Question

Calculate \(x\), the number of molecules of water in oxalic acid hydrate, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot x \mathrm{H}_{2} \mathrm{O},\) from the following data: \(5.00 \mathrm{~g}\) of the compound is made up to exactly \(250 \mathrm{~mL}\) solution and \(25.0 \mathrm{~mL}\) of this solution requires \(15.9 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{NaOH}\) solution for neutralization.

Short answer

The number of water molecules, \(x\), in the oxalic acid hydrate (H2C2O4.xH2O) is 2.

Step-by-step solution

Calculate the moles of NaOH

First, calculate the moles of NaOH used in the reaction using the equation: moles = Molarity * Volume. Therefore, moles of NaOH = 0.500 M * 15.9 mL = 7.95 mmol.

Determine the moles of Oxalic Acid

Considering that the stoichiometric ratio between NaOH and H2C2O4 is 2:1, the moles of H2C2O4 is calculated by dividing the moles of NaOH by 2. Hence, moles of H2C2O4 = 7.95 mmol / 2 = 3.975 mmol.

Calculate the molar mass of H2C2O4.xH2O

The mass of H2C2O4.xH2O is given as 5.00g. Dividing the mass by the moles of H2C2O4 calculated will give the molar mass. Thus, molar mass = 5.00 g / 3.975 mmol = 1.26 g/mmol.

Find the number of water molecules in the hydrate

The molar mass of H2C2O4 is 90.04 g/mol. Subtracting this from the molar mass of the hydrate gives the mass of the water molecules in the hydrate. Dividing this by the molar mass of water (18.015 g/mol) will give the number of water molecules, x. Thus, x = (1.26 g/mmol - 90.04 g/mol) / 18.015 g/mol

Key concepts

Molar Mass Calculation

Understanding molar mass is fundamental in chemistry, especially when dealing with compounds like hydrates. The molar mass is defined as the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms present in the formula. For hydrates, which contain a specific number of water molecules (\( \text{H}_2\text{O} \) ), we also include the mass of these water molecules in our calculation.

For a compound like oxalic acid hydrate (\( \text{H}_2\text{C}_2\text{O}_4 \cdot x \text{H}_2\text{O} \) ), the molar mass is determined by adding the molar masses of oxalic acid and water molecules. Once the total molar mass is calculated, we can identify the number of water molecules, represented by 'x,' by comparing it with the molar mass of anhydrous oxalic acid.

Say we have a compound with a known mass. To find its molar mass, we divide this mass by the number of moles of the compound, which can be figured out through reactions with other substances, like in the exercise involving a neutralization reaction.

By deducing the molar mass of the hydrated compound and subtracting the molar mass of the anhydrous part, we calculate the mass attributed to water. Then, dividing this by the molar mass of water (\(18.015 \text{ g/mol} \)) will reveal the stoichiometric number of water molecules in the hydrate.

Stoichiometry

Stoichiometry lies at the heart of chemical reactions. It involves the calculation of relative quantities of reactants and products in a chemical reaction. In stoichiometry, the law of conservation of mass dictates that matter is neither created nor destroyed; it is the quantities that change based on the balanced chemical equation.

In our hydrate example, stoichiometry allows us to use the volume and molarity of NaOH (a base) to find the moles of oxalic acid (an acid) in a neutralization reaction. The key to successful stoichiometry is a balanced chemical equation, which in the case of a neutralization, typically shows the acid and the base reacting in definite proportions to form water and a salt.

The molar ratio in the balanced equation informs us how to convert between moles of different substances. For instance, if the stoichiometric ratio between NaOH and H2C2O4 is 2:1, knowing the amount of NaOH used allows us to find the quantity of oxalic acid that reacted. This step is vitally important in determining the number of molecules of water in the hydrated form of oxalic acid, as seen in the provided exercise.

Neutralization Reaction

A neutralization reaction is a type of chemical reaction in which an acid and a base react to form water and a salt. This is an important class of reactions in chemistry, often used to determine concentrations through titration, which involves the gradual addition of a solution of known concentration (the titrant) to a known volume of a solution of unknown concentration until the reaction reaches completion, indicated by a color change from an indicator or some other means.

In the case of our exercise, neutralization was used to determine the amount of oxalic acid in solution. We know that sodium hydroxide (NaOH), a strong base, will react with oxalic acid (H2C2O4), a weak acid, in a 2:1 ratio respectively. By calculating the moles of NaOH required for the neutralization, we indirectly measure the moles of oxalic acid.

The information gained from a neutralization reaction is very useful in analyzing acid-base interactions and can lead to the determination of unknown concentrations or, as in our textbook problem, figuring out the stoichiometry of a hydrated compound. The balance between acid and base and the resulting products, often easily analyzed through simple stoichiometric relationships, are the essence of neutralization reactions.