A student is given a sample of a green nickel sulfate hydrate. She weights the s
ID: 1009811 • Letter: A
Question
A student is given a sample of a green nickel sulfate hydrate. She weights the sample in a dry covers crucible and obtains a mass of 22.326 g for the crucible, cover and sample. Earlier she had found that the crucible and cover weighted 21.244 g. She then heats the crucible to drive off the water of hydration, keeping the crucible at red heat for about 10 min with the cover slightly ajar. She then lets the crucible cool, and finds it had a lower mass; the crucible, cover and contents then wight 21.840 g. in the process the sample was converted to yellow anhydrous NiSO4.
A.) What was the mass of the hydrate sample? === 1.082 g
B.) what is the mass of the anhydrous NiSO4 === 0.596
C) how much water was driven off? == 0.486 g
D) what is the percentage of water in the hydrate? == 44.9?
E) How many grams of water would there be in 100.0 g hydrate? how many moles? (i don't know from here)
F) how many grams of NiSO4 are there in 100.0g of hydrate? How many moles? What percentage of the hydrate is NiSO4?
G) how many moles of water are present per mole NiSO4?
H) what is the formula of the hydrate?
Explanation / Answer
We shall work over the whole problem again.
A) Mass of hydrated sample = (22.326 – 21.244) g = 1.082 gm
B) Mass of anhydrous NiSO4 = (21.840 – 21.244) g = 0.596 gm
C) Mass of hydrate = (1.082 – 0.596) g = 0.486 gm
D) Percentage of hydrate in the sample = (0.486 gm/1.082 gm)*100 = 44.9168 44.917%
E) Since we derived the percentage of hydrate in the hydrated sample in the previous part, we can say that 100 percent hydrated sample contains 44.917 percent hydrate.
In other words, 100.0 gm hydrated sample will contain 44.917 gm water (ans).
The molar mass of water is 18.0 gm/mole.
Thus, number of moles of water in 44.917 gm hydrate = (44.917 gm)/(18.0 gm)*(1 mole) = 2.4953 2.495 moles (ans).
F) 100.0 gm hydrated sample will contain (100.0 – 44.917) gm = 55.083 gm NiSO4.
The molar mass of NiSO4 is 154.75 gm/mole.
Thus 55.08 gm NiSO4 = (55.083 gm NiSO4)/(154.75 gm NiSO4)*(1 mole) = 0.3559 mole 0.356 mole (ans)
G) 100.0 gm hydrated sample contains 0.356 mole NiSO4 and 2.495 moles water.
Therefore, 0.356 mole anhydrous NiSO4 combines with 2.495 moles of water.
Thus 1 mole of anhydrous NiSO4 will combine with (2.495 moles water)/(0.356 moles NiSO4)*(1 mole NiSO4) = 7.008 7.0 moles water (we take the positive integer value here since the number of moles in a compound cannot be fractional) (ans).
H) 1 mole of anhydrous NiSO4 combines with 7 moles of water; thus the formula of the hydrate is NiSO4.7H2O (ans)
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