Question

In 1983 the United States began coining the one-cent piece out of copper-clad zinc rather than pure copper. The mass of the old copper penny is 3.083 g and that of the new cent is 2.517 g. The density of copper is 8.920 g/cm3 and that of zinc is 7.133 g/cm3. The new and old coins have the same volume. Calculate the percent of zinc (by volume) in the new cent.

Answer 1

To solve this problem it is necessary to use the concepts related to density as a function of mass and volume. And apply this relationship to know the unknowns. In general, the density can be described as

$\rho = \frac{m}{V}$

Where,

m = mass

V= Volume

Our values are given as,

$m_c = 3.083g$

$\rho_c = 8.96g/cm^3$

$v_c = \frac{m_c}{\rho_c}\rightarrow v_c = \frac{3.83}{8.96}=0.4274cm^3$

The mass of the mixture between copper and zinc is

$m_{zc} = 2.617g$

Let mass of copper in the new coin be x and that of zinc be $(m_{zc} -x)$

$\rho_z = 7.133g/cm^3$

Thus,

\frac{m_{zc} -x}{\rho_z}+\frac{x}{\rho_c}=v_c

Since volume of new and old coin is the same:

$(\frac{2.517-x}{7.133})+(\frac{x}{8.96})=(\frac{3.083}{8.96})$

$2.517-x+0.796=2.454$

$x = 0.3088g$

Thus, mass of zinc is

$m_{zc} -x = 2.517-0.3088$

$m_{zc} -x = 2.2082g$

Thus, percentage of zinc by volume

$\% = \frac{m_{zc}-x}{\rho_{zc}}\frac{1}{V}*100$

$\% = (\frac{2.2082g}{7.133})(\frac{1}{3.083/8.96})(100)$

$\% = 89.97\%$

Therefore the percent of zinc in the new cent 89.97%