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How might one determine whether Ni is more dense than Cu using only the atomic masses of Ni and Cu, and the fact that Ni has a simple cubic unit cell, and Cu has a face-centered unit cell. You are not given the radii of each element. I feel like this question cannot be answered with the given information only, considering as Ni and Cu have very close densities (0.06 g/cm^3 difference).

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  • $\begingroup$ A quick look through the web suggests that both Ni and Cu crystallize in a ccp structure. The space group is the same: Fm-3m. Considering trends across the periodic table, you should have an answer now. $\endgroup$ – Abel Friedman Nov 23 '14 at 19:44
  • $\begingroup$ This was a question given on a test. You are supposed to use only the information given. You cannot use density trends in the periodic table. Also the problems states that Ni has simple cubic, Cu has face-centered. You must use these assumptions. This question is hypothetical. $\endgroup$ – Joshua Benabou Nov 23 '14 at 19:54
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    $\begingroup$ If what you say is true your teacher should be fired for gross incompetence. Instructions for an orgo class I was involved in were that if you saw a reaction or reagents that weren't in the answer key you had to give the student the opportunity to give a literature reference in their support. Chemistry is the real world, which is much larger than what is taught in a particular class. Your teacher seems intent on killing student independence and initiative. $\endgroup$ – Abel Friedman Nov 23 '14 at 20:00
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    $\begingroup$ This is the problem with high school - its not about learning, its about getting the answers correct. I encounter scenarios like this all the time. Of course, to answer the question with the given information only is a bit weird (how are we supposed to predict a difference if the densities are so close). Now, if we are given that Ni and Cu have ccp structure, the answer is obvious by examining relative radii of the two elements. Nonetheless, I believe the question is not totally invalid. Suppose ithe assumption about the unit cell structures was true, can the question be answered? $\endgroup$ – Joshua Benabou Nov 23 '14 at 20:10
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    $\begingroup$ In a sc packing there is much more empty space than in a fcc packing. That's why of the metals only polonium rystallizes in an sc structure, and it's a semi-metal. For a real world-example, compare the densities of lead (fcc) and polonium (sc). $\endgroup$ – Abel Friedman Nov 24 '14 at 2:47
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I think you need to make at least one more assumption along the way to determining the relative densities. But let’s take it stepwise:

  1. We are given a primitive cubic cell for nickel and a face-centred cube for copper.

  2. We therefore conclude that per unit cell one atom of nickel and three atoms of copper are present.

  3. We know from the periodic table that copper is a bit heavier than nickel ($63.5$ versus $58.7$)

  4. Therefore, it is clear that copper’s unit cell is more than four times as heavy as nickel’s. But we don’t know their sizes.

  5. From mathematical considerations we can also calculate how much of each unit cell is filled with atoms ($74~\%$ for fcc; somewhat less which I am too lazy to calculate now for primitive cubic) — this again means that there should be more weight per volume in the copper unit cell.

  6. What seems to be lacking in my eyes is a way to compare the unit cell sizes, whether by atom size or unit cell dimensions. However, since nickel and copper are right next to each other in the periodic table, it may be safe to assume their sizes to be almost identical.

  7. If that assumption is safe, we can use either route to clearly show how copper’s density is higher. But note that if copper’s atoms were significantly larger than nickel’s, this reasoning would break down extremely quickly.

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The density of a crystal is given by the equation

$\rho=\frac{Z\times M}{a^3\times N_0 \times 10^{-30}}~\mathrm{ g/cm^3}$, where $Z$ denotes the effective number of metal atoms per unit cell.

This means

$\frac{\rho_\ce{Cu}}{\rho_\ce{Ni}}=\frac{Z_\ce{Cu}\times M_\ce{Cu} \times {a_\ce{Ni}}^3}{Z_\ce{Ni}\times M_\ce{Ni}\times {a_\ce{Cu}}^3}$.

Just substitute the values to get the answer.

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