Question

A compound of \(\mathrm{P}\) and \(\mathrm{F}\) was analyzed as follows: Heating \(0.2324 \mathrm{~g}\) of the compound in a \(378-\mathrm{cm}^{3}\) container turned all of it to gas, which had a pressure of \(97.3 \mathrm{mmHg}\) at \(77^{\circ} \mathrm{C}\). Then the gas was mixed with calcium chloride solution, which converted all the \(\mathrm{F}\) to \(0.2631 \mathrm{~g}\) of \(\mathrm{CaF}_{2}\). Determine the molecular formula of the compound.

Short answer

The molecular formula of the compound is \( \text{PF}_5 \).

Step-by-step solution

Convert Pressure and Temperature to Standard Units

First, convert the pressure from mmHg to atm. \[1 \text{ atm} = 760 \text{ mmHg}\]Thus, \[97.3 \text{ mmHg} = \frac{97.3}{760} \text{ atm} \approx 0.1281 \text{ atm}\]Convert the temperature from Celsius to Kelvin.\[T(K) = 77 + 273.15 = 350.15 \text{ K}.\]

Calculate Moles of Gas Using Ideal Gas Law

Use the ideal gas law \( PV = nRT \) to find the number of moles, \( n \).Where:- \( P = 0.1281 \text{ atm} \)- \( V = 0.378 \text{ L} \)- \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \)- \( T = 350.15 \text{ K} \)Rearrange to find \( n \):\[n = \frac{PV}{RT} = \frac{0.1281 \times 0.378}{0.0821 \times 350.15} \approx 0.00164 \text{ mol}\]

Determine Mass of Fluorine from CaF2

Use the mass of \( \text{CaF}_2 \) to find the mass of \( \text{F} \) in the compound.The molar mass of \( \text{CaF}_2 \) is approximately 78.07 g/mol.Calculate moles of \( \text{CaF}_2 \):\[\text{Moles of CaF}_2 = \frac{0.2631}{78.07} \approx 0.00337 \text{ mol}\]Since there are two \( \text{F} \) per \( \text{CaF}_2 \), moles of \( \text{F} = 2 \times 0.00337 = 0.00674 \text{ mol}\).The molar mass of \( \text{F} \) is approximately 19.00 g/mol:\[\text{Mass of F} = 0.00674 \times 19.00 \approx 0.1281 \text{ g}\]

Determine Mass of Phosphorus

Subtract the mass of fluorine from the total mass of the compound to find the mass of phosphorus.\[\text{Mass of P} = 0.2324 - 0.1281 = 0.1043 \text{ g}\]The molar mass of phosphorus is approximately 30.97 g/mol, so calculate moles of phosphorus:\[\text{Moles of P} = \frac{0.1043}{30.97} \approx 0.00337 \text{ mol}\]

Determine Empirical Formula

Compare the mole ratio of \( \text{P} \) to \( \text{F} \).\[\text{Moles P} : \text{Moles F} = 0.00337 : 0.00674 \approx 1 : 2\]Thus, the empirical formula is \( \text{PF}_2 \).

Confirm Molecular Formula Using Molar Mass

From Step 2, the total moles are 0.00164 mol, and using the given mass (0.2324 g), find molar mass:\[\text{Molar Mass} = \frac{0.2324}{0.00164} \approx 141.7 \text{ g/mol}\]Recalculate \( \text{PF}_2 \)'s molar mass:\[30.97 + 2 \times 19.00 = 68.97 \text{ g/mol}\]The molar mass estimation suggests \( \text{PF}_5 \) (147.97 g/mol), thus the molecular formula is \( \text{PF}_5 \).

Key concepts

Ideal Gas Law

The Ideal Gas Law is an essential principle used to relate the pressure, volume, temperature, and number of moles of a gas. It is expressed as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. For this exercise, the Ideal Gas Law helps determine the number of moles of the gaseous compound formed by phosphorus (P) and fluorine (F).

Steps involved include:

• Converting the measured pressure from mmHg to atm using the conversion factor \( 1 \text{ atm} = 760 \text{ mmHg} \).
• Adjusting the temperature from Celsius to Kelvin, as calculations involving the gas constant \( R \) require absolute temperature.
• Using the rearranged equation \( n = \frac{PV}{RT} \) to solve for \( n \), representing the moles of gas.

This approach reveals the relation between the physical properties of the gas and emphasizes the importance of using consistent units, a fundamental aspect when working with gases.

Empirical Formula

The empirical formula represents the simplest whole-number ratio of elements in a compound. Determining this formula involves converting the masses of each element to moles and then comparing the mole ratios. In our given exercise, we utilize the empirical formula process to ascertain the ratio of phosphorus to fluorine atoms in the compound.
When analyzing the compound:

• First, calculate the moles of fluorine derived from the total formation of \( \text{CaF}_2 \).
• Next, subtract the mass of fluorine from the entire compound's mass to find the phosphorus mass, and convert it to moles.
• Finally, calculate the simplest mole ratio of phosphorus to fluorine, which forms the empirical formula \( \text{PF}_2 \).

The process underscores the significance of exact measurements and conversions, revealing the balance necessary to derive elemental ratios in compounds.

Molecular Weight Calculation

Calculating the molecular weight involves determining the actual mass per mole of the molecules, allowing confirmation of the compound's molecular formula. This exercise requires correlating the empirical formula with experimentally derived molar mass to find an accurate molecular formula.
Key steps in the exercise:

• Using total number of moles obtained from the Ideal Gas Law and the given mass of the compound to compute its molar mass.
• Comparing the calculated molar mass to the empirical formula's mass, verifying if multiples of the empirical formula are needed to match accurately.
• Confirming that \( \text{PF}_5 \) aligns closer to the experimentally derived molar mass, thus leading to the conclusion of \( \text{PF}_5 \) as the molecular formula.

This section highlights the interplay between observations and calculations, emphasizing the need for precision in concluding molecular identities.

Stoichiometry

Stoichiometry refers to the quantitative relationships in chemical reactions. It facilitates the calculation of reactants and products based on the mole concept, serving a crucial role in chemical analysis.
In this scenario, stoichiometry helps in:

• Determining the mass and moles of \( \text{CaF}_2 \), which stems from the amount of fluorine in the original compound.
• Establishing the relationship between the moles of calcium fluoride and fluorine, enabling back-calculation to quantify the fluorine content.
• Using these mole ratios to track and adjust phosphorus and fluorine atom ratios to deduce the empirical formula.

Through this disciplined approach, stoichiometry connects the dots between chemical formulas and exact substance proportions, demonstrating how chemistry relies on numeric precision to solve problems effectively.