A student found a new metal wrapped up in a  plastic bag with few information written on top of it.
Information:
1. It is a FCC lattice
2.Density = 1.54 g/cm^3
3.Radius of atom= 0.197nm
With the above information he wrote some equations to find the unknown metal but unfortunately while solving he has taken Avogadro Constant as Na=8.3848 * 10^23 .
Based on the above information ,what is the atomic number of Metal found by student ?

To determine the atomic number of the metal found by the student, we need to follow these steps:

1. **Calculate the volume of the unit cell:**
   - For an FCC (Face-Centered Cubic) lattice, the edge length \( a \) of the unit cell is related to the atomic radius \( r \) by the equation:
     \[
     a = \frac{4r}{\sqrt{2}}
     \]

2. **Convert the radius from nanometers to centimeters:**
   \[
   r = 0.197 \text{ nm} = 0.197 \times 10^{-7} \text{ cm}
   \]

3. **Calculate the edge length \( a \):**
   \[
   a = \frac{4 \times 0.197 \times 10^{-7} \text{ cm}}{\sqrt{2}} = \frac{0.788 \times 10^{-7} \text{ cm}}{1.414} \approx 0.557 \times 10^{-7} \text{ cm}
   \]

4. **Calculate the volume of the unit cell \( V \):**
   \[
   V = a^3 = (0.557 \times 10^{-7} \text{ cm})^3 \approx 1.729 \times 10^{-22} \text{ cm}^3
   \]

5. **Relate the density to the mass of atoms in the unit cell:**
   - The density \( \rho \) is given by:
     \[
     \rho = \frac{\text{mass of unit cell}}{\text{volume of unit cell}}
     \]
   - The mass of the unit cell is the mass of 4 atoms (since an FCC unit cell contains 4 atoms).

6. **Calculate the mass of the unit cell:**
   \[
   \text{Mass of unit cell} = \rho \times V = 1.54 \text{ g/cm}^3 \times 1.729 \times 10^{-22} \text{ cm}^3 \approx 2.664 \times 10^{-22} \text{ g}
   \]

7. **Calculate the mass of a single atom:**
   \[
   \text{Mass of one atom} = \frac{\text{Mass of unit cell}}{4} = \frac{2.664 \times 10^{-22} \text{ g}}{4} \approx 6.66 \times 10^{-23} \text{ g}
   \]

8. **Calculate the molar mass using Avogadro's constant:**
   \[
   \text{Molar mass} = \text{Mass of one atom} \times N_a = 6.66 \times 10^{-23} \text{ g} \times 8.3848 \times 10^{23} \approx 55.87 \text{ g/mol}
   \]

The molar mass of the metal is approximately 55.87 g/mol. This value closely matches the atomic mass of iron (Fe), which is 55.85 g/mol. Therefore, the atomic number of the metal found by the student is:

**Atomic number of the metal = 26 (Iron, Fe)**

Would you like more details on any part of the calculation or have any questions?

### Relative Questions:
1. What is the significance of the FCC lattice structure in determining the properties of metals?
2. How is the edge length of an FCC unit cell related to the atomic radius?
3. What is the importance of converting units properly in scientific calculations?
4. How does the density of a substance help in identifying it?
5. What are the steps to convert mass from one unit to another in chemistry?
6. Why is Avogadro's constant crucial in these types of calculations?
7. How can you experimentally determine the density of a metal?
8. What are some common applications of metals with FCC structures?

### Tip:
Always double-check the values used for constants in calculations, as incorrect values can lead to significant errors in the final results.

Learn how to calculate the atomic number of a metal found in an FCC lattice using density and atomic radius. Detailed step-by-step solution includes converting units, calculating unit cell volume, and determining molar mass using Avogadro's constant.

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