To complete homework I am working on, I need help finding the density of a penny. I'm not given the mass or the volume. All I know about this penny is it's zinc coated in copper, and I am given the percentage of copper in the penny: 3.0% (by mass). We are to use these values for the density of copper and zinc:

Copper = 8.93, Zinc = 7.14

My friends did this:

$$\mathrm{8.98 \times 0.03 + 7.14 \times 0.97 = average~density~of~a~penny}$$

I told them "I think you're wrong, you can only calculate the average density of something if you use the volume", but I couldn't explain it any better.

Here's what I did:

Let $M$ be the mass of this penny in grams.

$$\frac{0.03M~\mathrm{g~mL^{-1}}}{8.98~\mathrm{g}} = \frac{0.03}{8.98}M~\mathrm{ml}$$

$$\frac{0.97M~\mathrm{g~mL^{-1}}}{7.14~\mathrm{g}} = \frac{0.973}{7.14}M~\mathrm{ml}$$

Adding them together we get $7.1952M$.

I'm not sure what this number represents but we can use it to find the ratios of volume.

$$\frac{\frac{0.03}{8.98 M}}{7.1952 M} = 0.0374$$ $$\frac{\frac{0.97}{7.14 M}}{7.1952 M} = 0.963$$

Using these numbers, we do what my friends did:

$$8.98 \times 0.0374 + 7.14 \times 0.963 = \mathrm{average~density~of~a~penny}$$

Am I correct?

If I am, is there a formula or any other simpler way to do this.

This is my first question on this site, so let me know if I did anything wrong.

  • $\begingroup$ For a new penny, the mass is 2.5 g. The composition is slightly different, but acceptable. en.wikipedia.org/wiki/Penny_(United_States_coin) $\endgroup$
    – LDC3
    Commented Aug 30, 2014 at 15:45
  • $\begingroup$ The volume of a solution (or alloy) is not quite additive either. $\endgroup$ Commented Nov 2, 2015 at 7:36

1 Answer 1


You have the right idea, but somewhere your math is incorrect.

Let the mass of the penny be $M$.

The mass of the copper is $\pu{0.03M}$ and the mass of the zinc is $\pu{0.97M}$.

The volume of copper is $\pu{0.03M}/(\pu{8.93g/cm^3})$ and the volume of the zinc is $\pu{0.97M}/(\pu{7.14g/cm^3})$.

The total volume is

$$\pu{0.03M}/(\pu{8.93g/cm^3}) + \pu{0.97M}/(\pu{7.14g/cm^3})$$

The density of the penny is

$$\frac {M}{\pu{0.03M}/(\pu{8.93g/cm^3}) + \pu{0.97M}/(\pu{7.14g/cm^3})}$$

Factoring out M gives

$$\frac {1}{0.03/(\pu{8.93g/cm^3}) + 0.97/(\pu{7.14g/cm^3})}$$

So the density of a mixture works out to be:

$$D=\frac {1}{(\%_1/D_1)+(\%_2/D_2)}$$


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