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The active compound in Pepto-Bismol contains C, H, O, and Bi.
(a) When 0.22105 g of it was burned in excess O2, 0.1422 g of bismuth(III) oxide, 0.1880 g of carbon dioxide, and 0.02750 g of water were formed. What is the empirical formula of this compound?
(b) Given a molar mass of 1086 g/mol, determine the molecular formula. (c) Complete and balance the acid-base reaction between bismuth(III) hydroxide and salicylic acid (HC7H5O3), which is used to form this compound.
(d) A dose of Pepto-Bismol contains 0.600 mg of active ingredient. If the yield of the reaction in part (c) is 88.0%, what mass (in mg) of bismuth(III) hydroxide is required to prepare one dose?
The Correct Answer and Explanation is:
Given Data:
- Mass of compound burned: 0.22105 g
- Products formed:
- Bi₂O₃ (bismuth(III) oxide): 0.1422 g
- CO₂: 0.1880 g
- H₂O: 0.02750 g
(a) Find the empirical formula of the compound
Step 1: Determine the moles of each element from the products
- Bismuth (Bi):
From Bi₂O₃:
Molar mass Bi₂O₃ = (2 × 208.98) + (3 × 16.00) = 417.96 + 48 = 465.96 g/mol
Moles of Bi₂O₃ = 0.1422 g / 465.96 g/mol = 3.051 × 10⁻⁴ mol
Moles of Bi atoms = 2 × 3.051 × 10⁻⁴ = 6.102 × 10⁻⁴ mol
Mass of Bi = moles × atomic mass = 6.102 × 10⁻⁴ mol × 208.98 g/mol = 0.1275 g
- Carbon (C):
From CO₂:
Molar mass CO₂ = 44.01 g/mol
Moles of CO₂ = 0.1880 g / 44.01 g/mol = 4.273 × 10⁻³ mol
Moles of C = 4.273 × 10⁻³ mol
Mass of C = 4.273 × 10⁻³ mol × 12.01 g/mol = 0.0513 g
- Hydrogen (H):
From H₂O:
Molar mass H₂O = 18.02 g/mol
Moles of H₂O = 0.02750 g / 18.02 g/mol = 1.526 × 10⁻³ mol
Moles of H = 2 × 1.526 × 10⁻³ mol = 3.052 × 10⁻³ mol
Mass of H = 3.052 × 10⁻³ mol × 1.008 g/mol = 0.00308 g
- Oxygen (O):
Total mass of sample = 0.22105 g
Sum of masses of Bi, C, and H = 0.1275 + 0.0513 + 0.00308 = 0.1819 g
Mass of O = 0.22105 g – 0.1819 g = 0.03915 g
Moles of O = 0.03915 g / 16.00 g/mol = 2.447 × 10⁻³ mol
Step 2: Find the mole ratio of elements
| Element | Moles | Divide by smallest | Ratio (approx.) |
|---|---|---|---|
| Bi | 6.102 × 10⁻⁴ | 6.102 × 10⁻⁴ | 1 |
| C | 4.273 × 10⁻³ | 6.102 × 10⁻⁴ | 7.0 |
| H | 3.052 × 10⁻³ | 6.102 × 10⁻⁴ | 5.0 |
| O | 2.447 × 10⁻³ | 6.102 × 10⁻⁴ | 4.0 |
Step 3: Write empirical formula
The mole ratio approximates to:
Bi₁ C₇ H₅ O₄
(b) Molecular formula
Given molar mass = 1086 g/mol
Calculate molar mass of empirical formula:
- Bi: 208.98 × 1 = 208.98 g/mol
- C: 12.01 × 7 = 84.07 g/mol
- H: 1.008 × 5 = 5.04 g/mol
- O: 16.00 × 4 = 64.00 g/mol
Sum = 208.98 + 84.07 + 5.04 + 64.00 = 362.09 g/mol
Calculate multiplier:
n=1086362.09≈3n = \frac{1086}{362.09} \approx 3
So molecular formula is:
Bi₃ C₂₁ H₁₅ O₁₂
(c) Balanced acid-base reaction between bismuth(III) hydroxide and salicylic acid (HC₇H₅O₃)
- Bismuth(III) hydroxide: Bi(OH)₃
- Salicylic acid: HC₇H₅O₃ (one acidic H per molecule)
Reaction:
Bi(OH)3+3 HC7H5O3→Bi(C7H5O3)3+3 H2O\mathrm{Bi(OH)_3 + 3 \ HC_7H_5O_3 \rightarrow Bi(C_7H_5O_3)_3 + 3 \ H_2O}
Bismuth(III) hydroxide reacts with 3 moles of salicylic acid to form bismuth(III) salicylate and water.
