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Leviafan [203]
2 years ago
5

Formula of a copper (II)sulfate hydrate lab

Chemistry
1 answer:
s344n2d4d5 [400]2 years ago
7 0

Answer:

Weigh the empty crucible, and then weigh into it between 2 g and 3 g of hydrated copper(II) sulphate. Record all weighings accurate to the nearest 0.01 g.

Support the crucible securely in the pipe-clay triangle on the tripod over the Bunsen burner.

Heat the crucible and contents, gently at first, over a medium Bunsen flame, so that the water of crystallisation is driven off steadily. The blue colour of the hydrated compound should gradually fade to the greyish-white of anhydrous copper(II) sulfate. Avoid over-heating, which may cause further decomposition, and stop heating immediately if the colour starts to blacken. If over-heated, toxic or corrosive fumes may be evolved. A total heating time of about 10 minutes should be enough.

Allow the crucible and contents to cool. The tongs may be used to move the hot crucible from the hot pipe-clay triangle onto the heat resistant mat where it should cool more rapidly.

Re-weigh the crucible and contents once cold.

Calculation:

Calculate the molar masses of H2O and CuSO4 (Relative atomic masses: H=1, O=16, S=32, Cu=64)

Calculate the mass of water driven off, and the mass of anhydrous copper(II) sulfate formed in your experiment

Calculate the number of moles of anhydrous copper(II) sulfate formed

Calculate the number of moles of water driven off

Calculate how many moles of water would have been driven off if 1 mole of anhydrous copper(II) sulfate had been formed

Write down the formula for hydrated copper(II) sulfate.

#*#*SHOW FULLSCREEN*#*#

Explanation:

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Answer:

a) the wet density of the CL sample is 0.0453 lb/in³

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Explanation:

Given that;

diameter d = 2.83 in

length L = 6 in

weight m = 1.71 lbs

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wet density can be expressed as  p = M /v

V is volume of sample which is; π/4×d²×L

so p = M / π/4×d²×L

we substitute

p = 1.71 / (π/4 × (2.83)²× 6

p = 1.71 / 37.741

p = 0.0453 lbs/in³

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b)

water content of sample is taken as;

w =  (wet_weight - dry_weight) / dry_weight

we substitute

w = (140.9 - 85.2) / 85.2

w = 55.7 / 85.2

w = 0.6537 = 65.37%

therefore the water content in the sample is 65.37%

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dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

we substitute

Sd = 0.0453 / ( 1 + 0.6537)

Sd = 0.0453 /  1.6537

Sd = 0.0274 lb/in³

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