Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 30

Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing number of atoms: \(9.0 \times 10^{23}\) molecules of \(\mathrm{H}_{2} \mathrm{O}_{2}, 2.0 \mathrm{~mol} \mathrm{CH}_{4}, 16 \mathrm{~g} \mathrm{O}_{2}\)

Short Answer

Expert verified
The order of increasing number of atoms for the given samples, without doing detailed calculations, is: O2 < H2O2 < CH4.

Step by step solution

01

Identify atomic weights

Refer to the periodic table to find the atomic weights of the elements involved in the samples: - Hydrogen (H): 1 g/mol - Oxygen (O): 16 g/mol - Carbon (C): 12 g/mol
02

Calculating number of atoms in sample 1

Given: - \(9.0 \times 10^{23}\) molecules of H2O2 Each molecule of H2O2 contains 2 hydrogen atoms and 2 oxygen atoms: - Number of H atoms = \(9.0 \times 10^{23} \times 2\) - Number of O atoms = \(9.0 \times 10^{23} \times 2\) Total number of atoms in H2O2 = H atoms + O atoms
03

Calculating number of atoms in sample 2

Given: - 2.0 moles of CH4 Each molecule of CH4 contains 1 carbon atom and 4 hydrogen atoms. We know the number of atoms in a mole = Avogadro's Number (6.022 x 10^23). So, for 2 moles of CH4, atoms: - Number of H atoms = \(2 \times 4 \times (6.022 \times 10^{23})\) - Number of C atoms = \(2 \times 1 \times (6.022 \times 10^{23})\) Total number of atoms in CH4 = H atoms + C atoms
04

Calculating number of atoms in sample 3

Given: - 16 g of O2 Each molecule of O2 has 2 oxygen atoms. Oxygen has a molar mass of 16 g/mol. To find the number of moles in 16 g of O2, we can use the formula: Moles (O2) = (Mass of O2) / (Molar mass of O2) = \(16/32\) = 0.5 moles Now we can find the number of atoms: - Number of O atoms = \(0.5 \times 2 \times (6.022 \times 10^{23})\) Total number of atoms in O2 = O atoms
05

Rank the samples

Now, we'll compare the number of atoms in each sample: 1. H2O2: atoms = \(9.0 \times 10^{23} \times 2\) 2. CH4: atoms = \(2 \times 5 \times (6.022 \times 10^{23})\) 3. O2: atoms = \(0.5 \times 2 \times (6.022 \times 10^{23})\) The order of increasing number of atoms: O2 < H2O2 < CH4

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{CH}_{2}\), molar mass \(=84 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{NH}_{2} \mathrm{Cl}\), molar mass \(=51.5 \mathrm{~g} / \mathrm{mol}\)

An element \(X\) forms an iodide \(\left(\mathrm{Xl}_{3}\right)\) and a chloride \(\left(\mathrm{XCl}_{3}\right)\). The iodide is quantitatively converted to the chloride when it is heated in a stream of chlorine: $$ 2 \mathrm{XI}_{3}+3 \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{XCl}_{3}+3 \mathrm{I}_{2} $$ If \(0.5000 \mathrm{~g}\) of \(\mathrm{Xl}_{3}\) is treated, \(0.2360 \mathrm{~g}\) of \(\mathrm{XCl}_{3}\) is obtained. (a) Calculate the atomic weight of the element X. (b) Identify the element X.

The molecular formula of aspartame, the artificial sweetener marketed as NutraSweet \(^{\oplus}\), is \(\mathrm{C}_{14} \mathrm{H}_{18} \mathrm{~N}_{2} \mathrm{O}_{5} .\) (a) What is the molar mass of aspartame? (b) How many moles of aspartame are present in \(1.00 \mathrm{mg}\) of aspartame? (c) How many molecules of aspartame are present in \(1.00 \mathrm{mg}\) of aspartame? (d) How many hydrogen atoms are present in \(1.00 \mathrm{mg}\) of aspartame?

Give the empirical formula of each of the following compounds if a sample contains (a) \(0.0130 \mathrm{~mol} \mathrm{C}, 0.0390 \mathrm{~mol} \mathrm{H},\) and \(0.0065 \mathrm{~mol} \mathrm{O} ;\) (b) \(11.66 \mathrm{~g}\) iron and \(5.01 \mathrm{~g}\) oxygen; (c) \(40.0 \% \mathrm{C}, 6.7 \% \mathrm{H},\) and \(53.3 \% \mathrm{O}\) by mass.

The fizz produced when an Alka-Seltzer \(^{\circledast}\) tablet is dissolved in water is due to the reaction between sodium bicarbonate \(\left(\mathrm{NaHCO}_{3}\right)\) and citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\) $$ \begin{aligned} 3 \mathrm{NaHCO}_{3}(a q)+\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) \longrightarrow \\ & 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) \end{aligned} $$ In a certain experiment \(1.00 \mathrm{~g}\) of sodium bicarbonate and \(1.00 \mathrm{~g}\) of citric acid are allowed to react. (a) Which is the limiting reactant? (b) How many grams of carbon dioxide form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free