Question

Gold (19.3 g/cm3 ) and copper (8.96 g/cm3 ) can be blended to form an alloy called rose gold. Suppose a rose gold bar has a mass of 117 g and a volume of 7.00 cm3 . Calculate the mass percent of gold in the bar.

Answer 1

Answer on Question #80689, Chemistry / General Chemistry

Gold (19.3 g/cm³) and copper (8.96 g/cm³) can be blended to form an alloy called rose gold. Suppose a rose gold bar has a mass of 117 g and a volume of 7.00 cm³. Calculate the mass percent of gold in the bar.

Solution

Let mass of gold to be $x$ g, mass of copper to me y g.

Then $x + y = 117$

Volume of gold is $V = m/d = x/19.3$

Volume of copper is $V = m/d = y/8.96$, then $V(Au) + V(Cu) = V(\text{alloy})$, i.e., $x/19.3 + y/8.96 = 7$

We have the system of two equations:

$$\left\{ \begin{array}{l} x + y = 117 \\ \frac{x}{19.3} + \frac{y}{8.96} = 7 \end{array} \right.$$ $$\left\{ \begin{array}{l} x = 117 - y \\ \frac{(117 - y)}{19.3} + \frac{y}{8.96} = 7 \end{array} \right.$$ $$\left\{ \begin{array}{l} x = 117 - y \\ 8.96 \times (117 - y) + 19.3 \times y = 7 \times 19.3 \times 8.96 \end{array} \right.$$ $$\left\{ \begin{array}{l} x = 117 - y \\ 1048.32 - 8.96y + 19.3y = 1210.50 \end{array} \right.$$ $$\left\{ \begin{array}{l} x = 117 - y \\ 10.34y = 162.18 \end{array} \right.$$ $$\left\{ \begin{array}{l} x = 117 - 15.68 \\ y = 15.68 \end{array} \right.$$ $$\left\{ \begin{array}{l} x = 101.32 \\ y = 15.68 \end{array} \right.$$

So, $m(Au) = 101.32$ g, $m(Cu) = 15.68$ g

$$w(Au) = \frac{m(Au)}{m_{alloy}} \times 100\% = \frac{101.32}{117} \times 100\% = 86.6\%$$

Answer: 86.6%