In a similar analysis, a student determined that the percent of water in the hydrate was 25.3%. the instructor informed the student that the formula of the anhydrous compound was cuso4. calculate the formula for the hydrated compound.

Answer is: formula for the hydrated compound is CuSO₄·3H₂O.
ω(H₂O) = 25,3% = 0,253.
ω(CuSO₄) = 100% - 25,3%.
ω(CuSO₄) = 74,7% = 0,747.
ω(H₂O) : M(H₂O) = ω(CuSO₄) : M(CuSO₄).
0,253 : M(H₂O) = 0,747 : 159,6 g/mol.
M(H₂O) = (0,253 · 159,6 g/mol) ÷ 0,747.
M(H₂O) = 54 g/mol.
N(H₂O) = 54 g/mol ÷ 18 g/mol.
N(H₂O) = 3.

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kobenhavn kobenhavn · Answer 2 · 2019-09-23T09:11:52+02:00

Answer: [tex]CuSO_4.3H_2O[/tex]

Explanation:

Let us suppose there are x water molecules. Thus the formula of hydrated compound will be [tex]CuSO_4.xH_2O[/tex]

To calculate the mass percent of element in a given compound, we use the formula:

[tex]\text{Mass percent of water}=\frac{\text{Mass of water}}{\text{Molar mass of hydrated compound}}\times 100[/tex]

Given : Mass percent of water = 25.3 %

Mass of water = [tex](18\times x)[/tex] g

Molar mass of hydrated compound [tex](CuSO_4.xH_2O)[/tex] = [tex](159.6 +18\times x)[/tex] g

Putting in the values we get:

[tex]25.3=\frac{18\times x}{159.6 +18\times x}\times 100[/tex]

Solving for x we get:

[tex]x=3[/tex]

Thus the formula of hydrated compound will be [tex]CuSO_4.3H_2O[/tex]