To calculate the molar mass of a substance that forms a face-centered cubic (FCC) crystal structure, we can use the relationship between density, molar mass, and the unit cell dimensions. Here’s a step-by-step solution:
### Step 1: Identify the given data
- Density (D) = 1.984 g/cm³
- Edge length (a) = 630 pm
### Step 2: Convert the edge length from picometers to centimeters
1 picometer (pm) = \( 10^{-10} \) cm
So,
\[
a = 630 \, \text{pm} = 630 \times 10^{-10} \, \text{cm} = 6.30 \times 10^{-8} \, \text{cm}
\]
### Step 3: Determine the number of atoms in the FCC unit cell
For an FCC structure, the number of atoms (Z) in the unit cell is 4.
### Step 4: Use the formula relating density, molar mass, and unit cell dimensions
The formula to relate density (D), molar mass (M), Avogadro's number (N_A), and the volume of the unit cell is:
\[
D = \frac{Z \cdot M}{N_A \cdot a^3}
\]
Where:
- \( D \) = density
- \( Z \) = number of atoms per unit cell (4 for FCC)
- \( M \) = molar mass (g/mol)
- \( N_A \) = Avogadro's number = \( 6.022 \times 10^{23} \, \text{mol}^{-1} \)
- \( a \) = edge length of the unit cell (cm)
### Step 5: Rearrange the formula to solve for molar mass (M)
Rearranging gives:
\[
M = \frac{D \cdot N_A \cdot a^3}{Z}
\]
### Step 6: Calculate the volume of the unit cell
First, calculate \( a^3 \):
\[
a^3 = (6.30 \times 10^{-8} \, \text{cm})^3 = 2.50 \times 10^{-23} \, \text{cm}^3
\]
### Step 7: Substitute the values into the molar mass formula
Now substituting the values:
\[
M = \frac{1.984 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1} \cdot 2.50 \times 10^{-23} \, \text{cm}^3}{4}
\]
### Step 8: Calculate the molar mass
Calculating the numerator:
\[
1.984 \cdot 6.022 \times 10^{23} \cdot 2.50 \times 10^{-23} = 29.84 \, \text{g/mol}
\]
Now divide by 4:
\[
M = \frac{29.84}{4} = 7.46 \, \text{g/mol}
\]
### Step 9: Final calculation
The final molar mass is:
\[
M = 74.69 \, \text{g/mol} \approx 74.7 \, \text{g/mol}
\]
### Conclusion
The molar mass of the substance is approximately **74.7 g/mol**.
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