Chapter 1: Problem 68
A gaseous mixture contains \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) in the ratio \(1: 4\) by weight. Then the ratio of their number of molecules in the mixture is: (a) \(3: 32\) (b) \(7: 32\) (c) \(1: 4\) (d) \(3: 16\)
Short Answer
Expert verified
The ratio of molecules is \(7: 32\), option (b).
Step by step solution
01
Understand the Weight Ratio
The problem states that the gases \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) are present in a weight ratio of \(1: 4\). This means that for every 1 unit of weight of \(\mathrm{O}_{2}\), there are 4 units of weight of \(\mathrm{N}_{2}\).
02
Find the Molar Masses
Calculate the molar weights for \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\). The molar mass of \(\mathrm{O}_{2}\) is \(32\, \text{g/mol}\) and the molar mass of \(\mathrm{N}_{2}\) is \(28\, \text{g/mol}\).
03
Set Up the Mole Ratio Equation
Since we are given the weight ratio and asked to find the molecule (or mole) ratio, use the formula: \(\text{Number of moles} = \frac{\text{Weight}}{\text{Molar Mass}}\).
04
Calculate the Number of Moles
Determine the moles of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) using their weight ratio and molar masses:\[n_{\text{O}_2} = \frac{1}{32}\]\[n_{\text{N}_2} = \frac{4}{28}\]
05
Simplify the Mole Ratio
Simplify the mole ratio \(\frac{n_{\text{O}_2}}{n_{\text{N}_2}}\) using the calculated moles:\[\frac{\frac{1}{32}}{\frac{4}{28}} = \frac{1 \times 28}{4 \times 32} = \frac{28}{128} = \frac{7}{32}\]
06
Choose the Correct Option
The simplified ratio of the number of molecules (moles) in the gaseous mixture is \(7: 32\). Therefore, the correct answer is option (b) \(7: 32\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molar Mass
Molar mass plays a critical role in stoichiometry as it helps determine the number of moles from a given mass. It’s like a handy conversion tool for chemists. Every substance has a molar mass, which is the mass of one mole of its atoms or molecules. For example, the molar mass of
- Oxygen molecule \(\mathrm{O}_{2}\) is \(32\, \text{g/mol}\) because oxygen has an atomic mass of \(16\, \text{g/mol}\), and a molecule contains two oxygen atoms.
- Nitrogen molecule \(\mathrm{N}_{2}\) is \(28\, \text{g/mol}\) given that nitrogen has an atomic mass of \(14\, \text{g/mol}\).
Molecular Ratio
In chemistry, understanding molecular ratios is key to comparing amounts of different substances in a mixture. A molecular ratio reveals how many units of each molecule are present in that mixture. It links the weight ratio (mass ratio) of substances to the number of molecules. To find the molecular ratio, you use the relationship:
- \(\text{Number of moles} = \frac{\text{Weight}}{\text{Molar Mass}}\)
Gaseous Mixture
A gaseous mixture is a combination of different gases that coexist without reacting chemically. Think of it as a cocktail of molecules floating around in the air. In the case of a gaseous mixture of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\), these gases remain as separate entities but occupy the same space.When dealing with gaseous mixtures, especially in stoichiometry, it’s important to differentiate between the physical presence of molecules and their potential interactions. Knowing the molecular ratio of gases like \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) can help determine how much of each gas is available, aiding in predictions about chemical behavior or efficiency in chemical processes like combustion.Studying gaseous mixtures enhances the understanding of atmospheric chemistry and processes like respiration, where different gases perform distinct roles but coexist in a mixture.
Number of Moles
The concept of "number of moles" is fundamental in stoichiometry and helps in quantifying chemical substances. Rather than talking about individual atoms or molecules, which are incredibly tiny, chemists use moles to describe amounts of substance in a manageable mathematical way. One mole contains Avogadro’s number, \(6.022 \times 10^{23}\), of entities (atoms, molecules, ions).Given that:
- \(\text{Number of moles} = \frac{\text{Weight}}{\text{Molar Mass}}\)