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Chapter 1: Problem 48

The common Eastern mole, a mammal, typically has a mass of \(75 \mathrm{~g}\), which corresponds to about \(7.5\) moles of atoms. (A mole of atoms is \(6.02 \times 10^{23}\) atoms.) In atomic mass units (u), what is the average mass of the atoms in the common Eastern mole?

Short Answer

Expert verified
The average mass of the atoms is 10 atomic mass units.

Step by step solution

01

Calculate the Total Mass in Atomic Mass Units

Start by converting the mass of the common Eastern mole from grams to atomic mass units. Recall that 1 gram is approximately equivalent to \(6.02 \times 10^{23}\) atomic mass units. Therefore, the total mass in atomic mass units is \(75 \, \text{g} \times 6.02 \times 10^{23} \, \text{u/g}\). This equals \(4.515 \times 10^{25} \, \text{u}\).
02

Determine the Number of Atoms in the Mole

The problem states that the common Eastern mole corresponds to about \(7.5\) moles of atoms. Since one mole is \(6.02 \times 10^{23}\) atoms, \(7.5\) moles is \(7.5 \times 6.02 \times 10^{23}\). This equals \(4.515 \times 10^{24}\) atoms in the common Eastern mole.
03

Compute the Average Mass of the Atoms

Now that we have both the total mass in atomic mass units and the total number of atoms, we can determine the average mass of an atom. Divide the total mass by the total number of atoms: \(\frac{4.515 \times 10^{25} \, \text{u}}{4.515 \times 10^{24} \, \text{atoms}}\). This simplifies to \(10 \, \text{u/atom}\).

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Key Concepts

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

Moles of Atoms
The concept of "moles" is essential in chemistry to quantify substances at the atomic level. A "mole" bridges macroscopic amounts of matter with the atomic scale, providing a standard unit to count atoms, molecules, and other particles. One mole of any substance contains exactly Avogadro's number,
  • Standard Definition: A mole is defined as the amount of substance containing as many elementary entities (atoms, molecules) as there are atoms in 12 grams of pure carbon-12.
  • Relationship with Mass: The molar mass of an element (expressed in grams per mole) is numerically equal to its atomic mass (in atomic mass units, u).
Understanding moles helps translate atomic scale measurements to laboratory quantities. For the Eastern mole, even though its mass is in grams, we understand it contains about 7.5 moles of atoms, making it a convenient bridge for atomic calculations.
Average Atomic Mass
The average atomic mass of atoms in a sample gives an idea of the mass of an atom within a collection of various isotopes of the same element. This is typically calculated from the isotope abundances. In the context of the Eastern mole problem, the concept is simplified as follows:
  • Total Mass Calculation: Convert the mass of the Eastern mole from grams to atomic mass units (u) because atomic masses are expressed this way. Here, 1 gram equals approximately \(6.02 \times 10^{23}\) u, leading to a total mass of \(75 \, \text{g} \times 6.02 \times 10^{23} \, \text{u/g} = 4.515 \times 10^{25} \, \text{u}\).
  • Average Mass of Each Atom: By dividing the total atomic mass by the total number of atoms, \(4.515 \times 10^{25} \, \text{u} \div 4.515 \times 10^{24} \, \text{atoms}\), the average atomic mass is found as 10 u per atom.
This calculated average mass gives insights into the size of the atoms in a practical manner for complex mixtures.
Avogadro's Number
Avogadro's number is a fundamental constant that plays a crucial role in chemical calculations by providing the link between atomic and macroscopic scales. Avogadro's number is
  • Definition: Avogadro's number is \(6.02 \times 10^{23}\), representing the number of atoms, ions, or molecules in one mole of substance.
  • Utility in Chemistry: It allows conversion between number of atoms and moles, making it easy to carry out calculations involving chemical reactions and quantities.
For example, in the scenario of the Eastern mole, knowing the number of moles (7.5) allows us to use Avogadro's number to determine the total number of atoms, coming out to \(7.5 \times 6.02 \times 10^{23}\). This facilitates further understanding and calculations about the substance, enabling precise laboratory work.

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Most popular questions from this chapter

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In the United States, a doll house has the scale of \(1: 12\) of a real house (that is, each length of the doll house is \(\frac{1}{12}\) that of the real house) and a miniature house (a doll house to fit within a doll house) has the scale of \(1: 144\) of a real house. Suppose a real house (Fig. 1-7) has a front length of \(20 \mathrm{~m}\), a depth of \(12 \mathrm{~m}\), a height of \(6.0 \mathrm{~m}\), and a standard sloped roof (vertical triangular faces on the ends) of height \(3.0 \mathrm{~m}\). In cubic meters, what are the volumes of the corresponding (a) doll house and (b) miniature house?
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