Answer
a)$7.010\times10^{-4}\ mol$
$U_2O_5$
Uranium (V) oxide
$3.58\times10^{-4}\ mol$
b)238
c) 6
Work Step by Step
a) Number of moles of uranium:
$0.169\ g\div238.02891\ g/mol=7.010\times10^{-4}\ mol$
Number of moles of oxygen:
$(0.199-0.169)\div15.9994=1.876\times10^{-3}\ mol$
Dividing the oxygen value by uranium's:
$1.876\times10^{-3}/7.010\times10^{-4}=2.67\approx5/2$
The empirical formula is then : $U_2O_5$ which is called Uranium (V) oxide whose molecular weight is $556.06\ g/mol$
Number of moles obtained:
$0.199\ g/mol\div556.06\ g/mol=3.58\times10^{-4}\ mol$
b) The isotope with mass number of 238 because the atomic weight is closest to this mass number.
c) Molecular weight of: $UO_2(NO_3)_2$:$394.04\ g/mol$
$H_2O$: $18.015\ g/mol$
The number of moles must have been the same after the dehydration:
$0.679\ g\div394.04\ g/mol= 1.72\times10^{-3}\ mol$
Before:
$0.865\ g\div1.72\times10^{-3}\ mol=501.98\ g/mol$
$501.98=394.04+z\times18.015$
$z=6$
