1. mass of evaporating dish 56.17 g\n2. mass of hydrate 2.49 g\n3. mass of evaporating dish and hydrate…

1. mass of evaporating dish 56.17 g\n2. mass of hydrate 2.49 g\n3. mass of evaporating dish and hydrate (before heating) 58.65 g\n4. mass of evaporating dish and anhydrous compound (after heating) 57.71 g\n5. mass of water in hydrate 0.94 g\n6. % water in hydrate %\ndata analysis\n1. calculate the % water in cuso₄·5h₂o from the formula.\n2. did you determine the correct % water with your experimentation?\n3. calculate the % water (from the formula) in the following hydrates.\nmgso₄·7h₂o 51.2 %\nca(no₃)₂·4h₂o 30.5 %\n4. 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

1. mass of evaporating dish 56.17 g\n2. mass of hydrate 2.49 g\n3. mass of evaporating dish and hydrate (before heating) 58.65 g\n4. mass of evaporating dish and anhydrous compound (after heating) 57.71 g\n5. mass of water in hydrate 0.94 g\n6. % water in hydrate %\ndata analysis\n1. calculate the % water in cuso₄·5h₂o from the formula.\n2. did you determine the correct % water with your experimentation?\n3. calculate the % water (from the formula) in the following hydrates.\nmgso₄·7h₂o 51.2 %\nca(no₃)₂·4h₂o 30.5 %\n4. 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

Answer

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