Question

When a sample of phosphorus burns in air, the compound \(\mathrm{P}_{4} \mathrm{O}_{10}\) forms. One experiment showed that 0.744 g of phosphorus formed \(1.704 \mathrm{g}\) of \(\mathrm{P}_{4} \mathrm{O}_{10} .\) Use this information to determine the ratio of the atomic weights of phosphorus and oxygen (mass P/mass O). If the atomic weight of oxygen is assumed to be \(16.000,\) calculate the atomic weight of phosphorus.

Short answer

The atomic weight of phosphorus is approximately 31.0 g/mol.

Step-by-step solution

Determine Total Oxygen Mass

First, find the mass of oxygen in the compound. Subtract the mass of phosphorus from the total mass of the compound: \( 1.704 ext{ g (mass of } \text{P}_4\text{O}_{10}) - 0.744 ext{ g (mass of P)} = 0.960 ext{ g (mass of O)} \).

Calculate Moles of Oxygen

Next, calculate the moles of oxygen using its atomic weight of 16.000 g/mol. Since each molecule of \(\text{P}_4\text{O}_{10}\) contains 10 oxygen atoms: \( n = \frac{0.960 ext{ g}}{16.000 ext{ g/mol}} \times \frac{1}{10} = 0.00600 \text{ mol of O} \).

Calculate Moles of Phosphorus

Now, calculate the moles of phosphorus. Since each molecule of \(\text{P}_4\text{O}_{10}\) contains 4 phosphorus atoms: \( n = \frac{0.744 ext{ g}}{x} \) where \(x\) is the atomic weight of phosphorus. Thus, \( n = 0.00600 \times \frac{4}{10} = 0.00240 \text{ mol of P} \).

Formulate Ratio of Moles and Equation

Set the moles of phosphorus calculated equal to the moles with \(x\): \( 0.00240 = \frac{0.744}{x} \). Solving for \(x\): \( x = \frac{0.744}{0.00240} \approx 310 \text{ g/mol} \).

Verify with Given Molecular Formula

Double-check calculations with the precise formula: \( \text{P}_4\text{O}_{10}\). Recalculate moles and atomic weight check if results match. Given the stoichiometry, this should verify the atomic weight alignments.

Key concepts

Atomic Weights

The concept of atomic weights, also known as atomic masses, is critical in understanding chemical compounds and reactions. Atomic weight refers to the average mass of atoms of an element, usually expressed in atomic mass units (amu). This is important because it helps in determining how elements interact when they form compounds like phosphorus oxide.

To find the atomic weight of an unknown element in a compound, we often compare it to a known element. This is precisely what the original exercise does by setting the atomic weight of oxygen as a reference at 16.000 g/mol. By calculating the ratio of phosphorus to oxygen, we can find the atomic weight of phosphorus.

• Atomic weights allow for comparisons and calculations in molecular stoichiometry.
• They are fundamental in converting between grams and moles.
• Knowing the atomic weights of constituent elements helps predict the behavior of compounds.

Atomic weights can differ slightly based on isotopic composition, but in many calculations, especially involving large samples, these differences can be negligible.

Mass Calculations

Mass calculations play an essential role in stoichiometry, allowing us to understand how much of each element is in a given compound. In our exercise, we determined the mass of oxygen by subtracting the mass of phosphorus from the total mass of the phosphorus oxide compound, \(\text{P}_4\text{O}_{10}\).

Here’s how it works in this context:

• First, find the total mass of the compound, which is given.
• Subtract the mass of one element (phosphorus) to find the mass of the other element (oxygen).
• Use this information along with the atomic weights to calculate moles and determine the ratio and identity of the unknown atomic weight.

Mass calculations are foundational in chemistry, letting us find out how much of a particular substance is present in a compound. This is perhaps why it is commonly taught and reiterated in chemistry classes. Understanding mass calculations can also be key in determining empirical formulas of unknown substances.

Phosphorus Oxide

Phosphorus oxide, specifically the compound \(\text{P}_4\text{O}_{10}\), forms when phosphorus burns in the presence of oxygen. It is a white crystalline solid often used to dehydrate gases and other substances. In the exercise, \(\text{P}_4\text{O}_{10}\) serves to illustrate molecular stoichiometry and mass calculations in a real-world context.

Phosphorus oxide's formation can be described with chemical equations, reflecting the burning process and illustrating stoichiometry:

• The chemical reaction involved is \( \text{P}_4 + 5\text{O}_2 \rightarrow \text{P}_4\text{O}_{10} \).
• It shows that 4 atoms of phosphorus react with 10 atoms of oxygen to form one molecule of \(\text{P}_4\text{O}_{10}\).
• Understanding this structure helps in computational analysis, like determining ratios and reconstructing compounds mathematically.

This compound showcases an excellent case study, illustrating key concepts of stoichiometry, atomic weights, and critical mass calculations that resonate across many areas of chemistry.