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Homework Help: Easy percent composition question

  1. Sep 22, 2008 #1
    1. For a problem I require the atomic density of uranium atoms per cm^3 in a solution of water. The atomic fraction of natural uranium to water is 1 per 100 and the net density of the solution is 1 g/cm^3. (Natural uranium is assumed to be U-235 at 0.72% of uranium atoms and the rest is U-238) (assume the atomic weights of U-235, U-238 and water are 235, 238 and 18 respectively)

    What I require is the atomic density of U-235 atoms, the atomic density of U 238, and H2O atoms all in atoms/cm^3.

    I have tried to calculate the overall atomic weight of the solution weighted by the atom concentration as in:

    0.000072(235) + 0.009928(238) + 0.99(18) = 20.35 g/mol?

    then since the solution has net density 1 gm/cm^3:

    1 / 20.35 = 0.049 mol/cm^3

    then using avogadro's number convert to atoms/cm^3, then use the 1 in 100 again to find how many are U235, U238, water....is this a correct method. something seems wrong in this.

    thanks for the help
  2. jcsd
  3. Sep 22, 2008 #2


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    A simpler method is to find the mass percentages from the atomic or mole percentages that are given in the problem , then simply multiply this fraction to the solution density to find the density with respect to each substance. Then use Avogadros number to convert to the atomic density.
  4. Sep 22, 2008 #3
    how would i calculate the mass percentage from the atomic percentage?
  5. Sep 23, 2008 #4


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    Assume 1 g of mixture

    MW U 235 X moles U 235 + MW U 238 X moles U 238 + MW H2O X moles H2O = 1 g

    Assume x is the moles of U 235

    (x) X 9928 / 72 = moles of U 238

    (x) X 990000 / 72 = moles of H2O

    Solve for x

    Then simply use Avogadros number to convert to atoms and since you found x you are able to find the moles of U 238 and H2O.
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