# Find the number of copper spheres in a box

A box contains a mixture of small copper spheres and small lead spheres. The total volume of both metals is measured by the displacement of water to be $\pu{420 cm^3}$ and the total mass is $\pu{5.4 kg}$. What percentage of the spheres are copper?

I can't get the right answer. I did this:

Given:

$m_{\text{total}} = 5.4\;\mathrm{kg} =5400\;\mathrm{g}$

$V=420\;\mathrm{cm^3}$

$\rho_{\ce{Cu}} = 8.96\;\mathrm{g} ~\mathrm{cm^-3}$

$V\times{\mathrm{\rho_{\ce{Cu}}}} = 3763.2\;\mathrm{g}$

$\frac{m_{\ce{Cu}}}{m_{\text{total}}}=69.9\%$

Can someone tell me how to approach this problem?

• umm... the problem is underdefined. Are the spheres the same volume? The same mass? If it's the latter, you don't need the densities. It's possible that the box contains just two "small" spheres of the right size. – John Dvorak Sep 14 '13 at 14:57
• hint: $v_c + v_l = 420 cm^3; v_c*\rho_c + v_l*\rho_l = 5.4 kg$ - two equations, two unknowns, two table values. – John Dvorak Sep 14 '13 at 15:02

$m_{\ce{Cu}} + m_{\ce{Pb}} = m_{\text{tot}}$
$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}V_{\ce{Pb}} = m_{\text{tot}}$
$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}(V_{\text{tot}}-V_{\ce{Cu}}) = m_{\text{tot}}$
$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}(420-V_{\ce{Cu}}) = 5400$
$V_{\ce{Cu}}(\rho_{\ce{Cu}}-\rho_{\ce{Pb}}) + 420\rho_{\ce{Pb}}= 5400$
Put values for $\rho_{\mathrm{Cu}}$ and $\rho_{\mathrm{Pb}}$ in the equation, and rest is simple arithmetic.