(d) Mass of Bi(OH)₃ required to prepare one dose (0.600 mg active ingredient) at 88.0% yield
Step 1: Find mass of Bi(OH)₃ required theoretically for 0.600 mg of product
- Molecular mass of Bi(OH)₃:
Bi = 208.98 g/mol
O = 16.00 × 3 = 48.00 g/mol
H = 1.008 × 3 = 3.02 g/mol
Total = 208.98 + 48 + 3.02 = 260.0 g/mol - Molecular mass of active ingredient (bismuth salicylate) = 1086 g/mol
Step 2: Moles of active ingredient in 0.600 mg dose
Moles=0.600×10−31086=5.53×10−7 mol\text{Moles} = \frac{0.600 \times 10^{-3}}{1086} = 5.53 \times 10^{-7} \text{ mol}
Step 3: Moles of Bi(OH)₃ needed (1:1 mole ratio)
Same moles of Bi(OH)₃ required = 5.53 × 10⁻⁷ mol
Mass of Bi(OH)₃ theoretically required:
5.53×10−7×260.0=1.44×10−4 g=0.144 mg5.53 \times 10^{-7} \times 260.0 = 1.44 \times 10^{-4} \text{ g} = 0.144 \text{ mg}
Step 4: Adjust for 88.0% yield
Mass of Bi(OH)₃ required practically:
0.1440.88=0.164 mg\frac{0.144}{0.88} = 0.164 \text{ mg}
Final Answers:
(a) Empirical formula: BiC₇H₅O₄
(b) Molecular formula: Bi₃C₂₁H₁₅O₁₂
(c) Balanced reaction:
Bi(OH)3+3 HC7H5O3→Bi(C7H5O3)3+3 H2O\mathrm{Bi(OH)_3 + 3 \ HC_7H_5O_3 \rightarrow Bi(C_7H_5O_3)_3 + 3 \ H_2O}
(d) Mass of Bi(OH)₃ required = 0.164 mg
Explanation
The problem involves analyzing combustion data to determine the empirical formula of the active compound in Pepto-Bismol, followed by finding the molecular formula using the molar mass, balancing a key acid-base reaction, and calculating reagent mass for synthesis.
In part (a), the compound is combusted completely to produce bismuth(III) oxide, carbon dioxide, and water. By quantifying the amounts of these products, the moles of each element in the original compound can be calculated. Bismuth is determined from Bi₂O₃, carbon from CO₂, and hydrogen from H₂O. The remaining mass is attributed to oxygen. Dividing each elemental mole amount by the smallest mole quantity yields the mole ratio that forms the empirical formula BiC₇H₅O₄.
Part (b) compares the empirical formula mass (~362 g/mol) with the experimentally given molar mass (1086 g/mol). The molecular formula is a multiple (approximately 3) of the empirical formula, yielding Bi₃C₂₁H₁₅O₁₂.
Part (c) involves the acid-base reaction between bismuth(III) hydroxide and salicylic acid. Bismuth(III) hydroxide acts as a base, neutralizing three moles of salicylic acid to form bismuth(III) salicylate and water. The balanced equation shows the 1:3 molar ratio.
Finally, part (d) calculates the mass of Bi(OH)₃ needed to prepare one 0.600 mg dose of the active ingredient assuming 88% reaction yield. The moles of active compound are converted to the equivalent moles of Bi(OH)₃ (1:1 ratio), and then adjusted for the reaction efficiency. The result is 0.164 mg of Bi(OH)₃ required.
This detailed analysis combines stoichiometric calculations, empirical data interpretation, and reaction chemistry to characterize the compound and its synthesis accurately.
