3. A copper sulfate hydrate is taken in a crucible and heated until completely dry.
Mass of the crucible: 26,30g
Mass of the crucible + hydrate compound: 27.35 g
Mass of the crucible + compound (after 1" heating): 27.03 g
Mass of the crucible + compound (after 2nd heating): 26,98 g
Mass of the crucible + compound (after 3d heating): 26.97 g
Answer the following questions based on the above data. Show work.
(1) What is the mass of the anhydrous (fully dry) Cu(SO4)2?
How many moles of anhydrous Cu(SO4)2 is this?
(ii)
What is the mass of water (liberated during drying)?
How many moles of water is this?
(111)
What is the simplest molar ratio between anhydrous Cu(SO4)2 and H2O?
(iv)
What is the formula for the hydrate of copper sulfate?
1310: Experiment 3 - Study of Hydrates
36

Question

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Respuesta :

mickymike92 mickymike92 · Answer 1 · 2020-09-15T02:56:04+02:00

Answer:

i) mass of the anhydrous CuSO₄ = 0.67 g;

Number of moles of anhydrous CuSO₄ present = 0.004 moles

ii) mass of water liberated = 0.38 g;

Number of moles of water liberated = 0.02 moles

iii) simplest mole ratio = 0.02 / 0.004 = 5 : 1

iv) Formula of the hydrate of copper sulfate = CuSO₄.5H₂O

Explanation:

Molar mass of CuSO₄ = 159.6 g/mol

i) mass of the anhydrous CuSO₄ = (mass of crucible + compound after 3rd heating) - mass of crucible

= 26.97 g - 26.30 g = 0.67 g

Number of moles of anhydrous CuSO₄ present = mass/molar mass

= 0.67 g / 159.6 g/mol = 0.004 moles

ii) mass of water liberated = (Mass of the crucible + hydrate compound) - (mass of crucible + compound after 3rd heating)

mass of water liberated = 27.35 g - 26.97 g = 0.38 g

Number of moles of water liberated = mass/molar mas

Number of moles of water liberated = 0.38 g/ 18.0 g/mol = 0.02 moles

iii) simplest mole ratio = 0.02 / 0.004 = 5 : 1

iv) Formula of the hydrate of copper sulfate = CuSO₄.5H₂O