To calculate the theoretical percentage of water in manganese(II) sulfate monohydrate [tex]\( \text{MnSO}_4 \cdot \text{H}_2\text{O} \)[/tex], we need to follow these steps:
1. Determine the molar masses of the components involved:
- For [tex]\( \text{H}_2\text{O} \)[/tex] (water): We typically use the molar mass of water, which is [tex]\( 18.015 \, \text{g/mol} \)[/tex].
- For [tex]\( \text{MnSO}_4 \)[/tex] (manganese(II) sulfate): The molar mass is approximately [tex]\( 151.0 \, \text{g/mol} \)[/tex].
2. Combine the molar masses to find the total molar mass of the hydrate:
- The molar mass of [tex]\( \text{MnSO}_4 \cdot \text{H}_2\text{O} \)[/tex] is obtained by adding the molar masses of [tex]\( \text{MnSO}_4 \)[/tex] and [tex]\( \text{H}_2\text{O} \)[/tex]:
[tex]\[
151.0 \, \text{g/mol} + 18.015 \, \text{g/mol} = 169.015 \, \text{g/mol}
\][/tex]
3. Calculate the percentage of water in the hydrate:
- The percentage of water is found by dividing the molar mass of water by the total molar mass of the hydrate and then multiplying by 100 to get a percentage:
[tex]\[
\left( \frac{18.015 \, \text{g/mol}}{169.015 \, \text{g/mol}} \right) \times 100 = 10.658817264739817\%
\][/tex]
Hence, the theoretical percentage of water in manganese(II) sulfate monohydrate ([tex]\( \text{MnSO}_4 \cdot \text{H}_2\text{O} \)[/tex]) is approximately 10.66%.
