Problem 47 - Chemistry: The Central Science

Question

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(10.4 \% \mathrm{C}, 27.8 \% \mathrm{~S}\), and \(61.7 \% \mathrm{Cl}\) (b) \(21.7 \% \mathrm{C}, 9.6 \% \mathrm{O}\), and \(68.7 \% \mathrm{~F}\) (c) \(32.79 \% \mathrm{Na}, 13.02 \% \mathrm{Al}\), and the remainder \(\mathrm{F}\)

Step-by-Step Solution

Answer

The empirical formulas for the given compounds are: (a) \( CS_1Cl_2 \) (b) \( C_3O_1F_6 \) (c) \( Na_3Al_1F_6 \)

Step 1: Convert mass percentages to grams

Assume we have 100g of compound (a), then the mass of each element is: - C: 10.4g - S: 27.8g - Cl: 61.7g

Step 2: Find the moles of each element

Using the molar masses: - C: 12.01 g/mol - S: 32.07 g/mol - Cl: 35.45 g/mol We find the moles of each element: - Moles of C: \( \frac{10.4 \text{g}}{12.01 \text{g/mol}} = 0.866 \text{mol} \) - Moles of S: \( \frac{27.8 \text{g}}{32.07 \text{g/mol}} = 0.867 \text{mol} \) - Moles of Cl: \( \frac{61.7 \text{g}}{35.45 \text{g/mol}} = 1.74 \text{mol} \)

Step 3: Divide all moles by the smallest number of moles

We have the following mole ratios: - C: \( \frac{0.866}{0.866} = 1 \) - S: \( \frac{0.867}{0.866} \approx 1 \) - Cl: \( \frac{1.74}{0.866} \approx 2 \) The empirical formula for compound (a) is then approximately \( \mathrm{CS_1Cl_2} \). (b) \(21.7 \% \mathrm{C}, 9.6 \% \mathrm{O}\), and \(68.7 \% \mathrm{~F}\) Repeat the same steps for compound (b).

Step 1: Convert mass percentages to grams

Assume we have 100g of compound (b), then the mass of each element is: - C: 21.7g - O: 9.6g - F: 68.7g

Step 2: Find the moles of each element

Using the molar masses: - C: 12.01 g/mol - O: 16.00 g/mol - F: 19.00 g/mol We find the moles of each element: - Moles of C: \( \frac{21.7 \text{g}}{12.01 \text{g/mol}} = 1.81 \text{mol} \) - Moles of O: \( \frac{9.6 \text{g}}{16.00 \text{g/mol}} = 0.60 \text{mol} \) - Moles of F: \( \frac{68.7 \text{g}}{19.00 \text{g/mol}} = 3.61 \text{mol} \)

Step 3: Divide all moles by the smallest number of moles

We have the following mole ratios: - C: \( \frac{1.81}{0.60} = 3 \) - O: \( \frac{0.60}{0.60} = 1 \) - F: \( \frac{3.61}{0.60} \approx 6 \) The empirical formula for compound (b) is then approximately \( \mathrm{C_3O_1F_6} \). (c) \(32.79 \% \mathrm{Na}, 13.02 \% \mathrm{Al}\), and the remainder \(\mathrm{F}\) Finally, we do the same for compound (c).

Step 1: Convert mass percentages to grams

Assume we have 100g of compound (c), then the mass of each element is: - Na: 32.79g - Al: 13.02g - F: 100 - 32.79 - 13.02 = 54.19g

Step 2: Find the moles of each element

Using the molar masses: - Na: 22.99 g/mol - Al: 26.98 g/mol - F: 19.00 g/mol We find the moles of each element: - Moles of Na: \( \frac{32.79 \text{g}}{22.99 \text{g/mol}} = 1.43 \text{mol} \) - Moles of Al: \( \frac{13.02 \text{g}}{26.98 \text{g/mol}} = 0.483 \text{mol} \) - Moles of F: \( \frac{54.19 \text{g}}{19.00 \text{g/mol}} = 2.85 \text{mol} \)

Step 3: Divide all moles by the smallest number of moles

We have the following mole ratios: - Na: \( \frac{1.43}{0.483} \approx 3 \) - Al: \( \frac{0.483}{0.483} = 1 \) - F: \( \frac{2.85}{0.483} \approx 6 \) The empirical formula for compound (c) is then approximately \( \mathrm{Na_3Al_1F_6} \).

Key Concepts

Mass Percentage

Understanding mass percentage is essential when determining the empirical formula. Mass percentage refers to the proportion, by mass, of each element present in a compound. It is expressed as a percentage of the total mass. To simplify calculations, one often assumes a compound sample of 100g. This assumption allows you to directly use the given mass percentages as grams of each element, making it easier to progress to the next steps. This approach is particularly useful when you're only given percentage compositions without the actual sample size.
By expressing elements in mass percentages, one can systematically convert these percentages into grams, enabling calculations of moles, which are vital for determining empirical and molecular formulas.

Molar Mass Calculations

Calculating molar masses is a pivotal step when working through chemical composition problems. Molar mass is the mass of one mole of a substance, typically measured in grams per mole (g/mol). To find the number of moles of each element in a compound, you divide the mass of each element by its molar mass.
For example, in the case of carbon in compound (a), with a molar mass of 12.01 g/mol, you calculate moles from mass using the formula: \[ \text{Moles of C} = \frac{\text{mass in grams}}{12.01 \text{ g/mol}} \] These calculations help bridge the gap between mass percentages and identifying the formula by providing a numeric figure representing the quantity of each element in moles.

Mole Ratio

Determining the mole ratio is an important step that follows molar mass calculations. The mole ratio refers to the ratio of moles of each element in a compound. This ratio helps in identifying the simplest whole-number ratio of elements in the empirical formula. Once you have the moles of each element, you divide each by the smallest number of moles present in any element within the compound. This step transforms your moles into a simplified, more understandable form.
For instance, after finding the moles of C, S, and Cl, the ratios were simplified to approximate whole numbers: - For C: 1 - For S: 1 - For Cl: 2
This step is crucial because the empirical formula is derived directly from these whole-number mole ratios.

Chemical Composition Analysis

Chemical composition analysis culminates the process of finding the empirical formula. It involves systematic conversion and comparison, starting from mass percentage to mole ratio, to ultimately deduce the empirical formula. By understanding the chemical composition and structure of a compound, students can accurately predict reactivity, stability, and other chemical properties. This analysis is not only an academic exercise but also a crucial real-world application in chemistry.
Chemical composition analysis illustrates the interrelationship between different scientific concepts, showing the importance of empirical formulas in broader chemical understanding. This underscores the significance of attention to detail, precision in measurements, and comprehension of fundamental chemical principles.