A quartz nugget containing gold has a mass m = 51 g and an…

A quartz nugget containing gold has a mass $m = 51\ \mathrm{g}$ and an average density $\rho = 8500\ \mathrm{kg/m^3}$. How much gold does it contain? (Densities: quartz $\rho_1 = 2650\ \mathrm{kg/m^3}$, gold $\rho_2 = 19\,300\ \mathrm{kg/m^3}$.)

Explanation of Method

How much gold is in a quartz nugget.

Step 1: Volume of the nugget

\[ V = \frac{m}{\rho} = \frac{0.051}{8500} = 6.0 \cdot 10^{-6}\ \mathrm{m^3} = 6.0\ \mathrm{cm^3}. \]

Step 2: Two conditions

With $m_1$ the quartz and $m_2$ the gold, both the masses and the volumes add up: $m_1 + m_2 = m$.

\[ \frac{m_1}{\rho_1} + \frac{m_2}{\rho_2} = V. \]

Step 3: Solve for the gold

Substituting $m_1 = m - m_2$ and working in grams and cm³ ($\rho_1 = 2.65$, $\rho_2 = 19.3\ \mathrm{g/cm^3}$): $\frac{51 - m_2}{2.65} + \frac{m_2}{19.3} = 6.0$.

\[ m_2\left(\frac{1}{2.65} - \frac{1}{19.3}\right) = \frac{51}{2.65} - 6.0 \]

\[ m_2 = \frac{19.245 - 6.0}{0.37736 - 0.05181} = \frac{13.245}{0.32555} \]

\[m_2 \approx 40.7\ \mathrm{g} \ \text{of gold}\]

In general: $m_2 = \dfrac{\rho_2\,(m - \rho_1 V)}{\rho_2 - \rho_1}$.

Check

Quartz: $ 51 - 40.7 = 10.3\ \mathrm{g} $, volume $ 10.3/2.65 = 3.89\ \mathrm{cm^3} $

Gold volume $ 40.7/19.3 = 2.11\ \mathrm{cm^3} $; total $ 6.00\ \mathrm{cm^3}\ \checkmark $

$3.89\ \mathrm{cm^3}$ $2.11\ \mathrm{cm^3}$ $\text{volume}$ $V = 6.00\ \mathrm{cm^3}$ $10.3\ \mathrm{g}$ $40.7\ \mathrm{g}$ $\text{mass}$ $m = 51\ \mathrm{g}$

quartz (2650 kg/m³)
gold (19 300 kg/m³)

Bars to scale. Gold takes up only about a third of the volume but carries four fifths of the mass, because it is 7.3 times denser than quartz.

Result

\[\boxed{\,m_2 \approx 40.7\ \mathrm{g} \ \text{of gold}\,}\]

Remark

Note. The original result "≈ 40 g" is correct; the more precise value is 40.7 g. (Its intermediate numbers, such as 116 and 7.28, came from multiplying the volume condition by 19 300 with grams and cubic metres mixed; working in g and cm³ avoids that.)