Question

1. mass of evaporating dish 56.17 g
2. mass of hydrate 2.49 g
3. mass of evaporating dish and hydrate (before heating) 58.65 g
4. mass of evaporating dish and anhydrous compound (after heating) 57.71 g
5. mass of water in hydrate 0.94 g
6. % water in hydrate %

data analysis

1. calculate the % water in cuso₄·5h₂o from the formula.
2. did you determine the correct % water with your experimentation?
3. calculate the % water (from the formula) in the following hydrates.

mgso₄·7h₂o 51.2 %
ca(no₃)₂·4h₂o 30.5 %

1. if a student heats 5.00 g of cuso₄·5h₂o until all the water of crystallization is removed, how many grams of anhydrous compound should remain? 3.195 g

Explanation:

Step1: Calculate percentage of water in hydrate from data

The formula for percentage of water in hydrate is $\text{% water}=\frac{\text{mass of water in hydrate}}{\text{mass of hydrate}}\times100\%$. Given mass of water in hydrate is $0.94$ g and mass of hydrate is $2.49$ g.
$\text{% water}=\frac{0.94}{2.49}\times 100\% \approx 37.75\%$

Step2: Answer question 2

The experimental value of % water in hydrate is approximately $37.75\%$, while the theoretical value for $CuSO_4\cdot5H_2O$ is $36.1\%$. The experimental value is close but not exactly the same, so the answer is no, the experiment did not determine the exact correct % water.

Step3: Answer question 4

The molar mass of $CuSO_4\cdot5H_2O$ is $M_{CuSO_4\cdot5H_2O}=63.5 + 32+4\times16+5\times(2\times1 + 16)=249.5$ g/mol. The molar mass of $CuSO_4$ is $M_{CuSO_4}=63.5 + 32+4\times16 = 159.5$ g/mol.
The mass - ratio of $CuSO_4$ to $CuSO_4\cdot5H_2O$ is $\frac{M_{CuSO_4}}{M_{CuSO_4\cdot5H_2O}}=\frac{159.5}{249.5}$.
If we start with $m = 5.00$ g of $CuSO_4\cdot5H_2O$, the mass of anhydrous $CuSO_4$ is $m_{CuSO_4}=5.00\times\frac{159.5}{249.5}\approx3.195$ g

Answer:

1. Approximately $37.75\%$
2. No
3. (already given in the problem: $MgSO_4\cdot7H_2O$: $51.2\%$, $Ca(NO_3)_2\cdot4H_2O$: $30.5\%$)
4. $3.195$ g