Question

\begin{aligned} &\text { What mass of sodium oxalate }\left(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right) \text { is needed to prepare }\\\ &0.250 \mathrm{~L} \text { of a } 0.100-M \text { solution? } \end{aligned}

Short answer

To prepare 0.250 L of a 0.100-M solution of sodium oxalate (Na₂C₂O₄), you need 3.35 g of sodium oxalate.

Step-by-step solution

Determine the moles of sodium oxalate required to prepare the solution

We are given the volume of the solution (0.250 L) and the molarity (0.100 M). Use the formula to find the moles of sodium oxalate: Moles of sodium oxalate = Molarity × Volume $$ = 0.100\,M\times0.250\,L $$ $$ = 0.025\, moles $$

Calculate the molar mass of sodium oxalate

To find the mass of sodium oxalate, we need to calculate the molar mass of sodium oxalate (Na₂C₂O₄). The molar mass can be calculated by adding the molar masses of the elements present in it. Molar mass of Na₂C₂O₄ = 2 × (Molar mass of Na) + 2 × (Molar mass of C) + 4 × (Molar mass of O) From the periodic table, we can find the molar masses of the elements: Na = 22.99 g/mol, C = 12.01 g/mol, and O = 16.00 g/mol. Molar mass of Na₂C₂O₄ = 2(22.99 g/mol) + 2(12.01 g/mol) + 4(16.00 g/mol) $$ = 45.98\,g/mol + 24.02\,g/mol + 64.00\,g/mol $$ $$ = 134.00\,g/mol $$

Find the mass of sodium oxalate required

Now that we have the moles of sodium oxalate and its molar mass, we can find the mass of sodium oxalate required to prepare the solution: Mass of sodium oxalate = moles × molar mass $$ = 0.025\,moles \times 134.00\,g/mol $$ $$ = 3.35\,g $$ Thus, 3.35 g of sodium oxalate is needed to prepare 0.250 L of a 0.100-M solution.

Key concepts

Molarity Calculation

Molarity is a way to express the concentration of a solution. It tells us how many moles of solute, in this case sodium oxalate (\(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\)), are present in one liter of solution. It is denoted by the letter \(M\). Here’s the basic formula you need to perform a molarity calculation: \(\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\)In practice, if you need to determine how many moles of a substance are needed, you rearrange this formula:

• Multiply the given molarity by the volume of the solution (in liters) to find the number of moles.

In the original problem, with a molarity of 0.100 M and a volume of 0.250 L, the calculation is:\[0.100 \, \text{M} \times 0.250 \, \text{L} = 0.025 \, \text{moles}\] This tells us that \(0.025\) moles of sodium oxalate are needed to achieve this concentration.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It serves as a critical conversion factor in stoichiometry, allowing us to convert between moles and grams. To calculate the molar mass of a compound:

• Identify the elements in the compound and their amounts.
• Use the periodic table to find the molar mass of each element.
• Multiply each element’s molar mass by the number of atoms of that element in the compound.
• Add all these values together to determine the molar mass of the compound.

For sodium oxalate (\(\mathrm{Na}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\)), its molar mass is calculated as follows:

• Sodium (\(\mathrm{Na}\)): 2 atoms ✕ 22.99 g/mol = 45.98 g/mol
• Carbon (\(\mathrm{C}\)): 2 atoms ✕ 12.01 g/mol = 24.02 g/mol
• Oxygen (\(\mathrm{O}\)): 4 atoms ✕ 16.00 g/mol = 64.00 g/mol

Add them together to get the molar mass:\[45.98 + 24.02 + 64.00 = 134.00 \, \text{g/mol}\] So, the molar mass of sodium oxalate is \(134.00 \, \text{g/mol}\).

Solution Preparation

Preparing a solution is a common task in chemistry where we need specific concentrations for experiments and reactions. It involves dissolving a certain mass of solute in a specified volume of solvent to achieve the desired molarity. Here is a simple step-by-step guide for preparing a solution:

• First, calculate the number of moles of solute required using the planned molarity and volume.
• Next, convert the moles into grams using the molar mass of the solute.
• Measure the necessary mass of solute accurately on a balance.
• Slowly add the solute to a volume of solvent that is less than the desired final volume, allowing the solute to dissolve completely.
• Once dissolved, add additional solvent to achieve the total required volume.

In our specific example, after determining that \(0.025\) moles are needed, convert this into grams using the molar mass (\(134.00 \, \text{g/mol}\)):\[0.025 \, \text{moles} \times 134.00 \, \text{g/mol} = 3.35 \, \text{g}\]This means you need 3.35 g of sodium oxalate for the solution preparation.

Stoichiometry

Stoichiometry is the aspect of chemistry that involves calculating the relative quantities of reactants and products in chemical reactions. It helps us predict the amounts of substances consumed and produced in a reaction, ensuring no reactants are wasted. It primarily uses the concept of moles and molar masses from the periodic table. In the context of preparing a sodium oxalate solution, stoichiometry helps in determining the exact amount of the compound needed to create a solution of a specified concentration and volume.

• It begins with understanding the relationship between moles, molarity, and volume of solutions.
• We calculate moles from a given molarity and volume of the solution.
• Using molar mass, we convert these moles into a mass that can be practically measured.

These calculations ensure that we achieve a solution with the precise concentration of solute, which is crucial for scientific experiments where exact measurements are necessary for valid results. In our example, stoichiometry confirmed that exactly 3.35 g of sodium oxalate should be used to make the desired solution of 0.250 L with a concentration of 0.100 M .