Question
Uranium is used as a fuel, primarily in the form of uranium(IV) oxide, in nuclear power plants. A small
sample of uranium metal (0.169 g) is heated to between 800 and 900 0C in air to give 0.199 g of a dark
green oxide, UxOy. (a) How many moles of uranium metal were used? (b) What is the empirical formula of
the oxide, UxOy?
sample of uranium metal (0.169 g) is heated to between 800 and 900 0C in air to give 0.199 g of a dark
green oxide, UxOy. (a) How many moles of uranium metal were used? (b) What is the empirical formula of
the oxide, UxOy?
Answers
mols U metal = grams/atomic mass = 0.169/238 = 0.00071
U + O2 ==> UxOy
mass O is UxOy = mass UxOy - mass U = 0.199 - 0.169 = 0.030 g
mols U metal from above = 0.000710
mol O = 0.030/16 = 0.00187
Now you want to find the mole ratio. The easy way to do this is to divide both numbers of the smaller number; i.e.,
mols U = 0.000710/0.000710 = 1.000
mols O = 0.00187/0.000710 = 2.63
Compounds don't have ratios that are not whole numbers so multiply by whole numbers until we find BOTH whole numbers. Multiply by 2 gives
U = 1.000 x 2 = 2.000
O = 2.634 x 2 = 5.27. Doesn't work so try 3.
U = 1.000 x 3 = 3.000
O = 2.634 x 3 = 7.902 which is close enough to round to 8 so the empirical formula is U3O8
U + O2 ==> UxOy
mass O is UxOy = mass UxOy - mass U = 0.199 - 0.169 = 0.030 g
mols U metal from above = 0.000710
mol O = 0.030/16 = 0.00187
Now you want to find the mole ratio. The easy way to do this is to divide both numbers of the smaller number; i.e.,
mols U = 0.000710/0.000710 = 1.000
mols O = 0.00187/0.000710 = 2.63
Compounds don't have ratios that are not whole numbers so multiply by whole numbers until we find BOTH whole numbers. Multiply by 2 gives
U = 1.000 x 2 = 2.000
O = 2.634 x 2 = 5.27. Doesn't work so try 3.
U = 1.000 x 3 = 3.000
O = 2.634 x 3 = 7.902 which is close enough to round to 8 so the empirical formula is U3O8
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