Textbook Question
A cube of gold that is 1.00 cm on a side has a mass of 19.3 g. A single gold atom has a mass of 197.0 u. (a) How many gold atoms are in the cube?
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Ch.2 - Atoms, Molecules, and Ions
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232
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Table of contents
- Ch.1 - Introduction: Matter, Energy, and Measurement140
- Ch.2 - Atoms, Molecules, and Ions192
- Ch.3 - Chemical Reactions and Reaction Stoichiometry172
- Ch.4 - Reactions in Aqueous Solution156
- Ch.5 - Thermochemistry151
- Ch.6 - Electronic Structure of Atoms137
- Ch.7 - Periodic Properties of the Elements127
- Ch.8 - Basic Concepts of Chemical Bonding143
- Ch.9 - Molecular Geometry and Bonding Theories167
- Ch.10 - Gases147
- Ch.11 - Liquids and Intermolecular Forces97
- Ch.12 - Solids and Modern Materials129
- Ch.13 - Properties of Solutions126
- Ch.14 - Chemical Kinetics175
- Ch.15 - Chemical Equilibrium107
- Ch.16 - Acid-Base Equilibria127
- Ch.17 - Additional Aspects of Aqueous Equilibria133
- Ch.18 - Chemistry of the Environment88
- Ch.19 - Chemical Thermodynamics140
- Ch.20 - Electrochemistry137
- Ch.21 - Nuclear Chemistry99
- Ch.22 - Chemistry of the Nonmetals66
- Ch.23 - Transition Metals and Coordination Chemistry69
- Ch.24 - The Chemistry of Life: Organic and Biological Chemistry11
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232Not the one you use?Change textbook
Chapter 2, Problem 89b
A cube of gold that is 1.00 cm on a side has a mass of 19.3 g. A single gold atom has a mass of 197.0 u. (b) From the information given, estimate the diameter in Å of a single gold atom.
Verified step by step guidance1
Step 1: First, we need to calculate the volume of the gold cube. The volume of a cube is given by the formula \(V = a^3\), where \(a\) is the length of one side of the cube. In this case, \(a = 1.00\) cm.
Step 2: Next, we need to calculate the number of gold atoms in the cube. We know that the mass of the cube is 19.3 g and the mass of a single gold atom is 197.0 u. We can convert the mass of the cube to atomic mass units (u) by using the conversion factor 1 g = 6.022 x 10^23 u (Avogadro's number). Then, we divide the total mass of the cube by the mass of a single atom to find the number of atoms.
Step 3: Now, we can calculate the volume occupied by a single gold atom. We divide the total volume of the cube by the number of atoms to get the volume of a single atom.
Step 4: We know that atoms are roughly spherical, so we can use the formula for the volume of a sphere \(V = \frac{4}{3}πr^3\) to find the radius of a gold atom. We can solve this equation for \(r\) to get \(r = \left(\frac{3V}{4π}\right)^{\frac{1}{3}}\).
Step 5: Finally, we can find the diameter of a gold atom by multiplying the radius by 2. Remember to convert the diameter from cm to Å by using the conversion factor 1 cm = 10^8 Å.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Density
Density is defined as mass per unit volume and is a crucial property of materials. In this case, the density of gold can be calculated using the mass of the cube and its volume. Understanding density allows us to relate the macroscopic properties of a substance to its atomic structure, which is essential for estimating atomic dimensions.
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Density Concepts
Avogadro's Number
Avogadro's number, approximately 6.022 x 10²³, is the number of atoms or molecules in one mole of a substance. This concept is vital for converting between the macroscopic scale (grams) and the atomic scale (individual atoms). By using Avogadro's number, we can determine how many gold atoms are present in the given mass, which aids in estimating the size of a single atom.
Atomic Radius
The atomic radius is a measure of the size of an atom, typically defined as the distance from the nucleus to the outermost electron shell. In this context, estimating the diameter of a gold atom involves understanding its atomic radius and how it relates to the number of atoms in a given volume. This concept is fundamental for translating the mass and volume of gold into the dimensions of individual atoms.
Related Practice
Textbook Question
The diameter of a rubidium atom is 495 pm We will consider two different ways of placing the atoms on a surface. In arrangement A, all the atoms are lined up with one another to form a square grid. Arrangement B is called a close-packed arrangement because the atoms sit in the 'depressions' formed by the previous row of atoms: (a) Using arrangement A, how many Rb atoms could be placed on a square surface that is 1.0 cm on a side?
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Textbook Question
"The diameter of a rubidium atom is 495 pm We will consider two different ways of placing the atoms on a surface. In arrangement A, all the atoms are lined up with one another to form a square grid. Arrangement B is called a close-packed arrangement because the atoms sit in the 'depressions' formed by the previous row of atoms:
(c) If extended to three dimensions, which arrangement would lead to a greater density for Rb metal?"
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Textbook Question
"The diameter of a rubidium atom is 495 pm We will consider two different ways of placing the atoms on a surface. In arrangement A, all the atoms are lined up with one another to form a square grid. Arrangement B is called a close-packed arrangement because the atoms sit in the 'depressions' formed by the previous row of atoms:
(c) By what factor has the number of atoms on the surface increased in going to arrangement B from arrangement A?
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Textbook Question
The natural abundance of 3He is 0.000137%. (b) Based on the sum of the masses of their subatomic particles, which is expected to be more massive, an atom of 3He or an atom of 3H (which is also called tritium)?
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