Given MgSO4 x 7H20 What is the balanced chemical equation showing the evolution of water from your hydrate?
Given MgSO4 x 7H20 What is the balanced chemical equation showing the evolution of water from your hydrate?
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Given MgSO4 x 7H20
What is the balanced chemical equation showing the evolution of water from your hydrate?
![**Empirical Formula Post-Lab:**
*Completed after office hours*
1. **Mass of anhydrous salt:** 1.008 g
2. **Moles of anhydrous salt:**
\[
\frac{1.008 \, \text{g}}{120.36 \, \text{g/mol}} = 0.00837 \, \text{mol MgSO}_4
\]
3. **Mass of water removed:**
\( 2.008 \, \text{g} - 1.008 \, \text{g} = 1.000 \, \text{g H}_2\text{O} \)
4. **Moles of water removed:**
\[
\frac{1.000 \, \text{g H}_2\text{O}}{18.02 \, \text{g/mol}} = 0.0556 \, \text{mol H}_2\text{O}
\]
*(Note: In tutoring, an answer of 0.111 mol H₂O was provided, causing uncertainty)*
5. **% of water in hydrate:**
\[
\frac{\text{mass of H}_2\text{O}}{\text{mass of MgSO}_4} \times 100 = 49.800\% \, \text{or} \, 50\% \, \text{rounded}
\]
6. **Empirical Formula of hydrate:**
- Convert masses to moles:
\[
\frac{1.008 \, \text{g}}{120.37 \, \text{g/mol}} = 8.37 \times 10^{-3} \, \text{mol}
\]
\[
\frac{1.000 \, \text{g}}{18.00 \, \text{g/mol}} = 5.5 \times 10^{-2} \, \text{mol}
\]
- Divide each by the smallest number:
\[
\frac{8.37 \times 10^{-3}}{8.37 \times 10^{-3}} = 1
\]
\[
\frac{5.5 \times 10^{-2}}{8.37 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F21864e3f-d71b-470b-8315-da5e05623871%2F945f52dc-565f-43da-b7a0-155a67fd7e5b%2F03809ie_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Empirical Formula Post-Lab:**
*Completed after office hours*
1. **Mass of anhydrous salt:** 1.008 g
2. **Moles of anhydrous salt:**
\[
\frac{1.008 \, \text{g}}{120.36 \, \text{g/mol}} = 0.00837 \, \text{mol MgSO}_4
\]
3. **Mass of water removed:**
\( 2.008 \, \text{g} - 1.008 \, \text{g} = 1.000 \, \text{g H}_2\text{O} \)
4. **Moles of water removed:**
\[
\frac{1.000 \, \text{g H}_2\text{O}}{18.02 \, \text{g/mol}} = 0.0556 \, \text{mol H}_2\text{O}
\]
*(Note: In tutoring, an answer of 0.111 mol H₂O was provided, causing uncertainty)*
5. **% of water in hydrate:**
\[
\frac{\text{mass of H}_2\text{O}}{\text{mass of MgSO}_4} \times 100 = 49.800\% \, \text{or} \, 50\% \, \text{rounded}
\]
6. **Empirical Formula of hydrate:**
- Convert masses to moles:
\[
\frac{1.008 \, \text{g}}{120.37 \, \text{g/mol}} = 8.37 \times 10^{-3} \, \text{mol}
\]
\[
\frac{1.000 \, \text{g}}{18.00 \, \text{g/mol}} = 5.5 \times 10^{-2} \, \text{mol}
\]
- Divide each by the smallest number:
\[
\frac{8.37 \times 10^{-3}}{8.37 \times 10^{-3}} = 1
\]
\[
\frac{5.5 \times 10^{-2}}{8.37 \
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