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Chapter 8: Problem 9

In 26.98 g of aluminum, there are ______ aluminum atoms present.

Short Answer

Expert verified
In 26.98 g of aluminum, there are \(6.022 \times 10^{23}\) aluminum atoms present.

Step by step solution

01

Identify the given mass of aluminum and its molar mass

The given mass of aluminum is 26.98 g. The molar mass of aluminum (Al) is 26.98 g/mol.
02

Calculate the number of moles of aluminum

To calculate the number of moles, we will use the formula: Number of moles = (given mass) / (molar mass) Number of moles = (26.98 g) / (26.98 g/mol) = 1 mol of aluminum
03

Use Avogadro's number to find the number of aluminum atoms

Avogadro's number (N_A) is 6.022 x 10^23 atoms/mol. Now, we will use it to find the number of aluminum atoms present in 1 mol of aluminum. Number of aluminum atoms = (Number of moles) 脳 (Avogadro's number) Number of aluminum atoms = (1 mol) 脳 (6.022 x 10^23 atoms/mol) Number of aluminum atoms = 6.022 x 10^23 aluminum atoms So, there are 6.022 x 10^23 aluminum atoms present in 26.98 g of aluminum.

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Key Concepts

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

Molar Mass
The molar mass is a crucial concept in chemistry that connects the mass of a substance to the number of particles it contains. It is defined as the mass of one mole of a given element or compound. Measured in grams per mole (g/mol), the molar mass allows us to convert grams to moles or vice versa.
For example, the molar mass of aluminum (Al) is 26.98 g/mol. This means one mole of aluminum weighs 26.98 grams. This measurement is derived from the atomic mass of aluminum, found on the periodic table. Knowing the molar mass helps us understand how amounts of substances relate to the number of particles in a sample, which is a fundamental part of balancing chemical reactions.
Number of Moles
The number of moles in a substance is a concept that reflects the number of entities, such as atoms or molecules, present in a given mass. Calculated using the formula:\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \] It acts as a bridge between the macroscopic world we can measure and the atomic world.
  • Given mass is the mass of the substance you have and is typically measured in grams.
  • Molar mass is the mass of one mole of the substance, as explained earlier.
In the case of aluminum, if you have 26.98 grams, and the molar mass is also 26.98 g/mol, you can calculate:\[ \text{Number of moles} = \frac{26.98 \text{ g}}{26.98 \text{ g/mol}} = 1 \text{ mol} \] This result shows you have exactly one mole of aluminum atoms.
Aluminum Atoms
Aluminum atoms are linked to the larger picture of the atomic world by using Avogadro's number, a fundamental constant that expresses the number of atoms or molecules in one mole of a substance. This number is approximately \( 6.022 \times 10^{23} \) atoms/mol.
To find out how many aluminum atoms are in a given sample, you multiply the number of moles by Avogadro's number:
  • For 1 mole of aluminum: \( \text{Number of atoms} = 1 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} \)
  • This results in \( 6.022 \times 10^{23} \) aluminum atoms.
Understanding how to link moles to atomic quantities allows us to predict and calculate the outcomes of chemical reactions accurately, giving us insights into the microscopic compositions of samples.

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Most popular questions from this chapter

Calculate the number of moles of the indicated substance present in each of the following samples. a. \(21.4 \mathrm{mg}\) of nitrogen dioxide b. \(1.56 \mathrm{g}\) of copper(II) nitrate c. \(2.47 \mathrm{g}\) of carbon disulfide d. \(5.04 \mathrm{g}\) of aluminum sulfate e. \(2.99 \mathrm{g}\) of lead(II) chloride f. \(62.4 \mathrm{g}\) of calcium carbonate

The mass percentage of a given element in a compound must always be (greater/less) than \(100 \%\).

A compound has the following percentages by mass: barium, 58.84\%; sulfur, 13.74\%; oxygen, 27.43\%. Determine the empirical formula of the compound.

Calculate the number of moles of the indicated substance present in each of the following samples. a. \(1.28 \mathrm{g}\) of iron(II) sulfate b. \(5.14 \mathrm{mg}\) of mercury(II) iodide c. \(9.21 \mu g\) of tin(IV) oxide d. 1.26 lb of cobalt(II) chloride e. \(4.25 \mathrm{g}\) of copper(II) nitrate

Calculate the mass in grams of each of the following samples. a. 1.27 mmol of carbon dioxide b. \(4.12 \times 10^{3}\) mol of nitrogen trichloride c. 0.00451 mol of ammonium nitrate d. 18.0 mol of water e. 62.7 mol of copper(II) sulfate
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