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AP Chemistry Questions?
1. Mass of empty container 22.47
Initial mass of sample and container 25.29
Mass of sample and container after heating 24.43g
Student 1 is assigned the task of determining the number of moles of water in one mole of CuSO4 x nH2O
a) determine grams of anhydrous product
b) calculate total number of moles of water lost when the sample was heated
c) determine the formula of the hydrated compound
Student 2 does the lab as well. His data yields a different result. After doing his calculations, he determines that the formula of the hydrate is CuSO4 x 8H2O. The actual formula is CuSO4 x 5H2O.
The students were then asked to analyze their lab for error. One student said that as the copper sulfate hydrate was heating, small pieces of it popped out of the crucible and at the end of the lab, he noticed that there were several blue crystals on the table around the Bunsen burner. The other student thought that he didn’t hear the hydrate long enough.
B) identify the student (1 or 2) who did not beat the hydrate long enough. Explain how that error affected his calculations and final formula.
Sorry for the 2nd question I meant “heat” not beat
1 Answer
- Roger the MoleLv 77 months agoFavorite Answer
1. Supposing the missing units to be "grams":
a)
(24.43 g - 22.47 g) = 1.96 g CuSO4 (anhydrous)
b)
(25.29 g - 24.43 g) = 0.86 g H2O driven off
(0.86 g H2O) / (18.01532 g H2O/mol) = 0.047737 mol H2O
c)
(1.96 g CuSO4) / (159.6086 g CuSO4/mol) = 0.01228004 mol CuSO4 (anhydrous)
(0.047737 mol H2O) / (0.01228004 mol CuSO4) = 3.887 = 4
CuSO4·4H2O
B)
Student 1 got a mole ratio smaller than 5, so student 5 is the one who did not heat the hydrate long enough. By not heating long enough the "anhydrous" CuSO4 was not really anhydrous, so the mass and moles of "anhydrous" CuSO4 is too high and the calculated mass and moles of H2O are too low. So their ratio is too low.
