A silver ring contains 3.4 x 1034 silver atoms.

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A silver ring contains 3.4 x 1034 silver atoms. What is the mass of the silver ring? 2.) Calculate the mass of 1.23 x 1024 NaCl atoms. 3.) How many molecules are in a sample of water with a mass of 3.64g.

The Correct Answer and Explanation is:

1. What is the mass of a silver ring that contains 3.4×10343.4 \times 10^{34}3.4×1034 silver atoms?

Given:

  • Number of silver (Ag) atoms = 3.4×10343.4 \times 10^{34}3.4×1034
  • Molar mass of Ag = 107.87 g/mol
  • Avogadro’s number = 6.022×10236.022 \times 10^{23}6.022×1023 atoms/mol

Solution:

  1. Convert atoms to moles:

Moles of Ag=3.4×10346.022×1023=5.645×1010 mol\text{Moles of Ag} = \frac{3.4 \times 10^{34}}{6.022 \times 10^{23}} = 5.645 \times 10^{10} \text{ mol}Moles of Ag=6.022×10233.4×1034​=5.645×1010 mol

  1. Multiply moles by molar mass:

Mass=5.645×1010 mol×107.87 g/mol=6.087×1012 g\text{Mass} = 5.645 \times 10^{10} \text{ mol} \times 107.87 \text{ g/mol} = 6.087 \times 10^{12} \text{ g}Mass=5.645×1010 mol×107.87 g/mol=6.087×1012 g

2. Calculate the mass of 1.23×10241.23 \times 10^{24}1.23×1024 NaCl units.

Given:

  • Number of NaCl formula units = 1.23×10241.23 \times 10^{24}1.23×1024
  • Molar mass of NaCl = 58.44 g/mol

Solution:

  1. Convert to moles:

Moles of NaCl=1.23×10246.022×1023=2.043 mol\text{Moles of NaCl} = \frac{1.23 \times 10^{24}}{6.022 \times 10^{23}} = 2.043 \text{ mol}Moles of NaCl=6.022×10231.23×1024​=2.043 mol

  1. Multiply by molar mass:

Mass=2.043 mol×58.44 g/mol=119.45 g\text{Mass} = 2.043 \text{ mol} \times 58.44 \text{ g/mol} = 119.45 \text{ g}Mass=2.043 mol×58.44 g/mol=119.45 g

3. How many water (H₂O) molecules are in a 3.64 g sample?

Given:

  • Mass of water = 3.64 g
  • Molar mass of H₂O = 18.02 g/mol

Solution:

  1. Convert grams to moles:

Moles of H2O=3.6418.02=0.202 mol\text{Moles of H}_2\text{O} = \frac{3.64}{18.02} = 0.202 \text{ mol}Moles of H2​O=18.023.64​=0.202 mol

  1. Convert moles to molecules:

Molecules=0.202 mol×6.022×1023=1.216×1023 molecules\text{Molecules} = 0.202 \text{ mol} \times 6.022 \times 10^{23} = 1.216 \times 10^{23} \text{ molecules}Molecules=0.202 mol×6.022×1023=1.216×1023 molecules

📘 Textbook-Style Explanation

To solve problems involving atoms, molecules, and mass, chemists use the mole concept, which bridges the microscopic world of atoms and the macroscopic world of grams. One mole of any substance contains 6.022×10236.022 \times 10^{23}6.022×1023 representative particles (atoms, molecules, or formula units). This is known as Avogadro’s number.

In Problem 1, we are given the number of silver atoms. To find the mass of the silver ring, we first convert atoms to moles using Avogadro’s number. Once we have the number of moles, we multiply by the molar mass of silver (107.87 g/mol) to get the mass in grams. This is a direct application of the formula:mass=atoms6.022×1023×molar mass\text{mass} = \frac{\text{atoms}}{6.022 \times 10^{23}} \times \text{molar mass}mass=6.022×1023atoms​×molar mass

In Problem 2, we follow a similar procedure. Given the number of NaCl formula units (which are treated like molecules in ionic compounds), we convert them to moles and then multiply by the molar mass of NaCl (58.44 g/mol). This gives the mass of the salt sample.

In Problem 3, the situation is reversed. Instead of being given particles and finding mass, we’re given mass and asked to find the number of water molecules. We first convert grams to moles using the molar mass of water (18.02 g/mol), and then multiply by Avogadro’s number to find the number of molecules.

These calculations demonstrate the central importance of the mole concept in stoichiometry. Whether you’re scaling atomic quantities to measurable amounts or breaking down a sample to count individual particles, Avogadro’s number and molar mass serve as the critical conversion factors between the atomic scale and the laboratory scale.

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