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Calculate the density of gold (Au) given its FCC structure, edge length of 408 pm, and molar mass of 197 g/mol.

Chemistry

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Answer

19.27 g/cm3\text{19.27 g/cm}^3

Step-by-step solution

4 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}

  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 atoms/unit cell×197 g/mol(408×10−10 cm)3×6.022×1023 mol−1\rho = \frac{4 \text{ atoms/unit cell} \times 197 \text{ g/mol}}{(408 \times 10^{-10} \text{ cm})^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}} ρ=788 g(4083×10−30 cm3)×6.022×1023\rho = \frac{788 \text{ g}}{(408^3 \times 10^{-30} \text{ cm}^3) \times 6.022 \times 10^{23}} ρ=788 g(67917312×10−30 cm3)×6.022×1023\rho = \frac{788 \text{ g}}{(67917312 \times 10^{-30} \text{ cm}^3) \times 6.022 \times 10^{23}} ρ=788 g6.7917312×107×10−30 cm3×6.022×1023\rho = \frac{788 \text{ g}}{6.7917312 \times 10^7 \times 10^{-30} \text{ cm}^3 \times 6.022 \times 10^{23}} ρ=788 g6.7917312×6.022×107−30+23 cm3\rho = \frac{788 \text{ g}}{6.7917312 \times 6.022 \times 10^{7-30+23} \text{ cm}^3} ρ=788 g40.899×100 cm3\rho = \frac{788 \text{ g}}{40.899 \times 10^0 \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

    The density of gold is approximately 19.27 g/cm3\text{19.27 g/cm}^3.

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