The atomic fraction (d) of tin in bronze (fcc) with a density of 7717 kgm^-3 and a lattice parameter of3.903 $\stackrel{\mathrm{o}}{\mathrm{A}}$ is (A.t wt. Cu = 63.54, Sn = 118.7, 1 amu = 1.66x 10^-27kg)

$\mathrm{d}=\frac{\left[\begin{array}{l}\sum \left(\text{ Number of atom of each kind }\right)\times \\ \left({\mathrm{M}}_{\mathrm{w}}\text{ of each kind }\right)\times 1.66\times {10}^{-27}\mathrm{kg}\end{array}\right]}{{\mathrm{a}}^3}$
$7717{\mathrm{kgm}}^{-3}=\frac{\left[\begin{array}{l}\left(\text{ Number of Sn atoms }\right)\times \left(118.7\times 1.66\times {10}^{-27}\right)+\\ \left(\text{ Number of Cu atoms }\right)\times \left(63.54\times 1.66\times {10}^{-27}\right)\end{array}\right]}{{\left(3.903\times {10}^{-10}\right)}^3{\mathrm{m}}^3}$
$\begin{array}{l}276.4={\mathrm{n}}_{\mathrm{Sn}}\left(118.7\right)+{\mathrm{n}}_{\mathrm{Cu}}\left(63.54\right)\\ 4.35=1.86{\mathrm{n}}_{\mathrm{Sn}}+{\mathrm{n}}_{\mathrm{Cu}}\\ {\mathrm{n}}_{\mathrm{Cu}}=4\Rightarrow {\mathrm{n}}_{\mathrm{Sn}}=0.188\end{array}$
$\mathrm{Atomic} \mathrm{fraction} = \frac{{\mathrm{n}}_{\mathrm{Sn}}}{{\mathrm{n}}_{\mathrm{Sn}}+{\mathrm{n}}_{\mathrm{Cu}}}=0.05$