Part B: Comprehensive Questions 20 Points 36a 1345 G Samp ✓ Solved
Part B (Comprehensive Questions: 20 Points) 36a. 1345 G Sample Of A Co
Part B (Comprehensive Questions: 20 points) 36a. 1.345 g sample of a compound of barium and oxygen was dissolved in hydrochloric acid to give a solution of barium ion, which was then precipitated with an excess of potassium chromate to give 2.012 g of barium chromate, BaCrO4. What is the formula of the compound? (4 points)
37a. Phenol, commonly known as carbolic acid, was used by Joseph Lister as an antiseptic for surgery in 1865. Its principle use today is the manufacture of phenolic resins and plastics. Combustion of 5.23 mg of phenol yields 14.67 mg CO2 and 3.01 mg H2O. Phenol contains only C, H, and O. What is the percentage of each element in this substance? (2 points)
37b. How many milliliters of 0.250 M KMnO4 are needed to react with 4.40 g of iron(II) sulfate, FeSO4? (Hint: Volumetric analysis) The reaction is as follows: 10FeSO4(aq) + 2KMnO4(aq) + 8H2SO4(aq) → 5Fe2(SO4)3(aq) + 2MnSO4(aq) + K2SO4(aq) + 8H2O(l) (2 points)
38. Two compounds have the same composition: 85.62% C and 14.38% H.
a. Obtain the empirical formula corresponding to this composition.
b. One of the compounds has a molecular mass of 28.03 amu; the other, 56.06 amu. Obtain the molecular formulas of both compounds. (4 points)
39a. For the given aqueous reaction, write the molecular and net ionic equations, including phase labels. (2 points) The neutralization of acetic acid by calcium hydroxide, both in aqueous solution.
39b. Balance the following oxidation-reduction reaction by the half-reaction method: FeI3(aq) + Mg(s) → Fe(s) + MgI2(aq) (2 points)
40a. One step in the isolation of pure rhodium metal (Rh) is the precipitation of rhodium(III) hydroxide from a solution containing rhodium(III) sulfate according to the following balanced chemical equation: Rh2(SO4)3(aq) + 6NaOH(aq) → 2Rh(OH)3(s) + 3Na2SO4(aq). What is the theoretical yield of rhodium(III) hydroxide from the reaction of 0.590 g of rhodium(III) sulfate with 0.366 g of sodium hydroxide? Show detailed work. (3 points)
40b. Using questions 40a, calculate the percentage yield of rhodium(III) hydroxide, given that the actual yield is 0.320 g. (1 point)
Sample Paper For Above instruction
Introduction
The principles of stoichiometry and chemical reactions underpin much of inorganic chemistry. This paper comprehensively addresses the estimation of compound formulas, elemental composition, volumetric analysis, empirical and molecular formulas, ionic equations, redox balancing, and yield calculations in chemical synthesis. Through detailed calculations and chemical reasoning, the discussion elucidates essential quantitative methods used in modern chemistry laboratories.
Question 36a: Determining the Formula of a Barium Compound
A 1.345 g sample containing barium and oxygen was dissolved in hydrochloric acid, establishing a solution of barium ions. The subsequent addition of excess potassium chromate precipitated 2.012 g of barium chromate, BaCrO4. To find the formula of the original compound, we begin by calculating the moles of BaCrO4 formed.
The molar mass of BaCrO4 is calculated as:
\[
\text{M}_\text{BaCrO_4} = 137.33 (Ba) + 51.996 (Cr) + 4 \times 16.00 (O) = 137.33 + 51.996 + 64.00 = 253.33\, \text{g/mol}
\]
Number of moles of BaCrO4:
\[
n_{BaCrO_4} = \frac{2.012\, \text{g}}{253.33\, \text{g/mol}} \approx 7.94 \times 10^{-3}\, \text{mol}
\]
Since each mole of BaCrO4 contains one mole of Ba, the moles of Ba originally in the sample are equivalent:
\[
n_{Ba} \approx 7.94 \times 10^{-3}\, \text{mol}
\]
Corresponding mass of Ba in the original compound:
\[
m_{Ba} = n_{Ba} \times 137.33\, \text{g/mol} \approx 1.090\, \text{g}
\]
Initially, the total mass of the compound was 1.345 g, and the mass of oxygen:
\[
m_O = 1.345\, \text{g} - 1.090\, \text{g} \approx 0.255\, \text{g}
\]
Number of moles of oxygen:
\[
n_O = \frac{0.255\, \text{g}}{16.00\, \text{g/mol}} \approx 0.0159\, \text{mol}
\]
The molar ratio of Ba to O:
\[
\frac{n_{Ba}}{n_O} \approx \frac{7.94 \times 10^{-3}}{0.0159} \approx \frac{1}{2}
\]
Hence, the empirical formula ratio is approximately Ba1O2, leading to the empirical formula BaO2. To express the simplest whole-number ratio, the compound's formula is BaO2.
