Question

A \(2.20-\mathrm{g}\) sample of an unknown acid (empirical formula \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3}\) ) is dissolved in \(1.0 \mathrm{~L}\) of water. A titration required \(25.0\) \(\mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{NaOH}\) to react completely with all the acid present. Assuming the unknown acid has one acidic proton per molecule, what is the molecular formula of the unknown acid?

Short answer

The molecular formula of the unknown acid is \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\).

Step-by-step solution

Calculate the moles of NaOH used in the titration

First, we need to find out how many moles of NaOH were used in the titration. This can be done using the given volume and concentration of NaOH. Moles of NaOH = volume × concentration Moles of NaOH = \(25.0 \times 10^{-3} L \times 0.500 M\) Moles of NaOH = \(1.25 \times 10^{-2} mol\)

Calculate the moles of the unknown acid

Since one mole of the unknown acid reacts with one mole of NaOH, then the moles of the unknown acid will be equal to the moles of NaOH. Moles of unknown acid = moles of NaOH = \(1.25 \times 10^{-2} mol\)

Calculate the molar mass of the unknown acid

Now, we can find out the molar mass of the unknown acid using the given mass of the sample and the moles calculated before. Molar mass of unknown acid = mass of sample / moles of unknown acid Molar mass of unknown acid = \(2.20 g / (1.25 \times 10^{-2} mol)\) Molar mass of unknown acid = \(176 g/mol\)

Calculate the empirical mass of the unknown acid

Next, we will find the empirical mass of the unknown acid using the given empirical formula (\(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3}\)) by adding up the atomic masses of the constituent elements. Empirical mass = \(3 \times 12.01 g/mol_{C} + 4 \times 1.01 g/mol_{H} + 3 \times 16.00 g/mol_{O}\) Empirical mass = \(36.03 + 4.04 + 48.00\) Empirical mass = \(88.07 g/mol\)

Determine the molecular formula of the unknown acid

Now we can use the molar mass and empirical mass to determine the molecular formula of the unknown acid. Divide the molar mass by the empirical mass to find out how many empirical units are present in the molecular formula. Number of empirical units = molar mass / empirical mass Number of empirical units = \(176 g/mol / 88.07 g/mol\) Number of empirical units = 2 Since there are two empirical units in the molecular formula, we can multiply the subscripts in the empirical formula by 2 to find the molecular formula. Molecular formula = \((\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3})_{2}\) Molecular formula = \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\)

Key concepts

Mole Concept

The mole concept is a fundamental aspect of chemistry that connects the mass of a substance to the amount of particles it contains. Scientifically, one mole is defined as containing exactly Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\) entities (such as atoms, ions, or molecules). This concept allows chemists to count and measure quantities at the atomic scale.
In the context of titration, the mole concept is crucial for determining how many moles of a reactant are required to completely react with another substance. For example, in this exercise, titration was used to find out how many moles of sodium hydroxide (NaOH) completely reacted with the unknown acid. By multiplying the volume of NaOH by its concentration, we calculated the amount of moles used, which was \(1.25 \times 10^{-2}\) mol. This value represented the number of moles of the unknown acid as well, based on a 1:1 molar ratio of reaction between the acid and base.

Empirical Formula

An empirical formula represents the simplest whole-number ratio of elements in a compound. It does not provide information on the exact number of atoms, but rather the fundamental ratio of the elements present.
In the given exercise, the empirical formula of the unknown acid is \(\mathrm{C}_3\mathrm{H}_4\mathrm{O}_3\). This tells us that in every empirical unit of the compound, there are 3 carbon atoms, 4 hydrogen atoms, and 3 oxygen atoms. To determine the empirical mass, we add the masses of these atoms together: \(3 \times 12.01\, g/mol_{C} + 4 \times 1.01\, g/mol_{H} + 3 \times 16.00\, g/mol_{O}\), giving us an empirical mass of \(88.07\, g/mol\).
This empirical mass serves as a benchmark for further calculating the true "molecular formula" of the compound when combining with the molar mass.

Molecular Formula

The molecular formula reveals the actual number of atoms of each element in a molecule of the compound. Unlike the empirical formula, it provides an exact representation of the molecule's composition.
In the exercise, we calculated the molar mass of the unknown acid to be \(176\, g/mol\) and compared it to the empirical mass \(88.07\, g/mol\) obtained from the empirical formula. The division of the molar mass by the empirical mass provides the number of times the empirical unit is contained in the molecular structure of the compound. Here, \(\frac{176}{88.07} \approx 2\), showing that the molecule contains two empirical units.
Thus, each element's subscript in the empirical formula is doubled—yielding the molecular formula \(\mathrm{C}_6\mathrm{H}_8\mathrm{O}_6\), representing the true structure of the unknown acid.