Question

\(0.25 \mathrm{~mol}\) of \(\mathrm{P}_{4}\) molecules contains ___ atoms. (a) \(1.764 \times 10^{23}\) (b) \(6.02 \times 10^{19}\) (c) \(6.023 \times 10^{23}\) (d) \(8.086 \times 10^{23}\)

Short answer

\(0.25\) moles of \(\mathrm{P}_4\) molecules contain \(6.022 \times 10^{23}\) atoms. The correct answer is (c).

Step-by-step solution

Understand the Mole Concept

A "mole" of any substance contains Avogadro's number of entities (atoms, molecules, etc.), which is approximately \(6.022 \times 10^{23}\). Thus, one mole of \(\mathrm{P}_4\) molecules contains \(6.022 \times 10^{23}\) molecules of \(\mathrm{P}_4\).

Calculate the Number of \(\mathrm{P}_4\) Molecules

Since we have \(0.25\) moles of \(\mathrm{P}_4\), we find the number of \(\mathrm{P}_4\) molecules by multiplying the number of moles by Avogadro's number: \(0.25 \times 6.022 \times 10^{23} = 1.5055 \times 10^{23}\) molecules.

Determine the Number of Atoms in \(\mathrm{P}_4\) Molecules

Each \(\mathrm{P}_4\) molecule contains 4 atoms of phosphorus. Therefore, the total number of phosphorus atoms is given by: \(1.5055 \times 10^{23} \times 4 = 6.022 \times 10^{23}\) atoms.

Match with Provided Options

The calculated number of atoms in \(0.25\) moles of \(\mathrm{P}_4\) is \(6.022 \times 10^{23}\), matching option \(\text{(c)}\).

Key concepts

Mole Concept

The mole is a fundamental concept in chemistry that allows scientists to count entities at the atomic or molecular level, which is crucial because of their minute size. Essentially, one mole of any substance contains Avogadro's number of particles. This number, approximately \(6.022 \times 10^{23}\), is named after the Italian scientist Amedeo Avogadro.

• One mole of atoms, ions, or molecules contains the same number of entities as there are in 12 grams of carbon-12.
• This incredibly large number helps chemists to easily convert between atoms/molecules and grams.

By understanding the mole concept, we can relate the mass of a substance to the number of atoms or molecules it contains. This bridges the gap between the macroscopic and microscopic worlds in chemistry.

P4 Molecules

P4 refers to a specific molecular form of phosphorus, consisting of four phosphorus atoms bonded together. It is one of the common allotropes of elemental phosphorus.

• P4 exists as a tetrahedral structure, contributing to its chemical properties and reactivity.
• In calculations, it's essential to remember each P4 molecule consists of exactly four phosphorus atoms.

Knowing the composition of P4 molecules is vital for calculating the total number of atoms when given moles of P4. In the exercise, we started with 0.25 moles of P4 molecules, multiplying by Avogadro’s number gives us the number of complete P4 molecules.

Phosphorus Atoms

Phosphorus atoms, represented by the symbol \(\mathrm{P}\), are fundamental units comprising phosphorus molecules like \(\mathrm{P}_4\). When dealing with \(\mathrm{P}_4\), it is crucial to recognize that each molecule involves four atoms.

• This understanding allows us to determine total atoms by multiplying the number of \(\mathrm{P}_4\) molecules by four.
• The multiplication shows how straightforward chemical calculations can predict precise atom counts in a compound.

In this problem, even though we only have \(0.25\) moles of \(\mathrm{P}_4\), by multiplying by four after translating moles to the number of molecules, we correctly determine how many phosphorus atoms we truly have.

Chemistry Problem Solving

Solving chemistry problems typically involve a structured approach, moving from known values to unknowns. Here’s how we approach it:

• First, identify the given quantity, such as moles of a substance.
• Next, utilize fundamental constants like Avogadro's number to convert between moles and the number of molecules.
• Finally, assess the molecular composition to determine the total number of constituent atoms.

A clear understanding of formulas and their applications, as demonstrated by finding the number of atoms in \(\mathrm{P}_4\) from moles, illustrates how organized, step-by-step calculations simplify seemingly complicated chemistry problems. Practicing these steps helps build confidence and proficiency in problem-solving in chemistry.