Problem 101 - Introductory Chemistry Atoms First

Question

The compound \(\mathrm{P}_{4} \mathrm{O}_{10}\) has an empirical formula of \(\mathrm{P}_{2} \mathrm{O}_{5} .\) By what factor will the percent by mass composition differ for each element between these two formulas? Explain your answer.

Step-by-Step Solution

Answer

The percent by mass composition for each element in both the molecular formula (P4O10) and its empirical formula (P2O5) is the same (P - 43.64%, O - 56.36%). Therefore, the factor by which the percent by mass composition differs for each element between these two formulas is 1, meaning there is no difference.

Step 1: Determine the molar mass of each formula

First, we need to determine the molar mass of each formula: For P2O5: Molar mass of P: \(2\times30.97\text{ g/mol} = 61.94\text{ g/mol}\) Molar mass of O: \(5\times16.00\text{ g/mol} = 80.00\text{ g/mol}\) Total molar mass for P2O5: \(61.94\text{ g/mol} + 80.00\text{ g/mol} = 141.94\text{ g/mol}\) For P4O10: Molar mass of P: \(4\times30.97\text{ g/mol} = 123.88\text{ g/mol}\) Molar mass of O: \(10\times16.00\text{ g/mol} = 160.00\text{ g/mol}\) Total molar mass for P4O10: \(123.88\text{ g/mol} + 160.00\text{ g/mol} = 283.88\text{ g/mol}\)

Step 2: Calculate the percent by mass composition for each formula

Next, we need to calculate the percent by mass composition for both formulas: For P2O5: Percent by mass of P: \(\frac{61.94\text{ g/mol}}{141.94\text{ g/mol}} \times 100 = 43.64\%\) Percent by mass of O: \(\frac{80.00\text{ g/mol}}{141.94\text{ g/mol}} \times 100 = 56.36\%\) For P4O10: Percent by mass of P: \(\frac{123.88\text{ g/mol}}{283.88\text{ g/mol}} \times 100 = 43.64\%\) Percent by mass of O: \(\frac{160.00\text{ g/mol}}{283.88\text{ g/mol}} \times 100 = 56.36\%\)

Step 3: Determine the difference between the percent by mass composition

As we can see from the above calculations, the percent by mass composition for each element in both formulas is: P2O5: P - 43.64%, O - 56.36% P4O10: P - 43.64%, O - 56.36% Both the molecular formula (P4O10) and its empirical formula (P2O5) have the same percent by mass composition for each element (P and O). Therefore, the factor by which the percent by mass composition differs for each element between these two formulas is 1, meaning there is no difference.

Key Concepts

Molar Mass

The molar mass of a compound is essential in chemistry. It is the mass of a given substance divided by the amount of substance, measured in moles. To calculate molar mass, add up the atomic masses of all the atoms in a molecule.
For example, let's consider the compound \(\mathrm{P}_{4} \mathrm{O}_{10}\). To find the molar mass, follow these steps:

• Find the atomic mass of phosphorus (P), which is approximately 30.97 g/mol.
• Multiply this by the number of phosphorus atoms in the molecule. For \(\mathrm{P}_{4} \mathrm{O}_{10}\), there are 4 phosphorus atoms, giving \(4 \times 30.97 \text{ g/mol} = 123.88 \text{ g/mol}\).
• Do the same for oxygen (O), which has an atomic mass of 16.00 g/mol, then multiply by the 10 atoms of oxygen to obtain \(10 \times 16.00 \text{ g/mol} = 160.00 \text{ g/mol}\).
• Add these contributions from each type of atom to get the total molar mass: \(123.88 \text{ g/mol} + 160.00 \text{ g/mol} = 283.88 \text{ g/mol}\).

This molar mass is crucial for determining the proportions of each element present in a chemical reaction.

Percent Composition

Percent composition gives you insight into the relative amounts of each element in a compound. It's expressed as a percentage of the total mass of the compound.
To calculate the percent composition, you need the molar mass and the mass contributed by each element. Here's how you could do it for the compound \(\mathrm{P}_{2} \mathrm{O}_{5}\):

• Calculate the total mass of phosphorus in the compound: \(2 \times 30.97 \text{ g/mol} = 61.94 \text{ g/mol}\).
• Calculate the total mass of oxygen: \(5 \times 16.00 \text{ g/mol} = 80.00 \text{ g/mol}\).
• Add these together for the total molar mass of the compound: \(141.94 \text{ g/mol}\).
• To find the percent composition of phosphorus, divide its total mass in the compound by the total molar mass and multiply by 100: \(\frac{61.94}{141.94} \times 100 = 43.64\%\).
• Do the same for oxygen: \(\frac{80.00}{141.94} \times 100 = 56.36\%\).

This method gives you a clearer view of how much each element contributes to the compound's makeup by mass.

Chemical Formulas

Chemical formulas are shorthand for representing the composition of a compound. They show which elements are present and the number of atoms of each element in the smallest unit of the compound.
There are two primary types of chemical formulas: empirical formulas and molecular formulas.

• Empirical Formula: This tells you the simplest whole-number ratio of the elements in a compound. For example, the empirical formula of \(\mathrm{P}_{4} \mathrm{O}_{10}\) is \(\mathrm{P}_{2} \mathrm{O}_{5}\). It simplifies the ratio of atoms present in the compound.
• Molecular Formula: This provides the exact number of each type of atom in a molecule. \(\mathrm{P}_{4} \mathrm{O}_{10}\) is the molecular formula, indicating there are 4 phosphorus and 10 oxygen atoms in each molecule.

It's crucial to understand the distinction because the empirical formula captures the simplest ratio, while the molecular formula shows the precise composition of a molecule. Despite their differences, both can give the same percent composition by mass for the elements involved, as seen in the comparison between \(\mathrm{P}_{4} \mathrm{O}_{10}\) and \(\mathrm{P}_{2} \mathrm{O}_{5}\). This is because the molecular ratio is simply a scaled-up version of the empirical ratio.