Question

1.547 grams of hydrated mgso₄ is heated in a crucible. after heating, 0.7554 grams of anhydrous mgso₄ remains in the crucible. how many waters of hydration were attached to the mgso₄? round your answer to the nearest whole number. type answer: enter a number

Explanation:

Step1: Calculate mass of water

Mass of hydrated \( \text{MgSO}_4 = 1.547 \, \text{g} \), mass of anhydrous \( \text{MgSO}_4 = 0.7554 \, \text{g} \).
Mass of water \( = 1.547 - 0.7554 = 0.7916 \, \text{g} \).

Step2: Find moles of \( \text{MgSO}_4 \) and \( \text{H}_2\text{O} \)

Molar mass of \( \text{MgSO}_4 = 24.305 + 32.06 + 4 \times 16.00 = 120.365 \, \text{g/mol} \).
Moles of \( \text{MgSO}_4 = \frac{0.7554}{120.365} \approx 0.006276 \, \text{mol} \).

Molar mass of \( \text{H}_2\text{O} = 18.015 \, \text{g/mol} \).
Moles of \( \text{H}_2\text{O} = \frac{0.7916}{18.015} \approx 0.04394 \, \text{mol} \).

Step3: Determine ratio of \( \text{H}_2\text{O} \) to \( \text{MgSO}_4 \)

Ratio \( = \frac{\text{moles of } \text{H}_2\text{O}}{\text{moles of } \text{MgSO}_4} = \frac{0.04394}{0.006276} \approx 7 \).

Answer:

7