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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

If an average sodium atom weighs 22.99 amu, how many sodium atoms are contained in \(1.98 \times 10^{13}\) amu of sodium? What will \(3.01 \times 10^{23}\) sodium atoms weigh?

Hydrogen gas reacts with each of the halogen elements to form the hydrogen halides (HF, HCl, HBr, HI). Calculate the percent by mass of hydrogen in each of these compounds.

Find the item in column 2 that best explains or completes the statement or question in column 1. Column 1 (1) 1 amu (2) 1008 amu (3) mass of the "average" atom of an element (4) number of carbon atoms in \(12.01 \mathrm{g}\) of carbon (5) \(6.022 \times 10^{23}\) molecules (6) total mass of all atoms in 1 mol of a compound (7) smallest whole-number ratio of atoms present in a molecule (8) formula showing actual number of atoms present in a molecule (9) product formed when any carbon-containing compound is burned in \(\mathrm{O}_{2}\) (10) have the same empirical formulas, but different molecular formulas Column 2 (a) \(6.022 \times 10^{23}\) (b) atomic mass (c) mass of 1000 hydrogen atoms (d) benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\), and acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\) (e) carbon dioxide (f) empirical formula (g) \(1.66 \times 10^{-24} \mathrm{g}\) (h) molecular formula (i) molar mass (j) 1 mol

How does the molecular formula of a compound differ from the empirical formula? Can a compound's empirical and molecular formulas be the same? Explain.

What does an average magnesium atom weigh (in amu)? What would 345 magnesium atoms weigh? How many magnesium atoms are contained in a sample of magnesium that has a mass of \(2.071 \times\) \(10^{4}\) amu?
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