Question

Calculate \(x\), the number of molecules of water in oxalic acid hydrate \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot x \mathrm{H}_{2} \mathrm{O}\right),\) 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 oxalic acid hydrate is 2.

Step-by-step solution

Determine Moles of NaOH Used

The volume of NaOH used for neutralization is given as 15.9 mL, which we need to convert to liters (by dividing by 1000), giving us 0.0159 L. Then, use the concentration of the NaOH solution to find moles of NaOH: \[ \text{moles of NaOH} = \text{concentration} \times \text{volume} = 0.500 \times 0.0159 = 0.00795 \text{ moles} \]

Determine Moles of Oxalic Acid Neutralized

The balanced chemical equation for the reaction between oxalic acid (\(\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4\)) and \(\mathrm{NaOH}\) is:\[ \mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} + 2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{C}_2\mathrm{O}_4 + 2 \mathrm{H}_2\mathrm{O} \]From the equation, 1 mole of oxalic acid reacts with 2 moles of NaOH. Thus, moles of oxalic acid reacted is half the moles of NaOH:\[ \text{moles of oxalic acid} = \frac{0.00795}{2} = 0.003975 \text{ moles} \]

Calculate Moles of Oxalic Acid in Total Solution

The moles of oxalic acid calculated previously (0.003975 moles) was for a 25.0 mL aliquot of the 250 mL solution. To find the total moles in the entire solution, multiply by the dilution factor (250 mL/25 mL):\[ \text{total moles of oxalic acid} = 0.003975 \times \frac{250}{25} = 0.03975 \text{ moles} \]

Find Molar Mass of Oxalic Acid Hydrate

First, determine the molar mass of anhydrous oxalic acid \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\), which is:\[ \text{molar mass of } \mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} = 2(1.01) + 2(12.01) + 4(16.00) = 90.03 \text{ g/mol} \]Let the molar mass of water \(\mathrm{H}_2\mathrm{O}\) be 18.02 g/mol. Thus, the molar mass of the hydrate is:\[ 90.03 + x(18.02) \text{ g/mol} \]

Establish Equation and Solve for \(x\)

The compound's mass is 5.00 g containing 0.03975 moles of hydrate. Using this, form an equation:\[ 0.03975 \text{ moles} \times (90.03 + x(18.02)) \text{ g/mol} = 5.00 \text{ g} \]Rewriting:\[ 90.03 \times 0.03975 + x \times 18.02 \times 0.03975 = 5 \]Solving for \(x\), we first compute:\[ 3.577 \text{ g} + 0.7155x = 5 \]Rearranging gives:\[ 0.7155x = 1.423 \]Finally, solving for \(x\) gives:\[ x = \frac{1.423}{0.7155} \approx 1.99 \]Since \(x\) must be a whole number, round to nearest whole number: \(x = 2\).

Key concepts

Molar Mass Calculation

To understand how the molar mass is calculated, let's start with the concept of molar mass itself. Molar mass is the mass of one mole of a substance, and it's usually measured in grams per mole (g/mol). Each element in the periodic table has an atomic mass, which is used in the calculation. For oxalic acid \(\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4\), the molar mass calculation involves adding up the molar masses of hydrogen (H), carbon (C), and oxygen (O) present in the compound.

• Hydrogen has a molar mass of about 1.01 g/mol.
• Carbon has a molar mass of about 12.01 g/mol.
• Oxygen's molar mass is about 16.00 g/mol.

To find the molar mass of \(\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4\), multiply the atomic weights by the number of each atom:\[2(1.01) + 2(12.01) + 4(16.00) = 90.03 \text{ g/mol}\]When dealing with hydrates, such as in this exercise, you add the molar mass of the water \(\mathrm{H}_2\mathrm{O}\), which is 18.02 g/mol, multiplied by the number of water molecules, represented as \(x\) in \(\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4 \cdot x\mathrm{H}_2\mathrm{O}\).

Chemical Reaction Stoichiometry

Stoichiometry is all about the quantitative relationships in a chemical reaction based on the balanced chemical equation. In the exercise, we see the reaction of oxalic acid with sodium hydroxide (NaOH):\[\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4 + 2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{C}_2\mathrm{O}_4 + 2 \mathrm{H}_2\mathrm{O}\]This equation indicates that 1 mole of oxalic acid reacts with 2 moles of NaOH. This is crucial because stoichiometry allows us to convert between amounts of different chemical species using their molar ratios from the balanced equation.In practice, using stoichiometry:

• First, we calculate the moles of NaOH reacted, which is important for finding how many moles of oxalic acid were neutralized.
• We see that the moles of oxalic acid is half of the moles of NaOH used, thanks to the 1:2 stoichiometric ratio.

Once you know the moles of oxalic acid that reacted, you can work backwards using stoichiometric principles to find the mole information for the entire solution, an integral step for further calculations in the exercise.

Neutralization Reaction

Neutralization reactions are special types of chemical reactions where an acid reacts with a base to produce a salt and water. Such reactions are integral in calculating quantities in chemistry problems, especially those involving titrations.For oxalic acid \(\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4\) reacting with NaOH, the reaction is: \[\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4 + 2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{C}_2\mathrm{O}_4 + 2 \mathrm{H}_2\mathrm{O}\]Neutralization occurs when the oxalic acid donates its protons to the hydroxide ions (\(\mathrm{OH}^-\)) from the NaOH. This results in the formation of water and a salt, \(\mathrm{Na}_2\mathrm{C}_2\mathrm{O}_4\).The primary use of neutralization in this exercise was to determine how much NaOH was needed to react completely with a given volume of oxalic acid solution. By measuring the amount of NaOH used, we can then deduce the initial amount of oxalic acid present in the solution.

Mole Concept

The mole is a fundamental unit in chemistry representing a quantity of chemical substances. One mole contains Avogadro's number of entities, which is approximately \(6.022 \times 10^{23}\). This concept allows chemists to count particles by weighing them.In this exercise, the mole concept helps in several ways:

• Calculating the number of moles of sodium hydroxide used in the reaction.
• Determining the number of moles of oxalic acid in the solution from the reaction's stoichiometry.
• Finding the total moles of oxelic acid in the dilute solution.

Understanding the mole concept is vital because it underpins quantifying chemical reactions, allowing chemistry to move from simply descriptive to predictive.The concept simplifies dealing with the massive numbers of atoms and molecules by providing a manageable way to work with chemical quantities.This ultimately allows for accurate calculations needed to solve the given chemical problems.