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Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing numbers of atoms: \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}, 23 \mathrm{~g} \mathrm{Na}, 6.0 \times 10^{23} \mathrm{~N}_{2}\) molecules.

Short Answer

Expert verified
The samples can be ranked in order of increasing number of atoms as follows: 23 g of Na (\(6.022 \times 10^{23}\) atoms) < \(6.0 \times 10^{23}\) N2 molecules (\(1.20 \times 10^{24}\) atoms) < 0.50 mol H2O (\(1.81 \times 10^{24}\) atoms).

Step by step solution

01

Calculate the number of atoms in 0.50 mol H2O.

Using the Avogadro's constant, which is \(6.022 \times 10^{23}\) atoms/mol, we can calculate the number of atoms in 0.50 mol H2O. Since there are three atoms in each H2O molecule (two H atoms and one O atom), the calculation is as follows: Number of atoms in 0.50 mol H2O = (0.50 mol) * (6.022 x 10^23 atoms/mol) * 3 = \(1.81 \times 10^{24}\) atoms
02

Calculate the number of atoms in 23 g of Na.

Using the atomic weight of Na (23 g/mol), we can first calculate the number of moles and then convert it into the number of atoms using the Avogadro's constant: Number of moles of Na = (23 g) / (23 g/mol) = \(1 \mathrm{~mol}\) Number of atoms in 23 g of Na = (1 mol) * (6.022 x 10^23 atoms/mol) = \(6.022 \times 10^{23}\) atoms
03

Calculate the number of atoms in \(6.0 \times 10^{23}\) N2 molecules.

Since there are two N atoms in each N2 molecule, we can calculate the number of atoms as follows: Number of atoms in \(6.0 \times 10^{23}\) N2 molecules = (\(6.0 \times 10^{23}\) molecules) * (2 atoms/molecule) = \(1.20 \times 10^{24}\) atoms
04

Rank the samples in order of increasing numbers of atoms.

Based on our calculations, the samples have the following numbers of atoms: 1. 23 g of Na: \(6.022 \times 10^{23}\) atoms 2. \(6.0 \times 10^{23}\) N2 molecules: \(1.20 \times 10^{24}\) atoms 3. 0.50 mol H2O: \(1.81 \times 10^{24}\) atoms Thus, the order of increasing numbers of atoms is: 23 g of Na < \(6.0 \times 10^{23}\) N2 molecules < 0.50 mol H2O

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Key Concepts

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

Avogadro's Constant
Understanding Avogadro's constant is fundamental to comprehending the mole concept and comparing quantities of atoms in different substances. Avogadro's constant, denoted as \(6.022 \times 10^{23}\), represents the number of units (atoms, molecules, or ions) in one mole of a substance. In the context of our exercise,

when we talk about \(0.50 \text{ mol} \text{ H}_2\text{O}\) or \(23 \text{ g Na}\), we're referring to amounts that are related to each other through Avogadro's constant. For instance, if we have one mole of any substance, we inherently have \(6.022 \times 10^{23}\) units of that substance, be they atoms in elemental sodium (Na) or molecules in water (H2O).

Practical Use and Calculations

Avogadro's constant lets us translate between the microscopic world of atoms and the macroscopic world we can measure in a lab. For instance, if a student needs to find the number of atoms in a given sample, they'd use Avogadro's constant as a conversion factor. This constant is a bridge and makes stoichiometry and molecular calculations possible and meaningful in chemistry.
Atomic Weights
The atomic weight of an element, sometimes called its relative atomic mass, is the average mass of its atoms compared to 1/12th of the mass of a carbon-12 atom. It's a dimensionless quantity, commonly expressed in atomic mass units (amu) or grams per mole (g/mol). This value is crucial when comparing different elements and their amounts in terms of moles.

In our exercise, for example, sodium (Na) has an atomic weight of approximately 23 g/mol. This number indicates that one mole of sodium weighs 23 grams and contains \(6.022 \times 10^{23}\) atoms of sodium. When a problem, such as the one we're considering, provides a mass of an element, we need the atomic weight to transition from grams to moles, and then to the number of atoms using Avogadro's constant.

How Atomic Weights Facilitate Comparisons

Atomic weights enable the comparison between different elements and their respective amounts in a reaction. Without this knowledge, it would be nearly impossible to conduct precise chemical calculations or to compare one substance to another quantitatively, as we are doing between samples of Na, N2, and H2O.
Mole Concept
The mole is a cornerstone concept in chemistry that measures quantity. One mole is defined as exactly \(6.022 \times 10^{23}\) elementary entities, which may be atoms, molecules, ions, or other particles. This number, Avogadro's constant, is selected to provide an easy bridge between the atomic scale and the scale of chemical reactions at the macroscopic level.

When confronted with exercises involving different substances, the mole concept allows us to count particles by weighing them. For instance, in the provided exercise, we are not directly counting atoms or molecules; rather, we use the weight of Na and the given number of moles for H2O to infer the number of atoms present.

The mole is particularly helpful because it is independent of the type of substance. It merely represents a quantity, much like a 'dozen' represents twelve items, regardless of what those items are. This feature of the mole concept is why we can provide a straightforward ordering of various substances based on the number of atoms, as we did with our samples of Na, N2 molecules, and H2O.

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Most popular questions from this chapter

A key step in balancing chemical equations is correctly identifying the formulas of the reactants and products. For example, consider the reaction between calcium oxide, \(\mathrm{CaO}(s)\), and \(\mathrm{H}_{2} \mathrm{O}(l)\) to form aqueous calcium hydroxide. (a) Write a balanced chemical equation for this combination reaction, having correctly identified the product as \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)\). (b) Is it possible to balance the equation if you incorrectly identify the product as \(\mathrm{CaOH}(a q)\), and if so, what is the equation?

Write a balanced chemical equation for the reaction that occurs when (a) titanium metal undergoes a combination reaction with \(\mathrm{O}_{2}(g) ;\) (b) silver(I) oxide decomposes into silver metal and oxygen gas when heated; (c) propanol, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(l)\) burns in air; (d) methyl tert-butyl ether, \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}(l)\), burns in air.

Valproic acid, used to treat seizures and bipolar disorder, is composed of \(\mathrm{C}, \mathrm{H}\), and O. A 0.165-g sample is combusted in an apparatus such as that shown in Figure 3.14. The gain in mass of the \(\mathrm{H}_{2} \mathrm{O}\) absorber is \(0.166 \mathrm{~g}\), whereas that of the \(\mathrm{CO}_{2}\) absorber is \(0.403 \mathrm{~g}\). What is the empirical formula for valproic acid? If the molar mass is \(144 \mathrm{~g} / \mathrm{mol}\) what is the molecular formula?

(a) Combustion analysis of toluene, a common organic solvent, gives \(5.86 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(1.37 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\). If the compound contains only carbon and hydrogen, what is its empirical formula? (b) Menthol, the substance we can smell in mentholated cough drops, is composed of \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{O}\). A \(0.1005\) - \(g\) sample of menthol is combusted, producing \(0.2829 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.1159 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\). What is the empirical formula for menthol? If menthol has a molar mass of \(156 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?

Determine the formula weights of each of the following compounds: (a) nitric acid, \(\mathrm{HNO}_{3} ;\) (b) \(\mathrm{KMnO}_{4} ;\) (c) \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\); (d) quartz, \(\mathrm{SiO}_{2} ;\) (e) gallium sulfide, (f) chromium(III) sulfate, (g) phosphorus trichloride.

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