To calculate the mass of ammonium sulfide [tex]\(\left( NH_4 \right)_2 S\)[/tex] in 3.00 L of a 0.0200 M solution, we can follow these steps:
1. Determine the number of moles of [tex]\(\left( NH_4 \right)_2 S\)[/tex]:
- Molarity ([tex]\(M\)[/tex]) is defined as the number of moles of solute per liter of solution.
- The given molarity of the solution is 0.0200 M, which means there are 0.0200 moles of [tex]\(\left( NH_4 \right)_2 S\)[/tex] per liter of solution.
- Volume of the solution is 3.00 L.
- To find the total number of moles, multiply the molarity by the volume of the solution:
[tex]\[
\text{moles of} \ \left( NH_4 \right)_2 S = 0.0200 \ \text{mol/L} \times 3.00 \ \text{L} = 0.0600 \ \text{mol}
\][/tex]
2. Calculate the molar mass of [tex]\(\left( NH_4 \right)_2 S\)[/tex]:
- The molar mass is obtained by adding the atomic masses of all the atoms in the molecular formula.
- For ammonium sulfide, [tex]\(\left( NH_4 \right)_2 S\)[/tex]:
- Nitrogen [tex]\((N)\)[/tex] has an atomic mass of approximately [tex]\(14.01 \ \text{g/mol}\)[/tex].
- Hydrogen [tex]\((H)\)[/tex] has an atomic mass of approximately [tex]\(1.008 \ \text{g/mol}\)[/tex].
- Sulfur [tex]\((S)\)[/tex] has an atomic mass of approximately [tex]\(32.07 \ \text{g/mol}\)[/tex].
- The formula [tex]\(\left( NH_4 \right)_2 S\)[/tex] consists of:
- 2 nitrogen atoms: [tex]\(2 \times 14.01 \ \text{g/mol} = 28.02 \ \text{g/mol}\)[/tex]
- 8 hydrogen atoms: [tex]\(8 \times 1.008 \ \text{g/mol} = 8.064 \ \text{g/mol}\)[/tex]
- 1 sulfur atom: [tex]\(1 \times 32.07 \ \text{g/mol} = 32.07 \ \text{g/mol}\)[/tex]
- Adding these together:
[tex]\[
\text{Molar mass of} \ \left( NH_4 \right)_2 S = 28.02 \ \text{g/mol} + 8.064 \ \text{g/mol} + 32.07 \ \text{g/mol} = 68.154 \ \text{g/mol}
\][/tex]
3. Calculate the mass of [tex]\(\left( NH_4 \right)_2 S\)[/tex]:
- The mass of a substance can be calculated by multiplying the number of moles by the molar mass.
- Using the number of moles (0.0600 mol) and the molar mass (68.154 g/mol), we get:
[tex]\[
\text{mass of} \ \left( NH_4 \right)_2 S = 0.0600 \ \text{mol} \times 68.154 \ \text{g/mol} = 4.08924 \ \text{g}
\][/tex]
Hence, the mass of ammonium sulfide in 3.00 L of a 0.0200 M solution is approximately [tex]\(4.089 \ \text{g}\)[/tex].
