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Calculate the density of gold, given its FCC unit cell edge length is 408 pm, molar mass is 197 g/mol, and Avogadro's number is 6.022 imes 1023 ext mol⁻¹.

Chemistry

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Step-by-step solution

5 steps

  1. Identify the given values and constants.

    • Type of unit cell: Face-centred cubic (FCC). For an FCC structure, the number of atoms per unit cell (ZZ) is 4.
    • Edge length (aa) = 408 pm.
    • Molar mass (MM) of gold = 197 g/mol.
    • Avogadro's number (NAN_A) = 6.022×1023 mol−16.022 \times 10^{23} \text{ mol}^{-1}.

  2. Convert the edge length from picometers (pm) to centimeters (cm).

    a=408 pm×10−12 m1 pm×100 cm1 ma = 408 \text{ pm} \times \frac{10^{-12} \text{ m}}{1 \text{ pm}} \times \frac{100 \text{ cm}}{1 \text{ m}} a=408×10−10 cma = 408 \times 10^{-10} \text{ cm} a=4.08×10−8 cma = 4.08 \times 10^{-8} \text{ cm}

  3. Use the formula for the density (ρ\rho) of a unit cell.

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

  4. Substitute the values into the formula and calculate the density.

    ρ=4×197 g/mol(4.08×10−8 cm)3×6.022×1023 mol−1\rho = \frac{4 \times 197 \text{ g/mol}}{(4.08 \times 10^{-8} \text{ cm})^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}} ρ=788 g(4.083×(10−8)3 cm3)×6.022×1023\rho = \frac{788 \text{ g}}{(4.08^3 \times (10^{-8})^3 \text{ cm}^3) \times 6.022 \times 10^{23}} ρ=788 g(67.917312×10−24 cm3)×6.022×1023\rho = \frac{788 \text{ g}}{(67.917312 \times 10^{-24} \text{ cm}^3) \times 6.022 \times 10^{23}} ρ=788 g(67.917312×6.022)×10−24+23 cm3\rho = \frac{788 \text{ g}}{(67.917312 \times 6.022) \times 10^{-24+23} \text{ cm}^3} ρ=788 g408.990×10−1 cm3\rho = \frac{788 \text{ g}}{408.990 \times 10^{-1} \text{ cm}^3} ρ=788 g40.899 cm3\rho = \frac{788 \text{ g}}{40.899 \text{ cm}^3} ρ≈19.265 g/cm3\rho \approx 19.265 \text{ g/cm}^3

  5. Round the density to two decimal places and convert to kg/m3\text{kg/m}^3.

    ρ≈19.27 g/cm3\rho \approx 19.27 \text{ g/cm}^3

    To convert to kg/m3\text{kg/m}^3:

    1 g/cm3=1×10−3 kg(10−2 m)3=1×10−3 kg10−6 m3=103 kg/m31 \text{ g/cm}^3 = 1 \times \frac{10^{-3} \text{ kg}}{(10^{-2} \text{ m})^3} = 1 \times \frac{10^{-3} \text{ kg}}{10^{-6} \text{ m}^3} = 10^3 \text{ kg/m}^3 ρ=19.27×103 kg/m3\rho = 19.27 \times 10^3 \text{ kg/m}^3

    The density of gold is approximately 19.27 g/cm3 or 19.27×103 kg/m3\boxed{\text{19.27 g/cm}^3 \text{ or } 19.27 \times 10^3 \text{ kg/m}^3}.

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