Solid State: Speed-Solving Techniques (6)

hard • 2 min read
Tags Solid State

Question

A metal crystallises in a face-centred cubic (FCC) lattice with edge length a=400a = 400 pm. If the density is 10.5 g/cm³, find the atomic mass. (NA=6.022×1023N_A = 6.022 \times 10^{23}.)

Solution — Step by Step

For a unit cell:

ρ=Z⋅Ma3⋅NA\rho = \frac{Z \cdot M}{a^3 \cdot N_A}

where ZZ is the number of atoms per unit cell, MM is molar mass, aa is edge length.

For FCC, Z=4Z = 4.

a=400a = 400 pm =400×10−10= 400 \times 10^{-10} cm =4×10−8= 4 \times 10^{-8} cm.

a3=(4×10−8)3=64×10−24a^3 = (4 \times 10^{-8})^3 = 64 \times 10^{-24} cm³ =6.4×10−23= 6.4 \times 10^{-23} cm³.

M=ρ⋅a3⋅NAZM = \frac{\rho \cdot a^3 \cdot N_A}{Z}

M=10.5×6.4×10−23×6.022×10234M = \frac{10.5 \times 6.4 \times 10^{-23} \times 6.022 \times 10^{23}}{4}

M=10.5×6.4×6.0224=404.74≈101.2 g/molM = \frac{10.5 \times 6.4 \times 6.022}{4} = \frac{404.7}{4} \approx 101.2\text{ g/mol}

Final answer: M≈101M \approx 101 g/mol (likely ruthenium or close to silver, depending on source)

Why This Works

The unit cell is the smallest repeating unit of the crystal. Knowing how many atoms it contains (ZZ) and its volume (a3a^3), we can compute density from molar mass — or run the equation backward to find MM from density.

ZZ values to memorise:

  • Simple cubic: Z=1Z = 1
  • BCC: Z=2Z = 2
  • FCC (or CCP): Z=4Z = 4
  • HCP: Z=6Z = 6

Alternative Method

Compute the volume occupied by one atom: Vatom=a3/ZV_{atom} = a^3/Z. Then density =M/(Vatom⋅NA)= M/(V_{atom} \cdot N_A), rearrange for MM. Same answer; some students prefer this framing.

Common Mistake

Forgetting the unit conversion from pm to cm. 1 pm=10−101\text{ pm} = 10^{-10} cm (not 10−1210^{-12} cm — that would be metres). Mixing up gives a factor of 10610^6 error in density.

Also: students sometimes use Z=6Z = 6 for FCC (confusing it with HCP). FCC has 4 atoms per unit cell: 8 corners × 1/8 + 6 faces × 1/2 = 1 + 3 = 4.

JEE Main 2023 had a density-from-FCC question with edge length and atomic mass given. The same template, just rearranged. Always be ready to solve for whichever variable is unknown.

Want to master this topic?

Read the complete guide with more examples and exam tips.

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