How much water was in CuSO4.5H2O? That is
0.500g x (5mol H2O/1 mol CuSO4.5H2O) = xx g
0.500 g - 0.427 = yy g = mass H2O lost.
How much water still remains in the CuSO4.xH2O? That is xx g - yy g = zz g.
How much of that 0.427 g is CuSO4 with no water. 0.427 - water remaining so
0.427 - zz = w.
So you now have w g CuSO4 and zz g water remaining. Determine the ratio of each and the empirical formula from that. I get CuSO4.3H2O but check my thinking. Check my work.
When 0.500 g of copper(II) sulfate pentahydrate is heated at a temperature of 300°C for twenty minutes it loses some water of hydration. If the resulting sample weights 0.427 g then what is the formula of the resulting compound?
What equation(s) can I use to solve this?
2 answers
2008?
1 answer