Math Problem Statement

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 ?

Solution

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 aa of the unit cell is related to the atomic radius rr by the equation: a=4r2a = \frac{4r}{\sqrt{2}}
  2. Convert the radius from nanometers to centimeters: r=0.197 nm=0.197×107 cmr = 0.197 \text{ nm} = 0.197 \times 10^{-7} \text{ cm}

  3. Calculate the edge length aa: a=4×0.197×107 cm2=0.788×107 cm1.4140.557×107 cma = \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 VV: V=a3=(0.557×107 cm)31.729×1022 cm3V = 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 ρ\rho is given by: ρ=mass of unit cellvolume of unit cell\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: Mass of unit cell=ρ×V=1.54 g/cm3×1.729×1022 cm32.664×1022 g\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: Mass of one atom=Mass of unit cell4=2.664×1022 g46.66×1023 g\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: Molar mass=Mass of one atom×Na=6.66×1023 g×8.3848×102355.87 g/mol\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.

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Math Problem Analysis

Mathematical Concepts

Crystallography
Density Calculations
Unit Cell Volume

Formulas

Density formula
Volume of FCC unit cell

Theorems

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Suitable Grade Level

Advanced High School