Given: \( \mathrm{DyCl}_{3} \) Wanted: percent composition?

To determine the **percent composition** of \( \mathrm{DyCl}_3 \) (Dysprosium trichloride), follow these steps: ### 1. Determine the Molar Masses First, find the atomic masses of each element from the periodic table: - **Dysprosium (Dy):** Approximately **162.5 g/mol** - **Chlorine (Cl):** Approximately **35.45 g/mol** ### 2. Calculate the Molar Mass of \( \mathrm{DyCl}_3 \) The chemical formula \( \mathrm{DyCl}_3 \) indicates that there is **1 atom of Dy** and **3 atoms of Cl** in each molecule. \[ \text{Molar mass of } \mathrm{DyCl}_3 = (\text{1} \times 162.5\ \text{g/mol}) + (\text{3} \times 35.45\ \text{g/mol}) \] \[ \text{Molar mass of } \mathrm{DyCl}_3 = 162.5\ \text{g/mol} + 106.35\ \text{g/mol} = 268.85\ \text{g/mol} \] ### 3. Calculate the Percent Composition The percent composition is calculated by dividing the mass of each element in the compound by the total molar mass of the compound and then multiplying by 100%. #### a. **Percent Dysprosium (Dy)** \[ \% \text{Dy} = \left( \frac{162.5\ \text{g/mol}}{268.85\ \text{g/mol}} \right) \times 100\% \approx 60.45\% \] #### b. **Percent Chlorine (Cl)** \[ \% \text{Cl} = \left( \frac{106.35\ \text{g/mol}}{268.85\ \text{g/mol}} \right) \times 100\% \approx 39.55\% \] ### 4. Summary of Percent Composition - **Dysprosium (Dy):** **60.45%** - **Chlorine (Cl):** **39.55%** **Therefore, the percent composition of \( \mathrm{DyCl}_3 \) is approximately 60.45% Dysprosium and 39.55% Chlorine.**