Question

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 4. 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:

1.

Step1: Calculate molar mass of $CuSO_4\cdot5H_2O$

The molar mass of $Cu = 63.55\ g/mol$, $S=32.07\ g/mol$, $O = 16.00\ g/mol$, $H = 1.01\ g/mol$.
$M_{CuSO_4\cdot5H_2O}=63.55 + 32.07+4\times16.00+5\times(2\times1.01 + 16.00)$
$=63.55+32.07 + 64.00+5\times18.02$
$=63.55+32.07+64.00 + 90.10$
$=249.72\ g/mol$

Step2: Calculate molar mass of water in $CuSO_4\cdot5H_2O$

The molar mass of $5H_2O=5\times(2\times1.01 + 16.00)=90.10\ g/mol$

Step3: Calculate the percentage of water

$\text{Percent water}=\frac{90.10}{249.72}\times 100\% \approx 36.1\%$

Step1: Calculate molar mass of $MgSO_4\cdot7H_2O$

$M_{MgSO_4\cdot7H_2O}=24.31+32.07 + 4\times16.00+7\times(2\times1.01+16.00)$
$=24.31+32.07+64.00+7\times18.02$
$=24.31+32.07+64.00 + 126.14$
$=246.52\ g/mol$

Step2: Calculate molar mass of water in $MgSO_4\cdot7H_2O$

The molar mass of $7H_2O = 7\times(2\times1.01+16.00)=126.14\ g/mol$

Step3: Calculate the percentage of water

$\text{Percent water}=\frac{126.14}{246.52}\times 100\%\approx 51.2\%$

Step1: Calculate molar mass of $CuSO_4$

$M_{CuSO_4}=63.55 + 32.07+4\times16.00=159.62\ g/mol$
The molar mass of $CuSO_4\cdot5H_2O = 249.72\ g/mol$

Step2: Use proportion to find mass of anhydrous compound

Let $x$ be the mass of anhydrous $CuSO_4$.
We know the ratio of molar - masses is equal to the ratio of masses.
$\frac{M_{CuSO_4}}{M_{CuSO_4\cdot5H_2O}}=\frac{x}{5.00\ g}$
$x=\frac{159.62}{249.72}\times5.00\ g\approx 3.20\ g$

Answer:

$36.1\%$

3.