Problem 45 - 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) CS₂, (b) C₃OF₆, and (c) Na₃AlF₆.

Step 1: Convert mass percentages to grams

: Assume a total mass of 100g for each compound, as it will make it easier to work with the given mass percentages. This means each mass percentage becomes the mass in grams for each element. (a) 10.4 g C, 27.8 g S, 61.7 g Cl (b) 21.7 g C, 9.6 g O, 68.7 g F (c) 32.79 g Na, 13.02 g Al, and the remainder is F

Step 2: Convert grams to moles

: Use the molar mass of each element to convert the grams to moles. Molar mass of C = 12.01 g/mol, S = 32.07 g/mol, Cl = 35.45 g/mol, Na = 22.99 g/mol, Al = 26.98 g/mol, O = 16.00 g/mol, F = 19.00 g/mol. (a) \( \frac{10.4\,g}{12.01\,g/mol} = 0.866\,mol\) C, \( \frac{27.8\,g}{32.07\,g/mol} = 0.867\,mol\) S, \( \frac{61.7\,g}{35.45\,g/mol} = 1.74\,mol\) Cl (b) \( \frac{21.7\,g}{12.01\,g/mol} = 1.81\,mol\) C, \( \frac{9.6\,g}{16.00\,g/mol} = 0.600\,mol\) O, \( \frac{68.7\,g}{19.00\,g/mol} = 3.62\,mol\) F (c) \( \frac{32.79\,g}{22.99\,g/mol} = 1.425\,mol\) Na, \( \frac{13.02\,g}{26.98\,g/mol} = 0.4828\,mol\) Al,_remaining mass = 100 g - 32.79 g - 13.02 g = 54.19 g F, \( \frac{54.19\,g}{19.00\,g/mol} = 2.852\,mol\) F

Step 3: Calculate the simplest mole ratios

: Divide each of the moles calculated in step 2 by the smallest mole value in each compound. (a) \(\frac{0.866}{0.866} = 1\) C, \(\frac{0.867}{0.866} \approx 1\) S, \(\frac{1.74}{0.866} \approx 2\) Cl (b) \(\frac{1.81}{0.600} \approx 3\) C, \(\frac{0.600}{0.600} = 1\) O, \(\frac{3.62}{0.600} \approx 6\) F (c) \(\frac{1.425}{0.4828} \approx 3\) Na, \(\frac{0.4828}{0.4828} = 1\) Al, \(\frac{2.852}{0.4828} \approx 6\) F

Step 4: Write the empirical formula

: Combine the mole ratios obtained in step 3 to form the empirical formula for each compound. (a) CS₂ (b) C₃OF₆ (C) Na₃AlF₆ So, the empirical formulas for the compounds are (a) CS₂, (b) C₃OF₆, and (c) Na₃AlF₆.

Key Concepts

Molar Mass

Understanding molar mass is crucial for determining the amount of an element in terms of moles. Molar mass is the mass of one mole of a substance (usually in grams per mole). It serves as a bridge between the microscopic world of atoms and the macroscopic world we can measure in the lab.

For instance, if you're given a sample containing a specific mass of an element, like carbon, you can use its molar mass to figure out how many moles of that element are in the sample. The molar mass of carbon is 12.01 g/mol, which means one mole of carbon atoms weighs 12.01 grams.

To convert grams to moles, you simply divide the mass of the element (in grams) by its molar mass. This is a fundamental step in chemical analysis because it allows us to compare amounts of different substances on an equal footing based on moles rather than mass alone. This conversion is key in finding the empirical formula.

Mole Ratio

The concept of mole ratio involves comparing the number of moles of each element in a compound. It is a vital step in determining the simplest whole-number ratio of atoms in a compound, known as the empirical formula.

After converting the mass of each element in a sample to moles, you'll use these quantities to find the mole ratio. This involves dividing the number of moles of each element by the smallest number of moles calculated in the sample.

For example, if a sample contains 0.866 moles of carbon, 0.867 moles of sulfur, and 1.74 moles of chlorine, dividing each by 0.866 (the smallest value) yields a ratio of approximately 1:1:2. This ratio is used to write the empirical formula.

These calculations are essential because they help simplify the expression of chemical compounds, making it easier to communicate their composition and understand their properties.

Chemical Composition

Chemical composition refers to the types and amounts of elements that make up a compound. Analyzing the chemical composition is fundamental for identifying substances and understanding their chemical behavior.

In problems related to empirical formulas, compositions are often given in percentages. By assuming a 100-gram sample, you can directly convert these percentages to grams. For example, if a compound is 10.4% carbon by mass, you would have 10.4 grams of carbon in a 100-gram sample.

Understanding the composition allows us to determine which elements are present and their relative amounts, helping in identifying the simplest formula that represents the compound. This involves calculating the mass of each element and using their molar masses to proceed to further calculations, such as moles and mole ratios.

Chemical Formula Determination

Determining the chemical formula of a compound involves identifying the simplest ratio of atoms present. This is achieved through a series of steps that include using molar mass, converting percentages to moles, and calculating mole ratios.

The empirical formula represents the simplest whole-number ratio of elements in a compound. Once mole ratios are found by converting mass percentages to moles and simplifying these to the smallest whole numbers, you establish the empirical formula.

For example, given the mass percentages of elements, you first convert these to grams, then to moles using their molar masses. By finding the smallest mole value and using it to calculate the ratio for each element, you finalize the empirical formula, like CS₂ from C, S, and Cl compositions.

This process is critical because it gives a fundamental insight into the compound's nature, helping in various applications ranging from materials science to pharmaceutical development.