Cuso4 + 5h2o Essay
Calculations
Mass percent of water in CuSO4·5H2O using experimental data:
(mass of water lost / mass of hydrated salt) x 100
(0.658 / 2.009) x 100 = 32.75 %
Mass percent of water in CuSO4·5H2O using theoretical data:
(90 / 249) x 100 = 36.14 %
Percent error using experimental and theoretical mass percentages:
(experimental value – theoretical value / theoretical value) x 100
(32.75 – 36.14 / 36.14) x 100 = 9.38 %
Average percent water in CuSO4·5H2O using experimental values:
(trial 1 % water + trial 2 % water / number of trials)
(32.75 + 33.75 / 2) = 33.25 %
Amount of water lost in unknown hydrate CaSO4·XH2O (trial 1):
(mass of hydrated salt – mass of anhydrous salt)
(1.99 – 1.595) = 0.395 g
Mass percent of water in unknown hydrate CaSO4·XH2O (trial 1):
(mass of water / mass of hydrated salt) x 100
(0.395 / 1.99) x 100 = 19.85 %
Initial mole calculation for CaSO4 (trial 1):
(mass anhydrous salt / molecular weight of salt) …show more content…
One notable finding (tables 1-3) was, that although the formula for the known hydrate theoretically contained 5 water molecules, our experimental data indicated 4.3-4.5 water molecules. This suggests that the molar ratio is just above 4:1 rather than 5:1. Therefore, as 0.3 or 0.5 of a water molecule does not exist as a water molecule, these results were accepted as error. Either all of the water molecules were not evaporated or the hydrate sample bottle being opened frequently led to minor desiccation of the contents. For part 2, the mass of the anhydrous CaSO4 and water were also converted into moles and utilized to compute the molar ratio of the unknown sample. The ratio for these samples were calculated at a low 2:1 and a high of 12:1, water to calcium sulfate; the empirical formulas were CaSO4·2H2O and CaSO4·12H2O respectively. The determined percent mass of hydration was 19.85% and
Mass percent of water in CuSO4·5H2O using experimental data:
(mass of water lost / mass of hydrated salt) x 100
(0.658 / 2.009) x 100 = 32.75 %
Mass percent of water in CuSO4·5H2O using theoretical data:
(90 / 249) x 100 = 36.14 %
Percent error using experimental and theoretical mass percentages:
(experimental value – theoretical value / theoretical value) x 100
(32.75 – 36.14 / 36.14) x 100 = 9.38 %
Average percent water in CuSO4·5H2O using experimental values:
(trial 1 % water + trial 2 % water / number of trials)
(32.75 + 33.75 / 2) = 33.25 %
Amount of water lost in unknown hydrate CaSO4·XH2O (trial 1):
(mass of hydrated salt – mass of anhydrous salt)
(1.99 – 1.595) = 0.395 g
Mass percent of water in unknown hydrate CaSO4·XH2O (trial 1):
(mass of water / mass of hydrated salt) x 100
(0.395 / 1.99) x 100 = 19.85 %
Initial mole calculation for CaSO4 (trial 1):
(mass anhydrous salt / molecular weight of salt) …show more content…
One notable finding (tables 1-3) was, that although the formula for the known hydrate theoretically contained 5 water molecules, our experimental data indicated 4.3-4.5 water molecules. This suggests that the molar ratio is just above 4:1 rather than 5:1. Therefore, as 0.3 or 0.5 of a water molecule does not exist as a water molecule, these results were accepted as error. Either all of the water molecules were not evaporated or the hydrate sample bottle being opened frequently led to minor desiccation of the contents. For part 2, the mass of the anhydrous CaSO4 and water were also converted into moles and utilized to compute the molar ratio of the unknown sample. The ratio for these samples were calculated at a low 2:1 and a high of 12:1, water to calcium sulfate; the empirical formulas were CaSO4·2H2O and CaSO4·12H2O respectively. The determined percent mass of hydration was 19.85% and