Practice: Counting Atoms 8. Calculating Molar Mass 4(NH_(4))_(2)SO_(3) II OIN square II OIH square II of S square It of o square

To calculate the molar mass of the compound \(4(NH_{4})_{2}SO_{3}\), we need to determine the number of atoms of each element in the compound and then multiply by the atomic mass of each element.<br /><br />1. **Identify the number of atoms for each element:**<br /><br /> - \(N\): There are 2 ammonium ions (\(NH_{4}\)) in each \((NH_{4})_{2}\), and there are 2 \((NH_{4})_{2}\) units in the compound. So, there are \(2 \times 2 = 4\) nitrogen atoms.<br /> - \(H\): Each ammonium ion (\(NH_{4}\)) has 4 hydrogen atoms, and there are 2 ammonium ions in each \((NH_{4})_{2}\). So, there are \(4 \times 2 = 8\) hydrogen atoms.<br /> - \(S\): There is 1 sulfur atom in the sulfate ion (\(SO_{3}\)).<br /> - \(O\): There are 3 oxygen atoms in the sulfate ion (\(SO_{3}\)).<br /><br />2. **Calculate the molar mass:**<br /><br /> - Nitrogen (N): \(14.01 \, \text{g/mol}\)<br /> - Hydrogen (H): \(1.01 \, \text{g/mol}\)<br /> - Sulfur (S): \(32.07 \, \text{g/mol}\)<br /> - Oxygen (O): \(16.00 \, \text{g/mol}\)<br /><br /> Now, multiply the number of atoms by the atomic mass of each element:<br /><br /> - \(N\): \(4 \times 14.01 = 56.04 \, \text{g/mol}\)<br /> - \(H\): \(8 \times 1.01 = 8.08 \, \text{g/mol}\)<br /> - \(S\): \(1 \times 32.07 = 32.07 \, \text{g/mol}\)<br /> - \(O\): \(3 \times 16.00 = 48.00 \, \text{g/mol}\)<br /><br />3. **Sum the contributions:**<br /><br /> - Total molar mass = \(56.04 + 8.08 + 32.07 + 48.00 = 144.19 \, \text{g/mol}\)<br /><br />Therefore, the molar mass of \(4(NH_{4})_{2}SO_{3}\) is \(144.19 \, \text{g/mol}\).