Question

How many grams of \(\mathrm{Cl}\) are in \(38.0 \mathrm{~g}\) of each sample of chlorofluorocarbons (CFCs)? (a) \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) (b) \(\mathrm{CFCl}_{3}\) (c) \(\mathrm{C}_{2} \mathrm{~F}_{3} \mathrm{Cl}_{3}\) (d) \(\mathrm{CF}_{3} \mathrm{Cl}\)

Short answer

The grams of Cl in a 38.0 g sample of each CFC are: (a) 27.74 g, (b) 32.59 g, (c) 34.57 g, and (d) 13.52 g.

Step-by-step solution

- Calculate molar mass of compounds

Find the molar mass of each CFC compound by adding the atomic masses of its constituent atoms: carbon (C), fluorine (F), and chlorine (Cl). For example, \(\mathrm{CF}_{2}\mathrm{Cl}_{2}\) has 1 carbon (12.01 g/mol), 2 fluorines (2 x 19.00 g/mol), and 2 chlorines (2 x 35.45 g/mol).

- Determine the fraction of chlorine in each compound

Divide the total mass of chlorine in the formula by the molar mass of the compound to get the fraction of chlorine. For \(\mathrm{CF}_{2}\mathrm{Cl}_{2}\), the fraction of chlorine is \(\frac{2 \times 35.45 \mathrm{~g/mol}}{molar \ mass \ of \ \mathrm{CF}_{2}\mathrm{Cl}_{2}}\).

- Calculate grams of chlorine in the given sample

Multiply the mass of the given sample (38.0 g) by the fraction of chlorine obtained in Step 2 for each compound. The resulting value is the amount of chlorine in grams contained in the sample.

- Perform calculations for each compound

Repeat Steps 1 to 3 for each compound: \(\mathrm{CF}_{2}\mathrm{Cl}_{2}\), \(\mathrm{CFCl}_{3}\), \(\mathrm{C}_{2} \mathrm{~F}_{3} \mathrm{Cl}_{3}\), and \(\mathrm{CF}_{3}\mathrm{Cl}\). Calculate their molar masses, the fraction of chlorine, and finally the grams of chlorine in a 38.0 g sample.

Key concepts

Molar Mass Calculation

Understanding how to calculate molar mass is a fundamental skill in chemistry. Molar mass is the weight of one mole of a substance, which is the mass in grams equivalent to the atomic or molecular mass in atomic mass units (amu).

To calculate the molar mass of a compound, you sum the molar masses of all the atoms in its chemical formula. Let's take CF₂Cl₂ as an example. First, look up the atomic masses on the periodic table: carbon (C) is roughly 12.01 g/mol, fluorine (F) is about 19.00 g/mol, and chlorine (Cl) is approximately 35.45 g/mol. Since CF₂Cl₂ contains 1 carbon, 2 fluorines, and 2 chlorines, you would perform the following calculation:

1. Add the mass of carbon: 1 x 12.01 g/mol = 12.01 g/mol.
2. Add the mass of fluorines: 2 x 19.00 g/mol = 38.00 g/mol.
3. Add the mass of chlorines: 2 x 35.45 g/mol = 70.90 g/mol.

The molar mass of CF₂Cl₂ is then 12.01 g/mol + 38.00 g/mol + 70.90 g/mol = 120.91 g/mol. Repeat this process for each compound, and you will get the molar mass needed for further calculations.

Percent Composition

Once you've determined the molar mass of a compound, the next step in stoichiometry is to calculate the percent composition. This tells us the percentage by mass of each element in a compound.

For example, in CF₂Cl₂, chlorine's percent composition can be calculated by dividing the total mass of chlorine in the molecule by the molecule's total molar mass. You then multiply by 100 to get a percentage. In CF₂Cl₂, the calculation is (70.90 g/mol / 120.91 g/mol) x 100, equating to approximately 58.67%. This means that 58.67% of the mass of CF₂Cl₂ is due to chlorine. It's pivotal to average atomic masses accurately and keep track of the number of atoms of each element in the formula to mitigate errors in percent composition.

Chemical Formula

A chemical formula such as that of carbon tetrachloride, CCl₄, informs us about the types and quantities of atoms in a single molecule of a compound. It is an integral part of stoichiometry, as it serves as the basis for all the subsequent calculations.

When balancing chemical equations or calculating stoichiometric ratios, the accuracy of the chemical formula is essential. It's important to note that subscripts in the chemical formula represent the number of atoms of each element, while prefixes indicate the number of molecules. Misinterpreting these numbers can lead to significant errors, so care must be taken to ensure they are understood correctly. For example, the formula C₂F₄ would mean there are 2 carbon atoms and 4 fluorine atoms in each molecule. This distinct knowledge allows you to solve problems accurately in stoichiometry, like the one mentioned in the original exercise.