Avogadro's Number — Question 7

Question 7

A crystal connects microscopic spacing with macroscopic density

Use $N_A=6.02214076\times10^{23}\,\mathrm{mol^{-1}}$, exact. A hypothetical cubic crystal has four atoms per unit cell, cube edge $a=.400$ nm and assigned molar mass $M=.0400$ kg/mol. Use the supplied unit-cell count; neglect vacancies and surface effects.

1. Derive and calculate the crystal density.

2. A measured density is 2.00% below that ideal value while edge length is unchanged. If vacancies are the only cause, find the occupied-site fraction and mean atoms per cell.

Question 7 - Solution

A crystal connects microscopic spacing with macroscopic density

(a) Method and calculation

$$m_{\mathrm{cell}}=4M/N_A,\quad V_{\mathrm{cell}}=a^3=(.400\times10^{-9})^3=6.40\times10^{-29}\,\mathrm{m^3}.$$$$\rho=\frac{4M}{N_Aa^3}\approx4151.35\,\mathrm{kg/m^3}.$$

(b) Interpretation and further calculation

$$\rho_{\mathrm{measured}}=f_{\mathrm{occupied}}\rho_{\mathrm{ideal}}\ \Rightarrow\ f_{\mathrm{occupied}}=.980.$$$$\overline N_{\mathrm{cell}}=4(.980)=3.92.$$A noninteger mean count describes an ensemble of cells with integer occupancies, not a fractional atom in an individual cell.

Physical check

Density scales as $a^{-3}$; the vacancy inference depends on the explicitly unchanged lattice spacing.