Question

Empirical and molecular formulas. (a) Fluorocarbonyl hypofluorite is composed of \(14.6 \% \mathrm{C}, 39.0 \% \mathrm{O},\) 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 \(128.16 \mathrm{g} /\) mol. What are the empirical and molecular formulas of azulene?

Short answer

Fluorocarbonyl hypofluorite: empirical and molecular formula is CO2F2. Azulene: empirical formula is C5H4, molecular formula is C10H8.

Step-by-step solution

Convert percentage to moles for Fluorocarbonyl Hypofluorite

For each element, use the percentage composition to determine the moles of each element in a 100 g sample.- Carbon: \( \frac{14.6 \text{ g}}{12.01 \text{ g/mol}} = 1.215 \text{ moles of C} \)- Oxygen: \( \frac{39.0 \text{ g}}{16.00 \text{ g/mol}} = 2.4375 \text{ moles of O} \)- Fluorine: \( \frac{46.3 \text{ g}}{19.00 \text{ g/mol}} = 2.4368 \text{ moles of F} \)

Determine the empirical formula for Fluorocarbonyl Hypofluorite

Identify the smallest number of moles among the elements and divide each element's mole count by that smallest number:- Carbon: \( \frac{1.215}{1.215} = 1 \)- Oxygen: \( \frac{2.4375}{1.215} \approx 2 \)- Fluorine: \( \frac{2.4368}{1.215} \approx 2 \)This gives an empirical formula of \( \text{CO}_2\text{F}_2 \).

Calculate the molar mass of the empirical formula

The empirical formula molar mass is the sum of the molar masses of the constituent atoms: - Carbon: \( 1 \times 12.01 \text{ g/mol} = 12.01 \text{ g/mol} \)- Oxygen: \( 2 \times 16.00 \text{ g/mol} = 32.00 \text{ g/mol} \)- Fluorine: \( 2 \times 19.00 \text{ g/mol} = 38.00 \text{ g/mol} \)Total = \( 82.01 \text{ g/mol} \).

Verify the molecular formula for Fluorocarbonyl Hypofluorite

The molar mass of the empirical formula matches the given molar mass (\( 82 \text{ g/mol} \)), so the molecular formula is the same as the empirical formula: \( \text{CO}_2\text{F}_2 \).

Convert percentage to moles for Azulene

Since Azulene is hydrocarbon, carbon accounts for \(93.71\% \) of the compound. Assume rest are hydrogens, thus, hydrogen is \(100 - 93.71 = 6.29\% \). - Carbon: \( \frac{93.71 \text{ g}}{12.01 \text{ g/mol}} = 7.803 \text{ moles of C} \)- Hydrogen: \( \frac{6.29 \text{ g}}{1.008 \text{ g/mol}} = 6.24 \text{ moles of H} \)

Determine the empirical formula for Azulene

Divide each element's mole count by the smallest number of moles:- Carbon: \( \frac{7.803}{6.24} \approx 1.25 \)- Hydrogen: \( \frac{6.24}{6.24} = 1 \)Since we need integer subscripts, we multiply by 4 to get \(\text{C}_5\text{H}_4\) as the empirical formula.

Determine the molecular formula for Azulene

Calculate the molar mass of the empirical formula \( \text{C}_5\text{H}_4 \):- Carbon: \( 5 \times 12.01 \text{ g/mol} = 60.05 \text{ g/mol} \)- Hydrogen: \( 4 \times 1.008 \text{ g/mol} = 4.032 \text{ g/mol} \)The total is \(64.082 \text{ g/mol} \). Divide the compound's molar mass by empirical molar mass:\( \frac{128.16}{64.082} \approx 2 \).Therefore, the molecular formula is \(\text{C}_{10}\text{H}_8\).

Key concepts

Percentage Composition

Understanding percentage composition is crucial when dealing with chemical formulas. It represents the proportion by mass of each element within a compound. To find it, you need to know the mass of each element and the total mass of the compound. For example, if a compound is said to consist of 14.6% carbon, it means that in every 100 grams of the compound, 14.6 grams is carbon.
This is helpful in determining the empirical formula, as you can directly convert these percentages to mass if you assume you have a 100 gram sample.
Then, using the molar mass of each element, you convert mass to moles because formulas are based on the number of atoms, not their weights.

• Mass percentages provide a direct ratio and allow conversion to moles.
• Using these values, you can find the simplest whole number ratio of the elements, giving the empirical formula.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance and is typically expressed in grams per mole (g/mol). It's important because it lets you convert between the mass of a substance and the amount in moles.
To calculate the molar mass of a compound, sum the molar masses of all the atoms in its formula.
For instance, if you have CO₂F₂:

• Carbon: 1 carbon atom times 12.01 g/mol.
• Oxygen: 2 oxygen atoms times 16.00 g/mol each.
• Fluorine: 2 fluorine atoms times 19.00 g/mol each.

Add these up to get the total molar mass. Understanding how to calculate molar mass is key in verifying and determining both empirical and molecular formulas.

Chemical Formulas

Chemical formulas represent the elements present in a compound and their proportions. They can be empirical or molecular. The empirical formula is the simplest integer ratio of elements in a compound. For example, CO₂F₂ is an empirical formula.
On the other hand, a molecular formula represents the exact number of atoms of each element in a molecule. For azulene, the empirical formula is C₅H₄, but the molecular formula, based on its molar mass, is C₁₀H₈.
Understanding these distinctions helps in comprehending how compounds form and are represented.

• Empirical formulas provide the lowest whole-number ratio, useful for initial analysis.
• Molecular formulas give the precise number of each type of atom, indicating the true composition of molecules.

These formulas are foundational to the study and application of chemistry.

Hydrocarbons

Hydrocarbons are organic compounds made exclusively of hydrogen and carbon atoms. They are crucial in both natural processes and industrial applications, from fuels to the synthesis of everyday materials.
Azulene is a hydrocarbon, which, through analysis of its percentage composition, can be understood as consisting largely of carbon.
By assuming the remainder is hydrogen, you can determine its empirical formula by converting the percentages to moles, which then help in identifying its molecular formula via its molar mass.

• The simplicity of hydrocarbons makes it easier to determine ratios based on percentage composition.
• They serve as a basic but essential category of compounds in organic chemistry.

Understanding hydrocarbons is not only key to the realm of organic chemistry but also fundamental in fields like energy production and material manufacturing.