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Out of \(1.0 \mathrm{~g}\) dioxygen, \(1.0 \mathrm{~g}\) atomic oxygen and \(1.0 \mathrm{~g}\) ozone, the maximum number of oxygen atoms are contained in (a) \(1.0 \mathrm{~g}\) of atomic oxygen (b) \(1.0 \mathrm{~g}\) of ozone (c) \(1.0 \mathrm{~g}\) of oxygen gas (d) All contain the same number of atoms

Short Answer

Expert verified
The maximum number of oxygen atoms are contained in 1.0 g of atomic oxygen, since it has the lowest molar mass, and thus the highest number of moles of atoms per gram.

Step by step solution

01

Calculate the number of moles of atomic oxygen

Use the molar mass of atomic oxygen to convert grams to moles. The molar mass of atomic oxygen (O) is 16.00 g/mol. Using the formula: number of moles = mass / molar mass, calculate the number of moles of atomic oxygen in 1.0 g.
02

Calculate the number of moles of dioxygen

Dioxygen (O2) has a molar mass of 32.00 g/mol (16.00 g/mol per oxygen atom times 2). Use the same formula: number of moles = mass / molar mass, to calculate the number of moles of dioxygen in 1.0 g.
03

Calculate the number of moles of ozone

Ozone (O3) has a molar mass of 48.00 g/mol (16.00 g/mol per oxygen atom times 3). Use the formula: number of moles = mass / molar mass to calculate the number of moles of ozone in 1.0 g.
04

Calculate the number of atoms for each substance

Multiply the number of moles of each substance by Avogadro's number (approximately 6.022 x 10^23 atoms/mol) to obtain the total number of atoms for each substance.
05

Compare the number of atoms

Compare the total number of oxygen atoms for each form (atomic oxygen, ozone, and oxygen gas). The number of atoms is directly proportional to the number of moles since Avogadro's number is a constant that will be multiplied by each mole value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass Calculation
Understanding the molar mass of a substance is crucial for solving many chemical problems, including stoichiometry issues. The molar mass is defined as the mass of one mole of a substance, usually expressed in grams per mole (g/mol).

To calculate the molar mass, you need to sum the masses of all the atoms in a molecule. For instance, in dioxygen (O2), the molecule consists of two oxygen atoms. Since the atomic mass of oxygen is approximately 16.00 g/mol, you would multiply this by 2 to find the molar mass for dioxygen, which would be 32.00 g/mol. Similarly, for ozone (O3), you would multiply 16.00 g/mol by 3, giving a molar mass of 48.00 g/mol.

This is a critical step in stoichiometry as it allows you to convert grams of a substance to moles, which can then be related to other substances in a reaction using the coefficients in a balanced chemical equation.
Avogadro's Number
Avogadro's number, approximately 6.022 x 10^23, is another fundamental concept in chemistry that enables us to count particles on the atomic scale. It represents the number of atoms, ions, or molecules in one mole of a substance.

This constant is named after the scientist Amedeo Avogadro, who, through his hypothesis, established that equal volumes of gases at the same temperature and pressure contain an equal number of molecules. Avogadro's number allows us to answer questions like 'how many atoms are there in a given sample?' by relating a substance's macroscopic mass to its atomic-scale particle count.

For example, if you have one mole of oxygen gas, you have 6.022 x 10^23 oxygen molecules. In stoichiometry, this number helps bridge the gap between the mass of a substance and the number of particles it contains.
Moles to Atoms Conversion
Converting moles to atoms or molecules is an essential step in working through stoichiometry problems. It's performed with the help of Avogadro's number.

Let's say you have calculated the number of moles of a substance. To find out how many individual atoms that equates to, you would multiply the number of moles by Avogadro's number. The formula looks like this: number of atoms = (number of moles) x (Avogadro's number).

For instance, to find the number of atoms in 1.0 g of atomic oxygen, you first calculate the moles of oxygen using its molar mass. Then you multiply the moles by Avogadro's number to get the total number of atoms. This process is repeated for different molecular forms of oxygen, in this case, dioxygen and ozone, to compare the number of atoms contained in equal masses of different substances.

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Most popular questions from this chapter

The fractional abundance of \(\mathrm{Cl}^{35}\) in a sample of chlorine containing only \(\mathrm{Cl}^{35}\) (atomic weight \(=34.9\) ) and \(\mathrm{Cl}^{37}\) (atomic weight \(=36.9\) ) isotopes, is \(0.6\). The average mass number of chlorine is (a) \(35.7\) (b) \(35.8\) (c) \(18.8\) (d) \(35.77\)

To determine soluble (free) \(\mathrm{SiO}_{2}\) in a rock. an alkaline extraction was carried out, as a result of which there was found \(1.52 \%\) of \(\mathrm{SiO}_{2}\) in the extract and also \(1.02 \%\) of \(\mathrm{Al}_{2} \mathrm{O}_{3} .\) Considering that, apart from the free \(\mathrm{SiO}_{2}\), the extract also contained the \(\mathrm{SiO}_{2}\) that had passed into it from Kaolin \(\left(2 \mathrm{SiO}_{2} \cdot \mathrm{Al}_{2} \mathrm{O}_{3}\right)\), the percentage of free \(\mathrm{SiO}_{2}\) in the rock being analysed is \((\mathrm{Si}=28\), \(\mathrm{Al}=27\) ) (a) \(1.20\) (b) \(0.32\) (c) \(0.50\) (d) \(1.52\)

A volume of \(500 \mathrm{ml}\) of a \(0.1 \mathrm{M}\) solution of \(\mathrm{AgNO}_{3}\) added to \(500 \mathrm{ml}\) of \(0.1 \mathrm{M}\) solution of \(\mathrm{KCl}\). The concentration of nitrate ion in the resulting solution is (a) \(0.05 \mathrm{M}\) (b) \(0.1 \mathrm{M}\) (c) \(0.2 \mathrm{M}\) (d) Reduced to zero

An amount of \(1.0 \times 10^{-3}\) moles of \(\mathrm{Ag}^{+}\) and \(1.0 \times 10^{-3}\) moles of \(\mathrm{CrO}_{4}^{2-}\) reacts together to form solid \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\). What is the amount of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) formed? \((\mathrm{Ag}=108, \mathrm{Cr}=52)\) (a) \(0.332 \mathrm{~g}\) (b) \(0.166 \mathrm{~g}\) (c) \(332 \mathrm{~g}\) (d) \(166 \mathrm{~g}\)

A mixture is made equal volume of \(\mathrm{CO}\) and air. A spark passed through so that all the oxygen is converted to carbon dioxide. What will be fractional decrease in the total volume of system assuming pressure and temperature remain constant? Air contains \(20 \%\) oxygen by volume. (a) \(0.1\) (b) \(0.2\) (c) \(0.15\) (d) \(0.3\)

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