Problem 66 - Chemistry and Chemical Reactivity

Question

Analysis of a gaseous chlorofluorocarbon, \(\mathrm{CCl}_{x} \mathrm{F}_{y}\), shows that it contains \(11.79 \%\) C and \(69.57 \%\) Cl. In another experiment, you find that \(0.107 \mathrm{g}\) of the compound fills a 458 -mL. flask at \(25^{\circ} \mathrm{C}\) with a pressure of \(21.3 \mathrm{mm}\) Hg. What is the molecular formula of the compound?

Step-by-Step Solution

Answer

The molecular formula is \(\mathrm{C}_2\mathrm{Cl}_4\mathrm{F}_2\).

Step 1: Determine the Percent Composition

The compound is given as \(\text{CCl}_x\text{F}_y\) with \(11.79\%\) carbon and \(69.57\%\) chlorine. To determine the percentage of fluorine, subtract the known percentages from \(100\%\): \[\% ext{F} = 100\% - 11.79\% - 69.57\% = 18.64\%.\]

Step 2: Calculate Empirical Formula

Assume \(100\,\text{g}\) of compound: - \(11.79\,\text{g}\) of C,- \(69.57\,\text{g}\) of Cl,- \(18.64\,\text{g}\) of F.Convert grams to moles:- \(\text{Moles of C} = \frac{11.79}{12.01} = 0.981\text{ mol}\),- \(\text{Moles of Cl} = \frac{69.57}{35.45} = 1.963\text{ mol}\),- \(\text{Moles of F} = \frac{18.64}{19.00} = 0.981\text{ mol}\).The simplest ratio is 1 C : 2 Cl : 1 F, so the empirical formula is \(\text{CCl}_2\text{F}\).

Step 3: Use Ideal Gas Law to Find Molar Mass

Use the ideal gas equation to calculate the molar mass:\[ PV = nRT \]Convert given conditions:- \(\text{Pressure} = 21.3\,\text{mm Hg} = 0.028\,\text{atm}\) (by dividing by 760),- \(\text{Volume} = 458\,\text{mL} = 0.458\,\text{L}\),- \(\text{Temperature} = 25^{\circ}C = 298\,\text{K}\),- \(R = 0.0821\,\text{L atm}\,\text{mol}^{-1}\,\text{K}^{-1}\).Calculate moles:\[ n = \frac{PV}{RT} = \frac{(0.028)(0.458)}{(0.0821)(298)} = 5.26 \times 10^{-4}\text{ mol}. \]Calculate molar mass:\[ \text{Molar mass} = \frac{0.107}{5.26 \times 10^{-4}} = 203.6 \text{ g/mol}. \]

Step 4: Determine Molecular Formula

The empirical formula mass of \(\text{CCl}_2\text{F}\) is approximately:\[ 12.01 + 2(35.45) + 19.00 = 102.91 \text{ g/mol}.\]Molecular formula mass (from previous step) is \(203.6 \text{ g/mol}\).Divide molecular by empirical formula mass:\[ \frac{203.6}{102.91} \approx 2. \]Thus, the molecular formula is twice the empirical formula: \(\text{C}_2\text{Cl}_4\text{F}_2\).

Key Concepts

Empirical Formula

The empirical formula is a simplified representation of a compound's composition by indicating the ratio of its elements. To derive it, you must know the percentage composition of each element. In our case, we started with a compound hinting at chlorofluorocarbon attributes.

Here, analysis yielded that it contains 11.79% Carbon (C) and 69.57% Chlorine (Cl). Thus, the residual amount accounts for Fluorine (F), calculated by subtracting the known percentages from 100%, resulting in 18.64% Fluorine.

• This leads us to calculate the moles of each element assuming you have 100 grams of the compound:
• Moles of Carbon: 11.79 g / 12.01 g/mol
• Moles of Chlorine: 69.57 g / 35.45 g/mol
• Moles of Fluorine: 18.64 g / 19.00 g/mol

These values simplify into the smallest whole number ratio, providing the empirical formula: CCl\(_2\)F.

Ideal Gas Law

The Ideal Gas Law equation, represented as \( PV = nRT \), is crucial for calculations involving gaseous substances. It correlates pressure (P), volume (V), and temperature (T) of a gas to the number of moles (n) using a gas constant (R).

For calculation, it requires

• Pressure conversion: from mm Hg to atm by dividing by 760.
• Volume conversion: from mL to L.
• Temperature conversion: from Celsius to Kelvin.

In our example:

Pressure = 21.3 mm Hg = 0.028 atm

Volume = 458 mL = 0.458 L

Temperature = 25°C = 298 K

The equation is rearranged to find the number of moles by the formula \( n = \frac{PV}{RT} \). Subsequently, this helps in finding the substance's molar mass to derive the molecular formula.

Percent Composition

Percent composition helps to know how much of each element is present in a compound by mass. It is vital for calculating empirical formulas, as seen in the analysis of chlorofluorocarbons.

Let's break it further:

• Carbon composition: 11.79%
• Chlorine composition: 69.57%
• Fluorine composition: 18.64%

This knowledge enables calculating the respective masses if the sample size is assumed at 100 grams. The calculated moles of each element feed into finding the simplest mole ratio, establishing the empirical formula.

In summary, understanding percent composition is the key to unraveling the molecular makeup of compounds.

Chlorofluorocarbon Analysis

Chlorofluorocarbons (CFCs) are compounds composed of carbon, chlorine, and fluorine. Their analysis involves finding the percentage of each constituent and using these values to determine empirical and molecular formulas.

Our primary step entailed derisking the percentages, guiding us to the empirical formula (CCl\(_2\)F), followed by leveraging the Ideal Gas Law to deduce the molar mass of the compound.

Upon obtaining the molar mass, the empirical formula mass is calculated and compared with it. This comparison (empirical vs. molecular mass) ensured correctness in deducing the molecular formula, which for this case led to C\(_2\)Cl\(_4\)F\(_2\).

Thus, such analysis not only pinpoints the chemical structure of CFCs but also sheds light on the practical methodology used in determining the composition.