucible and hydrated salt Mass of crucible and salt after 1t heating Mass of crucible and salt after 2nd heating Mass of crucible and salt after 3rd heating 24.32 g 26.70 g 25.93 g 25.88 g 25.88 g Copper(ll) nitrate trihydrate was found to contain 37.5 g of anhydrous salt. What was the mass of the initial hydrated compound subjected to the same gravimetric method used in this lab? 3. esults and Conclusion a brief discussion of results, signifcant implications, errors, and their effect on the final ts. (Please note that simply stating "human error" will not be accepted. Explain important ausible errors and their effect on your final results.)Explanation / Answer
1. Mass of crucible and hydrated salt = 26.70 g.
Mass of empty crucible = 24.32 g.
Mass hydrated salt (taken initially) = 26.70-24.32
= 2.38g.
2. Mass of crucible and salt after 1st heating = 25.93g.
Mass of empty crucible = 24.32g
Mass of salt after 1st heating = 25.93-24.32
= 1.61 g
3. Mass of crucible and salt after 2 and 3 heating = 25.88g
Mass of empty crucible = 24.32g
Mass of salt after 2 and 3 heating = 25.88 - 24.32
= 1.56 g
As there is no change in mass after 2 and 3 heating so it may be assumed that whole water is removed from salt and what we get is anhydrous salt. So mass of anhydrous salt = 1.56g
Mass of water in hydrated salt = mass of hydrated salt - mass of anhydrous salt
= 2.38 - 1.56
= 0.82g
Moles of water in hydrated salt = 0.82/18. (18 being molecular mass of water)
= 0.0455g
Moles of anhydrous BaCl2 = 1.56/208.2 (208.2 being molecular mass of BaCl2)
= 0.0075g
Number of H2O in hydrated salt = 0.0455/ 0.0075
= 6
Formula of hydrated BaCl2 = BaCl2.6H2O
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