Question

What volume of \(0.112 M\) ammonium sulfate contains \(5.75 \mathrm{~g}\) of ammonium ion?

Short answer

1427 mL of 0.112 M ammonium sulfate is needed.

Step-by-step solution

Identify Relevant Information

The molecular formula of ammonium sulfate is \((NH_4)_2SO_4\). The molar mass of an ammonium ion \((NH_4^+)\) is \(14 + 4 = 18\) g/mol. The molar mass of ammonium sulfate is \(2(18) + 32 + 4(16) = 132\) g/mol. The concentration of the ammonium sulfate solution is \(0.112\ M\).

Calculate Moles of Ammonium Ion Needed

You know there are two ammonium ions per formula unit of ammonium sulfate. First, calculate the moles of ammonium ion from its mass: \[m = \frac{5.75\ g}{18\ g/mol} \approx 0.3194\ mol\]

Determine Moles of Ammonium Sulfate

Since there are two ammonium ions per mole of ammonium sulfate, the moles of ammonium sulfate, \(n\), needed is: \[n = \frac{0.3194\ mol}{2} \approx 0.1597\ mol\]

Calculate Volume of Solution Required

Use the concentration formula \(C = \frac{n}{V}\) to find the volume, \(V\), of the solution needed:\[V = \frac{n}{C} = \frac{0.1597\ mol}{0.112\ M} \approx 1.4268\ L\]

Convert Volume to Milliliters

Convert the volume from liters to milliliters since it's an easier unit for practical measurements: \[V = 1.4268\ L \times 1000\ mL/L = 1426.8\ mL\approx 1427\ mL.\]

Key concepts

Molecular Formula

The molecular formula helps you understand the composition of a chemical compound. For ammonium sulfate, this is denoted as \((NH_4)_2SO_4\), which tells us a lot about its composition. It has:

• Two ammonium ions (\(NH_4^+\)).
• One sulfate ion (\(SO_4^{2-}\)).

This molecular formula indicates that the compound consists of a complex iron combination of both nitrogen, hydrogen, sulfur, and oxygen. Understanding the molecular formula is the first step in any calculation involving chemical compounds since it provides the basis for calculating other properties such as molar mass.

Molar Mass

The molar mass is a measure of how much one mole of a substance weighs. It's calculated by adding the atomic masses of all atoms in a molecule. For an ammonium ion (\(NH_4^+\)), the molar mass is calculated as follows:

• Nitrogen (N) has an atomic mass of approximately 14 g/mol.
• Hydrogen (H) has an atomic mass of approximately 1 g/mol, and there are four hydrogen atoms, so 4 g/mol in total.

Adding these gives a molar mass of 18 g/mol for the ammonium ion.
For ammonium sulfate (\((NH_4)_2SO_4\)), the calculation is:

• Two ammonium ions contribute 2 x 18 g/mol = 36 g/mol.
• Sulfur (S) adds 32 g/mol.
• Oxygen (O), with four atoms present, contributes 4 x 16 g/mol = 64 g/mol.

Adding these together results in a molar mass of 132 g/mol for ammonium sulfate.
Understanding molar mass is essential to convert between grams and moles and is crucial for calculations involving solution concentrations.

Solution Concentration

Solution concentration is a way to express the amount of solute in a given volume of solution. It's often described in terms of molarity (\(M\)), which measures moles of solute per liter of solution.
For example, if the concentration of an ammonium sulfate solution is 0.112 M, it means there are 0.112 moles of ammonium sulfate in every liter of the solution.
When solving problems, knowing the molarity helps you calculate how much of the solution you will need to reach a desired amount of a particular compound (like ammonium ions in this case). The relationship between moles, volume, and molarity is given by the formula \(C = \frac{n}{V}\) where \(C\) is the concentration, \(n\) is the number of moles, and \(V\) is the volume.

Ammonium Ion

The ammonium ion (\(NH_4^+\)) is an important component in many compounds, including ammonium sulfate. This ion consists of:

• One nitrogen atom (\(N\)).
• Four hydrogen atoms (\(H\)).

In any given molecule of ammonium sulfate, there are two ammonium ions. They play a crucial role in the chemical properties and reactions of the compounds they are part of, because of their positive charge which can interact with negatively charged ions.
Calculating the amount of ammonium ions is necessary when determining concentrations or quantities in chemical solutions, especially if you know the mass of ammonium ions you need, for instance, 5.75 grams in the exercise provided above. By understanding how many moles of ammonium are present, you can determine the volume of solution necessary, making each calculation precise and effective.