Question

Determine the number of moles of \(\mathrm{Br}_{2}\) in a sample consisting of (a) \(8.08 \times 10^{22} \mathrm{Br}_{2}\) molecules; (b) \(2.17 \times 10^{24}\) Br atoms; (c) 11.3 kg bromine; (d) \(2.65 \mathrm{L}\) liquid bromine \((d=3.10 \mathrm{g} / \mathrm{mL})\)

Short answer

The answer will be the calculated value for each part (a), (b), (c), and (d) obtained after performing the calculations outlined in the steps above.

Step-by-step solution

- Calculate Moles from Molecules

Given that there are \(8.08 \times 10^{22}\) molecules of bromine \(\mathrm{Br}_{2}\), the number of moles can be calculated by dividing by Avogadro's number \(6.022 \times 10^{23}\) mol\(^{-1}\). Thus, Moles = \(\frac{8.08 \times 10^{22}}{6.022 \times 10^{23}}\)

- Calculate Moles from Atoms

Given that there are \(2.17 \times 10^{24}\) atoms of bromine, we have to keep in mind that one molecule of \(\mathrm{Br}_{2}\) has two atoms. So first calculate the number of molecules and then calculate the moles. Thus, Molecules = \(\frac{2.17 \times 10^{24}}{2}\), and Moles = \(\frac{Molecules}{6.022 \times 10^{23}}\)

- Calculate Moles from Mass

For 11.3 kg of bromine, first convert the mass from kg to g, then use the molar mass of \(\mathrm{Br}_{2} = 159.808 g/mol\). Thus, Mass = \(11.3 \times 10^{3}\) g, and Moles = \(\frac{Mass}{159.808}\)

- Calculate Moles from Volume

Given a volume 2.65 L of liquid bromine and the density \(3.10 g/cm^{3}\), convert the volume from L to mL, find the mass using the density, then use the molar mass of \(\mathrm{Br}_{2} = 159.808 g/mol\) to find the moles. Thus, Volume = \(2.65 \times 10^{3}\) mL, Mass = Volume \(\times\) Density, and Moles = \(\frac{Mass}{159.808}\)

Key concepts

Avogadro's number

Understanding Avogadro's number can make mole calculations easier. Avogadro's number is a fundamental constant, crucial in chemistry, and it represents the number of units (atoms, molecules, etc.) in one mole of a substance.
Avogadro's number is approximately \(6.022 \times 10^{23}\) mol\(^{-1}\).
This means that whether you have molecules of bromine, atoms of bromine, or any other particles, their quantity equals one mole when there are \(6.022 \times 10^{23}\) of them.

For example, to calculate the moles of \( \mathrm{Br}_{2} \) from \(8.08 \times 10^{22}\) molecules, we simply divide the number of molecules by Avogadro's number:

• Moles = \(\frac{8.08 \times 10^{22}}{6.022 \times 10^{23}}\) = approximately \(0.134\) moles.

This conversion is useful in chemistry because it allows you to relate masses of substances in the macroscopic world to numbers of atoms and molecules on the atomic level. This knowledge is fundamental for understanding stoichiometry, reactions, and yields in chemistry.

Molar Mass

Molar mass bridges the gap between the mass of a substance and the amount in moles. It's the mass of one mole of a substance and is expressed in grams per mole \((g/mol)\).
For diatomic bromine \(\mathrm{Br}_{2}\), the molar mass is \(159.808\) g/mol.
This value is derived from the atomic masses of the two bromine atoms in each molecule.

When calculating moles from mass, you first need to ensure that your mass is in grams. Here's how it's done:

• Convert the mass from kilograms to grams, for instance, \(11.3\, kg\) to \(11300\, g\).
• Use the molar mass formula: Moles = \(\frac{Mass}{Molar\: Mass}\). For \(11300\, g\:and\:159.808\, g/mol\), it results in approximately \(70.7\) moles.

Understanding molar mass is essential for converting between the mass of a substance you can measure and the moles, which relate to the amount of substance in chemical reactions. It's also key when you're dealing with concentrations and preparing solutions.

Bromine Density

Density links the volume of a substance to its mass and is described by \(d = \frac{mass}{volume}\). For bromine, the density is given as \(3.10\, g/mL\).
This property allows us to find the mass from a given volume, which is particularly helpful if a substance's physical state makes it hard to measure its mass directly.

Here's the step by step with liquid bromine:

• First, convert volume from liters to milliliters for easier calculations: \(2.65\, L = 2650\, mL\).
• Calculate mass using density: Mass = Volume \(\times\) Density. For \(2650\, mL\), this becomes \(2650\, mL\times 3.10\, g/mL = 8215\, g\).
• Convert mass to moles using the molar mass of \(\mathrm{Br}_{2}\) once more: Moles = \(\frac{8215}{159.808}\). This results in approximately \(51.4\) moles.

Mastering density allows you to work effectively with liquid chemicals, calculating mass and then transitioning to the more chemically useful measure of moles. Hence, density, along with molar mass and Avogadro's number, are invaluable in chemical data analysis, offering ways to calculate important quantities from practical measurements.