Question

data table

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 10 %

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
ca(no₃)₂·4h₂o

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?

Explanation:

Step1: Calculate molar masses

Molar mass of $CuSO_4\cdot5H_2O$:
$Cu: 63.55\ g/mol$, $S: 32.07\ g/mol$, $O: 16.00\ g/mol$, $H: 1.01\ g/mol$.
$M_{CuSO_4\cdot5H_2O}=63.55 + 32.07+4\times16.00 + 5\times(2\times1.01 + 16.00)=249.69\ g/mol$.
Molar mass of water in $CuSO_4\cdot5H_2O$: $5\times(2\times1.01 + 16.00)=90.10\ g/mol$.

Step2: Calculate % water in $CuSO_4\cdot5H_2O$

$\text{Percent water}=\frac{90.10}{249.69}\times100\%\approx36.1\%$.

Step3: Analyze experimental result

The experimental value of % water is not correct as it is given as $6\%$ which is far from the calculated $36.1\%$.

Step4: Calculate % water in $MgSO_4\cdot7H_2O$

Molar mass of $MgSO_4\cdot7H_2O$: $Mg: 24.31\ g/mol$, $S: 32.07\ g/mol$, $O: 16.00\ g/mol$, $H: 1.01\ g/mol$.
$M_{MgSO_4\cdot7H_2O}=24.31+32.07 + 4\times16.00+7\times(2\times1.01 + 16.00)=246.47\ g/mol$.
Molar mass of water in $MgSO_4\cdot7H_2O$: $7\times(2\times1.01 + 16.00)=126.14\ g/mol$.
$\text{Percent water}=\frac{126.14}{246.47}\times100\%\approx51.2\%$.

Step5: Calculate % water in $Ca(NO_3)_2\cdot4H_2O$

Molar mass of $Ca(NO_3)_2\cdot4H_2O$: $Ca: 40.08\ g/mol$, $N: 14.01\ g/mol$, $O: 16.00\ g/mol$, $H: 1.01\ g/mol$.
$M_{Ca(NO_3)_2\cdot4H_2O}=40.08+2\times14.01+6\times16.00 + 4\times(2\times1.01 + 16.00)=236.15\ g/mol$.
Molar mass of water in $Ca(NO_3)_2\cdot4H_2O$: $4\times(2\times1.01 + 16.00)=72.08\ g/mol$.
$\text{Percent water}=\frac{72.08}{236.15}\times100\%\approx30.5\%$.

Step6: Calculate mass of anhydrous $CuSO_4$

Since % water in $CuSO_4\cdot5H_2O$ is $36.1\%$, % anhydrous $CuSO_4$ is $100 - 36.1=63.9\%$.
Mass of anhydrous $CuSO_4$ in $5.00\ g$ of $CuSO_4\cdot5H_2O$ is $5.00\times0.639 = 3.195\ g$.

Answer:

1. $36.1\%$
2. No
3. $MgSO_4\cdot7H_2O$: $51.2\%$, $Ca(NO_3)_2\cdot4H_2O$: $30.5\%$
4. $3.195\ g$