Question

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 1

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.

Answer 2

Answer:
`CuSO_4.3H_2O`

Step-by-step explanation:

Let us suppose there are x water molecules. Thus the formula of hydrated compound will be
`CuSO_4.xH_2O`

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

`\text{Mass percent of water}=\frac{\text{Mass of water}}{\text{Molar mass of hydrated compound}}* 100`

Given : Mass percent of water = 25.3 %

Mass of water =
`(18* x)` g

Molar mass of hydrated compound
`(CuSO_4.xH_2O)` =
`(159.6 +18* x)` g

Putting in the values we get:

`25.3=(18* x)/(159.6 +18* x)* 100`

Solving for x we get:

`x=3`

Thus the formula of hydrated compound will be
`CuSO_4.3H_2O`