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Calculating Bond Length

  1. Jul 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that the bond length for sodium chloride is 0.28nm given that it has a relative density of 2.2, sodium has an atomic weight of 23, and chlorine 35.5. Avogadro's number is 6.0x10^23

    2. Relevant equations

    3. The attempt at a solution

    I've tried many things but here's one:
    molar mass of NaCl = 23 + 35.5 = 58.5g. Relative density = 2.2g/cm^3.
    Therefore 58.5g= 26.59cm^3 of NaCl. One side = cube root of 26.59 = 2.98cm = 42171633 = 7.066x 10^-8cm per molecule. So divide through by 2 again to find bond length = 0.35nm. which isn't 0.28nm.
  2. jcsd
  3. Jul 20, 2009 #2


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    Staff: Mentor

    Nice try, but that's incorrect.

    Imagine smallest possible cube with Na and Cl ions on the vertices. How many NaCl molecules INSIDE of this cube? What volume must it have containing this number of molecules?
  4. Jul 20, 2009 #3
    Smallest cube lattice is a 3x3 interlocking Face centred cubic structures. Having 27 atoms, i.e. 13.5 NaCl molecules.
    Since there are 6x10^23 molecules in one mol.
    If we divide this number by 13.5. We get the fractional volume of our 26.58cm^3 mol of NaCl.

    6x10^23/13.5 = 4.44x10^22

    Divide 26.58 by this to get 5.98x10^-22 cm^3. i.e. the volume of our smallest cube lattice. One edge of our cube lattice = 2 atomic bond distances since it goes sodium/chlorine/sodium. So if we cube root our volume we should get the length of one side.

    this gives: 8.42 x10^-8cm. Divide by 2 giving 4.21x10^-8cm = 4.21x10^-10 m = 0.42 nm.
    Still not getting the right answer....help me :(
  5. Jul 21, 2009 #4

    What I think Borek meant was considering the smallest possible enclosed Sodium Chloride structure possible. Mainly a simple cube with either a Sodium or Chlorine atom at each vertex. Such a structure will have eight atoms in total, but more importantly four NaCl molecules.

    Then you have to determine what fraction of the NaCl molecule is actually INSIDE the enclosed cube. This is where diagram may help in case you have difficulty visualizing this. Eventually you should realize that only one eighth of each atom is really inside the cube. This of course means as there are four molecules in the structure, the total number of NaCl molecules INSIDE the structure is 4 x 1/8 = 1/2

    Fortunately, this initial step was as challenging as it will get. From here you can calculate the number of NaCl moles this structure contains (simply a fraction of the number of molecules over Avogadro's constant). Undoubtedly this will be a tiny number.

    Since you know the molar mass of the particle as well, you can calculate the mass of the structure. (Moles= mass/molar mass). Using the density, you can calculate the volume of the cube. Cube root that number and you should get the length of one side of the cube, which in this instance is the bond length! Be wary of units of course as it looks like you are working in cm.

    The key really was to consider the molecules INSIDE the structure, and not as a whole. All credit to Borek for reminding me of that fact.
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