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Easy percent composition question

by thenewbosco
Tags: composition, percent
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thenewbosco
#1
Sep22-08, 09:41 PM
P: 187
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
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GCT
#2
Sep22-08, 10:34 PM
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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.
thenewbosco
#3
Sep22-08, 10:48 PM
P: 187
how would i calculate the mass percentage from the atomic percentage?

GCT
#4
Sep23-08, 10:34 AM
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Easy percent composition question

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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