---
Question 37a: Elemental Analysis of Phenol
Combustion of 5.23 mg phenol yields 14.67 mg CO2 and 3.01 mg H2O. First, determine the moles of carbon and hydrogen:
\[
\text{Moles of C} = \frac{14.67\, \text{mg}}{44.00\, \text{mg/mmol}} \approx 0.333\, \text{mmol}
\]
\[
\text{Moles of H} = \frac{3.01\, \text{mg}}{18.00\, \text{mg/mmol}} \approx 0.167\, \text{mmol}
\]
Masses:
\[
\text{Mass of C} = 0.333\, \text{mmol} \times 12.01\, \text{g/mol} \approx 4.00\, \text{mg}
\]
\[
\text{Mass of H} = 0.167\, \text{mmol} \times 1.008\, \text{g/mol} \approx 0.168\, \text{mg}
\]
Total mass of phenol combusted:
\[
5.23\, \text{mg}
\]
Mass of oxygen in phenol:
\[
m_O = 5.23\, \text{mg} - (4.00\, \text{mg} + 0.168\, \text{mg}) \approx 1.062\, \text{mg}
\]
Percentage by mass:
\[
\% C = \frac{4.00}{5.23} \times 100 \approx 76.5\%
\]
\[
\% H = \frac{0.168}{5.23} \times 100 \approx 3.2\%
\]
\[
\% O = \frac{1.062}{5.23} \times 100 \approx 20.3\%
\]
Thus, phenol is approximately 76.5% C, 3.2% H, and 20.3% O.
---
Question 37b: Volumetric Titration of Iron(II) Sulfate
Given:
\[
\text{Mass of FeSO}_4 = 4.40\, \text{g}
\]
Calculate moles of FeSO4:
\[
\text{M}_\text{FeSO}_4 = 151.91\, \text{g/mol}
\]
\[
n_{FeSO_4} = \frac{4.40}{151.91} \approx 0.02896\, \text{mol}
\]
From the balanced reaction, the molar ratio is:
\[
10\, \text{FeSO}_4 : 2\, \text{KMnO}_4
\]
Thus, moles of KMnO4 required:
\[
n_{KMnO_4} = \frac{2}{10} \times n_{FeSO_4} = 0.005792\, \text{mol}
\]
Volume of 0.250 M KMnO4 solution needed:
\[
V = \frac{n}{C} = \frac{0.005792}{0.250} \approx 0.02317\, \text{L} = 23.17\, \text{mL}
\]
Therefore, approximately 23.2 mL of KMnO4 is required.
---
Question 38: Empirical and Molecular Formulas from Composition
Given the compound's mass percentages:
\[
\text{C} = 85.62\%
\]
\[
\text{H} = 14.38\%
\]
Calculate moles per 100 g:
\[
n_C = \frac{85.62}{12.01} \approx 7.13
\]
\[
n_H = \frac{14.38}{1.008} \approx 14.26
\]
Divide by the smallest:
\[
\frac{7.13}{7.13} = 1,\quad \frac{14.26}{7.13} \approx 2
\]
Empirical formula is CH2.
Molecular mass:
\[
\text{For } 28.03\, \text{amu}: \text{Molecular formula} = \text{CH}_2
\]
\[
\text{For } 56.06\, \text{amu}: \text{Molecular formula} = (\text{CH}_2)_2 = C_2H_4
\]
---
Question 39a: Ionic Equations of Acid-Base Neutralization
The molecular equation:
\[
\mathrm{CH}_3\mathrm{COOH} + \mathrm{Ca(OH)}_2 \rightarrow \mathrm{Ca(CH}_3\mathrm{COO)}_2 + 2\mathrm{H}_2\mathrm{O}
\]
Net ionic equation:
\[
\mathrm{CH}_3\mathrm{COOH} + \mathrm{OH}^- \rightarrow \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_2\mathrm{O}
\]
Question 39b: Balancing Redox Reaction
Half-reactions:
Oxidation:
\[
\mathrm{Mg} \rightarrow \mathrm{Mg}^{2+} + 2e^-
\]
Reduction:
\[
\mathrm{Fe}^{3+} + 3e^- \rightarrow \mathrm{Fe}
\]
Multiply oxidation half-reaction by 3:
\[
3\mathrm{Mg} \rightarrow 3\mathrm{Mg}^{2+} + 6e^-
\]
Multiply reduction half-reaction by 2:
\[
2\mathrm{Fe}^{3+} + 6e^- \rightarrow 2\mathrm{Fe}
\]
Combined balanced equation:
\[
3\mathrm{Mg} + 2\mathrm{Fe}^{3+} \rightarrow 3\mathrm{Mg}^{2+} + 2\mathrm{Fe}
\]
---
Question 40a: Theoretical Yield of Rhodium(III) Hydroxide
Given:
\[
\text{Mass of Rh}_2(\text{SO}_4)_3 = 0.590\, \text{g}
\]
Molar mass:
\[
\text{M}_\text{Rh}_2(\text{SO}_4)_3 = (2 \times 102.906) + 3 \times (32.065 + 4 \times 16.00) = 205.812 + 3 \times 96.065 = 205.812 + 288.195 = 493.997\, \text{g/mol}
\]
Moles of Rh2(SO4)3:
\[
n = \frac{0.590}{493.998} \approx 1.194 \times 10^{-3}\, \text{mol}
\]
From the balanced equation, 1 mol Rh2(SO4)3 yields 2 mol Rh(OH)3:
\[
n_{Rh(OH)_3} = 2 \times n = 2.388 \times 10^{-3}\, \text{mol}
\]
Mass of Rh(OH)3:
\[
\text{M}_\text{Rh(OH)}_3 = 102.906 + 3 \times 17.008 = 102.906 + 51.024 = 153.93\, \text{g/mol}
\]
\[
m_{Rh(OH)_3} = 2.388 \times 10^{-3} \times 153.93 \approx 0.368\, \text{g}
\]
The theoretical yield is approximately 0.368 g.
---
Question 40b: Percentage Yield Calculation
Actual yield:
\[
0.320\, \text{g}
\]
Percentage yield:
\[
\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 = \frac{0.320}{0.368} \times 100 \approx 86.96\%
\]
The percentage yield is approximately 87%.
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