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 formula of anthraquinone is C_{14}H_8O_2 and its molecular formula is also C_{14}H_8O_2.

Step-by-step solution

Determine mole ratios of carbon, hydrogen, and oxygen

First, we need to find the moles of carbon, hydrogen, and oxygen in anthraquinone. Moles of carbon can be determined from the moles of \(CO_2\) produced: Moles of \(C = \frac{14.2\ mg\ CO_2}{44.01\ g/mol}\) Similarly, moles of hydrogen can be determined from the moles of \(H_2O\) produced: Moles of \(H = 2\times\frac{1.65\ mg\ H_2O}{18.02\ g/mol}\) (The factor of 2 is because each water molecule contains 2 hydrogen atoms) To find the moles of oxygen, first find the moles of anthraquinone: Moles of anthraquinone = \(\frac{4.8\ mg}{1000\ mg/g}\) Then subtract the moles of carbon and hydrogen from the moles of anthraquinone to find the moles of oxygen.

Find the empirical formula

To find the empirical formula, divide the moles of each element by the smallest mole value, and if necessary, multiply to the nearest whole number to obtain mole ratios. This will give us the empirical formula.

Determine the molecular weight of anthraquinone using the freezing point depression data

We can use the following formula for freezing point depression: \(\Delta T = K_f \times m \) where \(\Delta T\) is the freezing point depression, \(K_f\) is the cryoscopic constant of camphor (37.7 °C kg/mol), and m is the molality (moles of solute / kg of solvent) We have: \(22.3^\circ C = 37.7 \frac{^\circ C \ kg/mol}{g} \times m \) Now we can solve for molality and then use the mass of anthraquinone dissolved to find the molecular weight.

Find the molecular formula

Compare the molecular weight determined in Step 3 to the empirical formula weight calculated in Step 2 to find a whole number ratio (n). Multiply the empirical formula by this ratio to find the molecular formula.

Key concepts

Mole Ratio in Compounds

Understanding the mole ratio in compounds is crucial when it comes to identifying the composition of substances. Every compound is made up of elements in a specific ratio, known as the mole ratio. It represents the proportion of moles of each element within the compound. This ratio can be deduced from experimental data, such as from a combustion analysis, where the amounts of products formed during combustion are used to determine the mole ratio of the elements in the original compound.

For instance, in the provided exercise, the mole ratios are found by comparing the moles of carbon to carbon dioxide and hydrogen to water, both products of combustion. This process involves mathematical conversions, because the amount of a substance is often given in grams or milligrams, and moles are the standard unit in chemistry for counting particles like atoms and molecules.

Combustion Analysis

The technique of combustion analysis allows us to determine the elemental composition of a compound, especially those containing carbon and hydrogen. During a combustion reaction, the compound is burnt in excess oxygen to produce carbon dioxide and water. These products are then measured, and through stoichiometric calculations, we can deduce the amount of carbon and hydrogen in the original compound.

The exercise provided is an excellent example of combustion analysis. Here’s an essential tip: Always keep in mind that the mass of carbon dioxide produced can be directly used to find the moles of carbon in the initial compound, as each molecule of carbon dioxide contains exactly one atom of carbon. On the other hand, since water contains two hydrogen atoms, you must remember to double the number of moles of water to get the moles of hydrogen in the compound.

Freezing Point Depression

The concept of freezing point depression is a colligative property, meaning it depends on the number of particles dissolved in a solvent, regardless of their nature. When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent. This phenomenon is used in laboratories to calculate the molecular weight of a substance.

In our exercise, the lowering of camphor's freezing point upon dissolving anthraquinone is used to find its molecular weight. The extent of the freezing point depression is directly proportional to the molality of the solution, which in turn can be used to determine the molar mass of the solute (anthraquinone). The formula given in the step-by-step solution demonstrates this relationship and how it's applied. Always remember to convert milligrams to grams to maintain consistent units throughout the calculation.

Stoichiometry

The field of stoichiometry is all about the quantitative relationships between the reactants and products in a chemical reaction. Stoichiometry uses the concept of the mole ratio to convert quantities from one substance to another. It's fundamental in many types of chemical analyses, including determining empirical and molecular formulas.

The application of stoichiometry in the exercise allows us to first find the empirical formula by comparing the mole ratios of the elements and then deduce the molecular formula using the molar mass. It is essential to comprehend that the empirical formula is the simplest whole-number ratio of elements in a compound, and the molecular formula is a multiple of the empirical formula that reflects the actual number of atoms in a molecule of that compound. Thus, accurate stoichiometric calculations are key to resolving the empirical and molecular formulas of a compound.