Problem 6 - Chemistry A Guided Inquiry

Question

Verify that when \(10.0 \mathrm{~g}\) of sodium sulfate dissolves in water: a) there are \(7.04 \times 10^{-2}\) moles of sodium sulfate in the water. b) there are \(7.04 \times 10^{-2}\) moles of sulfate in the water. c) there are \(14.1 \times 10^{-2}\) moles of sodium in the water.

Step-by-Step Solution

Answer

Yes, when 10.0 g of sodium sulfate dissolves in water, there are \(7.04 \times 10^{-2}\) moles of sodium sulfate, \(7.04 \times 10^{-2}\) moles of sulfate, and \(14.1 \times 10^{-2}\) moles of sodium.

Step 1: Calculate the molar mass of sodium sulfate

First off, find the molar mass of sodium sulfate (Na2SO4). This requires the values of atomic masses for Sodium (Na), Sulfur (S), and Oxygen (O), which are approximately 23 g/mol, 32 g/mol, and 16 g/mol, respectively. The molar mass of Na2SO4 will be: \(2 * 23 + 32 + 4*16 = 142 \mathrm{~g/mol}\)

Step 2: Verify the amount of moles of sodium sulfate

Now that we know the molar mass of sodium sulfate, we can calculate how many moles are in 10.0 g. The formula for calculating moles is to divide the given mass by molar mass. So for this case, it will be: \(10.0 g / 142 g/mol \approx 7.04 * 10^{-2}\), which is match the value we were given.

Step 3: Confirm the amount of moles of sulfate

As stated before, the compound sodium sulfate dissolves into two Na atoms and one SO4 atom in water. Thus, the amount of moles of sodium sulfate (Na2SO4) is equal with moles of sulfate (SO4) which is \(7.04 \times 10^{-2}\) moles.

Step 4: Confirm the amount of moles of sodium

Similar to the previous step, we note that each molecule of sodium sulfate contains two sodium atoms. As such, the number of moles of sodium will be twice the number of moles of sodium sulfate. Therefore we get: \( 2 * 7.04 \times 10^{-2} = 14.1 \times 10^{-2}\) moles.

Key Concepts

Molar Mass Calculation

Understanding the molar mass of a compound is the cornerstone of many calculations in chemistry. Molar mass allows us to translate between the mass of a substance and the number of moles, which is a measure of the amount of substance in terms of particles, like atoms or molecules.

To calculate the molar mass, we sum the atomic masses of all the atoms that make up the compound. For example, sodium sulfate ((Na_2SO_4)) has two sodium atoms, one sulfur atom, and four oxygen atoms. Knowing that the atomic masses for sodium (Na), sulfur (S), and oxygen (O) are approximately 23 g/mol, 32 g/mol, and 16 g/mol respectively, we can calculate the molar mass of sodium sulfate by multiplying the number of each type of atom by its atomic mass and adding these values together. This gives us the molar mass of (Na_2SO_4) as (2 * 23 + 32 + 4 * 16 = 142 g/mol).

With this information, we can perform various calculations to understand the composition of a sample, including how many grams of the substance are in one mole, or conversely, how many moles are in a given mass of the substance.

Stoichiometry

Stoichiometry is essentially the mathematics of chemistry. It involves using the balanced chemical equation to understand the quantitative relationships between the reactants and products in a chemical reaction.

In stoichiometry, the coefficients in a balanced equation indicate the ratio of moles of each substance involved. For instance, in the dissolution of sodium sulfate in water, we deal with the stoichiometry of the dissolution process. Considering the molar ratios, we can infer how many moles of ions are produced when sodium sulfate dissolves.

Example:

If we dissolve one mole of sodium sulfate, we would end up with two moles of sodium ions (Na^+) because each formula unit of sodium sulfate ((Na_2SO_4)) yields two sodium ions upon dissolution. It is through stoichiometry that we can also confirm the number of moles of sodium and sulfate ions that result from this process.

Dissolution of Ionic Compounds

Ionic compounds, such as sodium sulfate ((Na_2SO_4)), dissociate into ions when they dissolve in water due to water's polar nature. The process of dissolution involves the separation of the ions in the solid compound as they become solvated by water molecules.

During the dissolution, the ionic bonds holding the compound together are broken, allowing the individual cations and anions to disperse throughout the solution. For sodium sulfate, one unit of the compound separates into two sodium ions (Na^+) and one sulfate ion (SO_4^{2-}).

Thus, when calculating the amount of ions formed from a certain mass of ionic compound, it is important to consider not only the molar mass but also the ratio of the ions produced according to the chemical formula. This concept is crucial when determining the conductivity, the ability to form precipitates, and the chemical reactivity of the solution.