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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.

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  • $\begingroup$ 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 Aug 30 '14 at 15:45
  • $\begingroup$ The volume of a solution (or alloy) is not quite additive either. $\endgroup$ – Ivan Neretin Nov 2 '15 at 7:36
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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 0.03M and the mass of the zinc is 0.97M.
The volume of copper is $0.03M/(8.93g/cm^3)$ and the volume of the zinc is $0.97M/(7.14g/cm^3)$.
The total volume is $0.03M/(8.93g/cm^3) + 0.97M/(7.14g/cm^3)$.
The density of the penny is $\frac {M}{0.03M/(8.93g/cm^3) + 0.97M/(7.14g/cm^3)}$.
Factoring out M, gives $\frac {1}{0.03/(8.93g/cm^3) + 0.97/(7.14g/cm^3)}$.

Added:
So the density of a mixture works out to be: $$D=\frac {1}{(\%_1/D_1)+(\%_2/D_2)}$$

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