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Mobile App Chapter 7: Problem 61
Given 1.00-g samples of each of the compounds \(\mathrm{CO}_{2}, \mathrm{O}_{2}\), \(\mathrm{H}_{2} \mathrm{O}\), and \(\mathrm{CH}_{3} \mathrm{OH}\), (a) which sample will contain the largest number of molecules? (b) which sample will contain the largest number of atoms? Show proof for your answers.
Short Answer
Expert verified
(a) \(\mathrm{H}_2 \mathrm{O}\) (b) \(\mathrm{CH}_3 \mathrm{OH}\)
Step by step solution
01
- Calculate the Molar Masses
First, calculate the molar masses of each compound. 1. For \(\mathrm{CO}_2\): Carbon (C): 12.01 g/mol Oxygen (O): 16.00 g/mol Molar mass: 12.01 + 2(16.00) = 44.01 g/mol 2. For \(\mathrm{O}_2\): Molar mass: 2(16.00) = 32.00 g/mol 3. For \(\mathrm{H}_2 \mathrm{O}\): Hydrogen (H): 1.01 g/mol Oxygen (O): 16.00 g/mol Molar mass: 2(1.01) + 16.00 = 18.02 g/mol 4. For \(\mathrm{CH}_3 \mathrm{OH}\): Carbon (C): 12.01 g/mol Hydrogen (H): 4(1.01) = 4.04 g/mol Oxygen (O): 16.00 g/mol Molar mass: 12.01 + 4.04 + 16.00 = 32.05 g/mol
02
- Convert Mass to Moles
Convert 1.00 g of each compound to moles using their respective molar masses. 1. For \(\mathrm{CO}_2\): \(\text{{Moles}} = \frac{{1.00 \text{{ g}}}}{{44.01 \text{{ g/mol}}}} = 0.02272 \text{{ moles}}\) 2. For \(\mathrm{O}_2\): \(\text{{Moles}} = \frac{{1.00 \text{{ g}}}}{{32.00 \text{{ g/mol}}}} = 0.03125 \text{{ moles}}\) 3. For \(\mathrm{H}_2 \mathrm{O}\): \(\text{{Moles}} = \frac{{1.00 \text{{ g}}}}{{18.02 \text{{ g/mol}}}} = 0.05549 \text{{ moles}}\) 4. For \(\mathrm{CH}_3 \mathrm{OH}\): \(\text{{Moles}} = \frac{{1.00 \text{{ g}}}}{{32.05 \text{{ g/mol}}}} = 0.03120 \text{{ moles}}\)
03
- Determine the Largest Number of Molecules
Use Avogadro's number (\(6.022 \times 10^{23}\) molecules/mol) to find the number of molecules in each sample. 1. For \(\mathrm{CO}_2\): \(0.02272 \text{{ moles}} \times 6.022 \times 10^{23} = 1.37 \times 10^{22} \text{{ molecules}}\) 2. For \(\mathrm{O}_2\): \(0.03125 \text{{ moles}} \times 6.022 \times 10^{23} = 1.88 \times 10^{22} \text{{ molecules}}\) 3. For \(\mathrm{H}_2 \mathrm{O}\): \(0.05549 \text{{ moles}} \times 6.022 \times 10^{23} = 3.34 \times 10^{22} \text{{ molecules}}\) 4. For \(\mathrm{CH}_3 \mathrm{OH}\): \(0.03120 \text{{ moles}} \times 6.022 \times 10^{23} = 1.88 \times 10^{22} \text{{ molecules}}\) Thus, \(\mathrm{H}_2 \mathrm{O}\) has the largest number of molecules.
04
- Determine the Largest Number of Atoms
Multiply the number of molecules by the number of atoms per molecule for each substance. 1. For \(\mathrm{CO}_2\): 3 atoms/molecule \(1.37 \times 10^{22} \text{{ molecules}} \times 3 = 4.11 \times 10^{22} \text{{ atoms}}\) 2. For \(\mathrm{O}_2\): 2 atoms/molecule \(1.88 \times 10^{22} \text{{ molecules}} \times 2 = 3.76 \times 10^{22} \text{{ atoms}}\) 3. For \(\mathrm{H}_2 \mathrm{O}\): 3 atoms/molecule \(3.34 \times 10^{22} \text{{ molecules}} \times 3 = 1.00 \times 10^{23} \text{{ atoms}}\) 4. For \(\mathrm{CH}_3 \mathrm{OH}\): 6 atoms/molecule \(1.88 \times 10^{22} \text{{ molecules}} \times 6 = 1.13 \times 10^{23} \text{{ atoms}}\) Thus, \(\mathrm{CH}_3 \mathrm{OH}\) has the largest number of atoms.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molar Mass Calculation
In chemistry, calculating the molar mass of a compound is the first step in many problems. The molar mass is the mass of one mole of a substance (atoms, molecules, etc.) in grams. To find it, simply add up the atomic masses of all the atoms in the molecule. For example:
- For \(\text{CO}_2\): The atomic mass of carbon is 12.01 g/mol, and oxygen is 16.00 g/mol. The molar mass is calculated as 12.01 + 2(16.00) = 44.01 g/mol.
- For O\(_2\): Since there are two oxygen atoms, the molar mass is 2(16.00) = 32.00 g/mol.
- For H\(_2\)O: Hydrogen has an atomic mass of 1.01 g/mol, so you add 2(1.01) + 16.00 = 18.02 g/mol.
- For CH\(_3\)OH: Add the atomic masses 12.01 (carbon) + 4(1.01) (hydrogen) + 16.00 (oxygen) to get 32.05 g/mol.
Avogadro's Number
Avogadro's Number is a fundamental constant in chemistry, represented as 6.022 \times 10^{23} molecules/mol. This constant provides the number of molecules or atoms in one mole of a substance. For example, 1 mole of any substance contains exactly 6.022 \times 10^{23} entities (atoms, molecules, ions, etc.). To use it in calculations:
- If you have 0.05549 moles of H\(_2\)O, multiply it by Avogadro's Number: 0.05549 \times 6.022 \times 10^{23} = 3.34 \times 10^{22} molecules of H\(_2\)O.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It involves using balanced chemical equations to calculate masses, moles, and volumes. Here, we use stoichiometry to convert between different units such as grams and moles:
- If you have 1.00 g of CO\(_2\), you can convert this to moles using its molar mass (44.01 g/mol): 1.00 g / 44.01 g/mol = 0.02272 moles.
Molecular Composition
Molecular composition describes the number and types of atoms in a molecule. For example, in CH\(_3\)OH (methanol), the molecule consists of 1 carbon atom, 4 hydrogen atoms, and 1 oxygen atom. Analyzing molecular composition is important for understanding the properties and reactions of substances. To determine which sample in the given exercise contains the largest number of atoms:
- Count the atoms in each molecule.
- For CH\(_3\)OH: There are 6 atoms per molecule (1 C + 4 H + 1 O).
- Multiply the number of atoms per molecule by the total number of molecules calculated using Avogadro's Number.
- CH\(_3\)OH: 1.88 \times 10^{22} molecules \times 6 atoms/molecule = 1.13 \times 10^{23} atoms, which is the largest number of atoms among all samples in the problem.
