Question

Anthraquinone contains only carbon, hydrogen, and oxygen. When 4.80 \(\mathrm{mg}\) anthraquinone is burned, 14.2 \(\mathrm{mg} \mathrm{CO}_{2}\) and 1.65 \(\mathrm{mg} \mathrm{H}_{2} \mathrm{O}\) are produced. The freezing point of camphor is lowered by \(22.3^{\circ} \mathrm{C}\) when 1.32 \(\mathrm{g}\) anthraquinone is dissolved in 11.4 \(\mathrm{g}\) camphor. Determine the empirical and molecular formulas of anthraquinone.

Short answer

The empirical and molecular formulas of anthraquinone are both C14H8O2. This was determined by calculating the percent composition of carbon, hydrogen, and oxygen from the combustion data, finding the molecular weight of anthraquinone using freezing point depression data, and comparing the molecular weight to the empirical formula weight.

Step-by-step solution

Find the mass of carbon and hydrogen in anthraquinone

Given that 14.2 mg of CO2 and 1.65 mg of H2O are produced when burning 4.80 mg of anthraquinone, we can use the stoichiometry of these reactions to find out the mass of carbon and hydrogen in anthraquinone. For Carbon(C): CO2 produced: 14.2 mg Moles of CO2 produced: \(\frac{14.2}{44.01} = 0.3224\) moles (using the molar mass of CO2: 44.01 g/mol) Moles of Carbon in anthraquinone: 0.3224 moles (1 mole of C in 1 mole of CO2) Mass of Carbon in anthraquinone: \(0.3224 \times 12.01 = 3.870\) mg (using the molar mass of C: 12.01 g/mol) For Hydrogen (H): H2O produced: 1.65 mg Moles of H2O produced: \(\frac{1.65}{18.02} = 0.0916\) moles (using the molar mass of H2O: 18.02 g/mol) Moles of Hydrogen in anthraquinone: \(0.0916 \times 2 = 0.1832\) moles (2 moles of H in 1 mole of H2O) Mass of Hydrogen in anthraquinone: \(0.1832 \times 1.01 = 0.185\) mg (using the molar mass of H: 1.01 g/mol)

Find the mass of oxygen in anthraquinone

Since anthraquinone contains only carbon, hydrogen, and oxygen, we can find the mass of oxygen by subtracting the mass of carbon and hydrogen from the initial mass of anthraquinone: Mass of Oxygen: 4.80 mg - 3.870 mg - 0.185 mg = 0.745 mg

Calculate the percent composition

Once we have the mass of carbon, hydrogen and oxygen in anthraquinone, we can find the percent composition by dividing the respective mass by the total mass of the compound (4.80 mg): Percent of Carbon: \(\frac{3.870}{4.80} \times 100 = 80.63\% \) Percent of Hydrogen: \(\frac{0.185}{4.80} \times 100 = 3.854\% \) Percent of Oxygen: \(\frac{0.745}{4.80} \times 100 = 15.52\% \)

Calculate the molecular weight from freezing point depression data

We are given that the freezing point of camphor is lowered by 22.3°C when 1.32 g of anthraquinone is dissolved in 11.4 g of camphor. Using the formula: ΔTf = Kf . \( \frac{moles \, of\, solute}{mass \, of\, solvent} \) where ΔTf is the freezing point depression, Kf is the cryoscopic constant for camphor (40.7 K kg/mol) and moles of solute (anthraquinone) can be calculated using the molecular weight (M). We can rewrite the formula to solve for M: M = \(\frac{Kf \times moles \,of\, solute}{ΔTf \times mass \,of\, solvent}\) Plugging in the given values: M = \(\frac{40.7 \times \frac{1.32}{M}}{22.3 \times 11.4}\) Solving for M gives, M = 208.2 g/mol (molecular weight of anthraquinone)

Determine the empirical and molecular formulas

To find the empirical formula, we need to determine the lowest whole number ratio of C, H, and O atoms in the compound using their moles: Moles of Carbon: \(\frac{3.870}{12.01} = 0.3224\) moles Moles of Hydrogen: \(\frac{0.185}{1.01} = 0.1832\) moles Moles of Oxygen: \(\frac{0.745}{16.00} = 0.0466\) moles To find the lowest whole number ratio: \( \frac{0.3224}{0.0466} = 6.92 \approx 7\) \( \frac{0.1832}{0.0466} = 3.93 \approx 4\) \( \frac{0.0466}{0.0466} = 1\) Thus, the empirical formula is C14H8O2. To find the molecular formula, we will compare the molecular weight (M) to the empirical formula weight (EFW): EFW = 14(12.01) + 8(1.01) + 2(16.00) = 208.3 g/mol Since the molecular weight of anthraquinone (208.2 g/mol) is approximately equal to the empirical formula weight, the molecular formula of anthraquinone is the same as its empirical formula: C14H8O2.

Key concepts

Molecular Formula

A molecular formula represents the actual number of each type of atom in a molecule. It provides a detailed composition of the molecule, encompassing the number of atoms of each element present in the compound. This is different from the empirical formula, which merely shows the ratio of elements.
For instance, in the case of anthraquinone, the empirical formula was determined to be \(\text{C}_{14}\text{H}_8\text{O}_2\).
This means there are 14 carbon atoms, 8 hydrogen atoms, and 2 oxygen atoms in one molecule of anthraquinone.
In some cases, the molecular formula can be the same as the empirical formula, especially if the empirical formula weight matches the molecular weight, as seen here with the molecular weight being 208.2 g/mol, matching the empirical formula weight. This alignment confirms that no additional scaling of the formula is needed.

Stoichiometry

Stoichiometry is the study of the quantitative relationships within a chemical reaction. This involves calculations of reactants and products in chemical processes.
In the context of this problem, stoichiometry helps determine the mass of carbon and hydrogen in anthraquinone by analyzing combustion products.

• When anthraquinone is burned, it produces CO₂ and H₂O. By calculating the moles of these products, one can back-calculate to find the moles of carbon and hydrogen in the original anthraquinone compound.
• For instance, the moles of CO₂ allow us to determine the moles of carbon because CO₂ contains one carbon atom per molecule.

This quantitative method enables accurate assessment of substance quantities, pivotal in calculating the empirical and molecular formulas.

Freezing Point Depression

Freezing point depression describes the lowering of a solvent's freezing point due to the presence of a solute. It's an essential colligative property, depending only on the number of solute particles, not their identity.
The formula used is \(\Delta T_f = K_f \cdot \frac{moles \; of \; solute}{mass \; of \; solvent}\).
Here, \(K_f\) is a constant characteristic of the solvent, camphor in this case, with a value of 40.7 K kg/mol.

• By knowing the freezing point depression (22.3°C), and the masses of solute and solvent, we calculated the molecular weight of anthraquinone as part of the exercise.
• It's important as the calculation provides the actual mass of the molecular formula, helping in determining the true structure of anthraquinone.

Percent Composition

Percent composition explains the percentage by mass of each element within a compound. This is crucial for understanding how much of each element is present in a substance, which in turn aids in deriving the empirical formula.
In order to calculate percent composition for anthraquinone, we used the masses of carbon, hydrogen, and oxygen determined from stoichiometric calculations.

• The mass of each element (in mg) is divided by the total mass of the compound to yield a percentage.
• In anthraquinone, carbon takes up about 80.63%, hydrogen 3.85%, and oxygen 15.52% of the total mass.

Understanding percent composition helps confirm the reliability of calculated formulas and offers insight into the material's properties and reactivity.