Problem 129 - Chemistry and Chemical Reactivity

Question

Empirical and molecular formulas. (a) Fluorocarbonyl hypofluorite is composed of \(14.6 \% \mathrm{C}, 39.0 \%\) O, and \(46.3 \%\) F. The molar mass of the compound is \(82 \mathrm{g} / \mathrm{mol}\). Determine the empirical and molecular formulas of the compound. (b) Azulene, a beautiful blue hydrocarbon, is \(93.71 \%\) C and has a molar mass of \(128.16 \mathrm{g} / \mathrm{mol} .\) What are the empirical and molecular formulas of azulene?

Step-by-Step Solution

Answer

(a) Empirical and molecular formulas are CO2F2. (b) Empirical formula is C5H4, molecular formula is C10H8.

Step 1: Convert Percentages to Grams for Fluorocarbonyl Hypofluorite

Assume you have 100 g of the compound, so the percentages can be directly converted to grams: 14.6 g of C, 39.0 g of O, and 46.3 g of F.

Step 2: Convert Grams to Moles for Each Element in Fluorocarbonyl Hypofluorite

Calculate the moles for each element: For C: \( \frac{14.6 \text{ g}}{12.01 \text{ g/mol}} = 1.216 \text{ mol} \)For O: \( \frac{39.0 \text{ g}}{16.00 \text{ g/mol}} = 2.438 \text{ mol} \)For F: \( \frac{46.3 \text{ g}}{19.00 \text{ g/mol}} = 2.437 \text{ mol} \)

Step 3: Determine the Simplest Whole Number Ratio for Fluorocarbonyl Hypofluorite

Divide each number of moles by the smallest value: C: \( \frac{1.216}{1.216} = 1 \)O: \( \frac{2.438}{1.216} \approx 2 \)F: \( \frac{2.437}{1.216} \approx 2 \)The empirical formula is \(\text{CO}_2\text{F}_2\).

Step 4: Calculate the Empirical Formula Mass for Fluorocarbonyl Hypofluorite

Calculate the molar mass of the empirical formula \(\text{CO}_2\text{F}_2\):12.01 (C) + 2*16.00 (O) + 2*19.00 (F) = 82.01 g/mol.

Step 5: Derive Molecular Formula for Fluorocarbonyl Hypofluorite

Divide the compound's molar mass by the empirical formula mass: \(\frac{82}{82.01} \approx 1\). This confirms that the empirical formula \(\text{CO}_2\text{F}_2\) is also the molecular formula.

Step 6: Convert Percentages to Grams for Azulene

Assume you have 100 g of the compound: 93.71 g of C and 6.29 g of H (since it's a hydrocarbon).

Step 7: Convert Grams to Moles for Each Element in Azulene

For C: \( \frac{93.71 \text{ g}}{12.01 \text{ g/mol}} = 7.804 \text{ mol} \)For H: \( \frac{6.29 \text{ g}}{1.008 \text{ g/mol}} = 6.24 \text{ mol} \)

Step 8: Determine the Simplest Whole Number Ratio for Azulene

Divide each number of moles by the smallest value:C: \( \frac{7.804}{6.24} \approx 1.25 \)H: \( \frac{6.24}{6.24} = 1 \)To get whole numbers, multiply both by 4: C = 5, H = 4. The empirical formula is \(\text{C}_5\text{H}_4\).

Step 9: Calculate the Empirical Formula Mass for Azulene

Calculate the molar mass of the empirical formula \(\text{C}_5\text{H}_4\): 5*12.01 (C) + 4*1.008 (H) = 64.084 g/mol.

Step 10: Derive Molecular Formula for Azulene

Divide the compound's molar mass by the empirical formula mass: \(\frac{128.16}{64.084} \approx 2\). So the molecular formula is \(\text{C}_{10}\text{H}_8\).

Key Concepts

Chemical Composition Analysis

Chemical composition analysis is the process of identifying the percentage of each element present in a compound. Think of this as breaking down the whole into parts. When solving problems related to empirical and molecular formulas, the first step is to understand what each component of the compound contributes. This process typically involves:

• Converting percentage composition into grams, which allows for easier calculations of moles.
• Assuming a convenient sample size, often 100 grams, as this simplifies converting percentages directly into grams.

For example, with fluorocarbonyl hypofluorite, since the compound is composed of 14.6% carbon, 39.0% oxygen, and 46.3% fluorine, converting these percentages directly into grams results in 14.6 g C, 39.0 g O, and 46.3 g F. This step is fundamental because it sets the groundwork for further stoichiometric calculations that follow. Analysis at this level ensures that subsequent steps, such as calculating moles, can be executed accurately.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions, here specifically focusing on the relationship between the quantities of different elements in a compound. It involves calculating the exact amounts (in moles) of elements involved.

• Once you convert grams to moles using molar masses from the periodic table, you establish the basis for finding simplest ratios of atoms in the compound.
• For instance, converting 14.6 g of C to moles involves dividing by carbon’s molar mass, 12.01 g/mol, resulting in approximately 1.216 moles.

This step is repeated for each element in a compound. Understanding stoichiometry is essential in chemistry as it allows prediction of the quantities of reactants needed and products formed in a given reaction, or in our case, determining the simplest formula of a compound based on elemental composition. It goes beyond simplicity to provide precise quantification necessary for every chemical reaction or formula derivation.

Chemical Formula Derivation

Chemical formula derivation is the method where from a given set of data, such as percentages or compositions, the empirical formula (simplest ratio) and molecular formula (actual number of atoms) of a compound is defined. This involves a few key steps:

• The simplest ratio of the elements is found by dividing the moles of each element by the smallest number of moles calculated.
• Which allows the derivation of the empirical formula as a whole-number ratio.

If further calculations or data, like molar mass, are available, the molecular formula can be established. Essentially this involves comparing the molar mass of the empirical formula with that of the given compound's molar mass.
For azulene, after finding the empirical formula \[ \text{C}_5\text{H}_4 \], the empirical formula mass is calculated as 64.084 g/mol. By comparing it with the compound's molar mass, which is 128.16 g/mol, it turns out that the molecular formula is double the empirical one, \[ \text{C}_{10}\text{H}_8 \]. Hence, mastering chemical formula derivation is crucial because it reveals the true nature and size of the molecules involved in the substance.