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レ31 2- The mass percent of water in a hydrate of MnCl2 is 36.41%. What is the empirical formųla of the hydrate? [Mn 54.9 ; Cl 35.5]

レ31 2- The mass percent of water in a hydrate of MnCl2 is 36.41%. What is the empirical formųla of the hydrate? [Mn 54.9 ; Cl 35.5]

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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**Title: Empirical Formula Determination of a Hydrate of MnCl\(_{\text{2}}\)** **Objective:** To determine the empirical formula of a hydrate of MnCl\(_{\text{2}}\) based on its mass percent of water. **Given Data:** - Mass percent of water in MnCl\(_{\text{2}}\) hydrate: 36.41% - Molar masses: - Mn: 54.9 g/mol - Cl: 35.5 g/mol **Procedure:** 1. **Calculate Mass of Anhydrous Compound:** - Assume a sample mass of 100 g. - Mass of water = 36.41 g. - Mass of anhydrous MnCl\(_{\text{2}}\) = 100 g - 36.41 g = 63.59 g. 2. **Calculate Moles of Anhydrous Compound:** - Molar mass of MnCl\(_{\text{2}}\) = 54.9 + 2(35.5) = 126.9 g/mol. - Moles of MnCl\(_{\text{2}}\) = 63.59 g / 126.9 g/mol = 0.501 moles. 3. **Calculate Moles of Water:** - Molar mass of water (H\(_2\)O) = 18 g/mol. - Moles of water = 36.41 g / 18 g/mol = 2.023 moles. **Determine Empirical Formula:** - Ratio of water to anhydrous MnCl\(_{\text{2}}\): - Water : Anhydrous = 2.023 : 0.501 ≈ 4:1 - Empirical formula: MnCl\(_{\text{2}}\)•4H\(_2\)O **Conclusion:** The empirical formula of the hydrate of MnCl\(_{\text{2}}\) is MnCl\(_{\text{2}}\)•4H\(_2\)O. **Additional Notes:** - Calculations illustrate a thorough understanding of stoichiometric relationships and the concept of hydration in ionic compounds. - The work involves numeric and arithmetic skills critical for precise chemical analysis.
Transcribed Image Text:**Title: Empirical Formula Determination of a Hydrate of MnCl\(_{\text{2}}\)** **Objective:** To determine the empirical formula of a hydrate of MnCl\(_{\text{2}}\) based on its mass percent of water. **Given Data:** - Mass percent of water in MnCl\(_{\text{2}}\) hydrate: 36.41% - Molar masses: - Mn: 54.9 g/mol - Cl: 35.5 g/mol **Procedure:** 1. **Calculate Mass of Anhydrous Compound:** - Assume a sample mass of 100 g. - Mass of water = 36.41 g. - Mass of anhydrous MnCl\(_{\text{2}}\) = 100 g - 36.41 g = 63.59 g. 2. **Calculate Moles of Anhydrous Compound:** - Molar mass of MnCl\(_{\text{2}}\) = 54.9 + 2(35.5) = 126.9 g/mol. - Moles of MnCl\(_{\text{2}}\) = 63.59 g / 126.9 g/mol = 0.501 moles. 3. **Calculate Moles of Water:** - Molar mass of water (H\(_2\)O) = 18 g/mol. - Moles of water = 36.41 g / 18 g/mol = 2.023 moles. **Determine Empirical Formula:** - Ratio of water to anhydrous MnCl\(_{\text{2}}\): - Water : Anhydrous = 2.023 : 0.501 ≈ 4:1 - Empirical formula: MnCl\(_{\text{2}}\)•4H\(_2\)O **Conclusion:** The empirical formula of the hydrate of MnCl\(_{\text{2}}\) is MnCl\(_{\text{2}}\)•4H\(_2\)O. **Additional Notes:** - Calculations illustrate a thorough understanding of stoichiometric relationships and the concept of hydration in ionic compounds. - The work involves numeric and arithmetic skills critical for precise chemical analysis.
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