To determine the atomic mass of the metal in the trivalent metal oxide, we can start by understanding the composition of the compound. Let's represent the metal as M. Since the oxide is trivalent, the formula can be represented as \( M_2O_3 \).
In \( M_2O_3 \), there are 2 atoms of metal (M) and 3 atoms of oxygen (O). The molar mass of oxygen (O) is approximately 16 g/mol.
Step 1: Calculate the mass of oxygen in the compound
The mass contribution from oxygen in \( M_2O_3 \) can be calculated as follows:
\[ \text{Mass of 3 Oxygen atoms} = 3 \times 16 , \text{g/mol} = 48 , \text{g/mol} \]
Step 2: Calculate the total mass of the compound
Let \( M_m \) be the molar mass of the trivalent metal (M). The total molar mass of \( M_2O_3 \) is:
\[ \text{Total molar mass} = 2M_m + 48 \]
Step 3: Use the information about the mass percentage of oxygen
According to the problem, the oxide contains 31.6% oxygen. This can be represented as:
\[ \frac{\text{Mass of O}}{\text{Total mass}} \times 100 = 31.6 \]
Substituting the known values:
\[ \frac{48}{2M_m + 48} \times 100 = 31.6 \]
Step 4: Solve for \( M_m \)
Rearranging the equation gives:
\[ \frac{48}{2M_m + 48} = 0.316 \]
Cross-multiplying:
\[ 48 = 0.316(2M_m + 48) \]
Expanding the right side:
\[ 48 = 0.632M_m + 15.168 \]
Now, isolate \( M_m \):
\[ 48 - 15.168 = 0.632M_m \] \[ 32.832 = 0.632M_m \] \[ M_m = \frac{32.832}{0.632} \approx 51.91 , \text{g/mol} \]
Conclusion
The atomic mass of the trivalent metal is approximately 51.91 g/mol.
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