1. A student is given a sample of a blue nickel sulfate hydrate. She weighs the

Question

1. A student is given a sample of a blue nickel sulfate hydrate. She weighs the sample in a dry covered cru-

cible and obtains a mass of 23.711 g for the crucible, cover, and sample. Earlier she had found that the

crucible and cover weighed 21.594 g. She then heats the crucible to drive off the water of hydration, keep-

ing the crucible at red heat for about 10 minutes with the cover slightly ajar. She then lets the crucible

cool, and finds it has a lower mass; the crucible, cover and contents then weigh 22.840 g. In the process

the sample was converted to greenish-yellow anhydrous NiSO4.

a. What was the mass of the hydrate sample?

______________ g hydrate

b. What is the mass of the anhydrous NiSO4?

______________ g NiSO4

c. How much water was driven off?

______________ g H2O

d. What is the percent by mass of water in the hydrate?

______________ %mass H2O

Formula : % water = Mass of water in sample / mass hydrate sample x 100%

e. How many grams of water would there be in 100.0 g hydrate? How many moles?

______________ g H2O; ______________ moles H2O

f. How many grams of NiSO4 are there in 100.0 g hydrate? How many moles? (What percentage of the

hydrate is NiSO4? Convert the mass of NiSO4 to moles. The molar mass of NiSO4 is 154.8 g.)

____________ g NiSO4; ____________ moles NiSO4

g. How many moles of water are present per mole NiSO4?

____________

h. What is the formula of the hydrate?

____________

**Thank you!

Explanation / Answer

(a) Mass of hydrate = initial mass - mass of crucible and cover

= 23.711 - 21.594 = 2.117 g hydrate


(b) Mass of anhydrous NiSO4 = final mass - mass of crucible and cover

= 22.840 - 21.594 = 1.246 g NiSO4


(c) Mass of water = mass of hydrate - mass of NiSO4

= 2.117 - 1.246 = 0.871 g H2O


(d) %Mass H2O = mass of H2O/mass of hydrate x 100%

= 0.871/2.117 x 100% = 41.14% H2O


(e) Mass of H2O in 100.0 g of hydrate = 41.14/100 x 100 = 41.14 g H2O

Moles of H2O in 100.0 g of hydrate = mass/molar mass of H2O

= 41.14/18.02 = 2.283 moles H2O


(f) %Mass NiSO4 = 100 - %Mass H2O = 100 - 41.14 = 58.86%

Mass of NiSO4 in 100.0 g of hydrate = 58.86/100 x 100 = 58.86 g NiSO4

Moles of NiSO4 in 100.0 g of hydrate = mass/molar mass of NiSO4

= 58.86/154.8 = 0.3802 moles NiSO4


(g) Moles of H2O per mole of NiSO4 = moles of H2O/moles of NiSO4

= 2.283/0.3802 = 6


(h) Formula of hydrate: NiSO4.6H2O

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