U=238; O=16
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To solve this problem, we need to assume that every uranium atom is oxidized, and that all the oxide molecules have the same formula. We'll also ignore any experimental uncertainty in the measured masses and atomic masses.
How many grams of oxygen are added to the original sample? We just subtract the masses.
Omass = UOmass - Umass
Omass = 2.806 g - 2.38 g
Omass = 0.426 g
You give the atomic masses of uranium and oxygen. Those numbers tell you the number of grams that a 1 mole (6.02 x 10^23 atom) sample weighs - the atomic mass of uranium is 238 grams / mole. So, you can figure out how many uranium atoms are in the original sample, and how many oxygen atoms are added.
Umole = Umass / Uatomicmass
Umole = 2.38 g / (238 g/mol)
Umole = 0.01 mol
Omole = Omass / Oatomicmass
Omole = 0.426 g / (16 g/mol)
Omole = 0.026625 mol
So, the ratio of oxygen atoms to uranium atoms in the oxide is O:U = 0.026625 : 0.01. Now, there has to be a whole number of atoms in each oxide molecule, so we need to find a way to express that ratio that uses whole numbers.
0.026625 : 0.01 = 26625 : 10000
0.026625 : 0.01 = 5325 : 2000
0.026625 : 0.01 = 1065 : 400
0.026625 : 0.01 = 213 : 80
That's the simplest form. So, using the numbers you gave, a uranium oxide molecule has 80 atoms of uranium for every 213 atoms of oxygen; the chemical formula is U_n*80 O_n*213 (the integer constant n means that we only know the ratio of atoms, not the exact number, in the oxide).
Now, 293 atoms (at least!) is a pretty big molecule, and that tips me off that there's something wrong with our calculation. Maybe you made a typo - I think you meant to write that the resulting oxide weighed 2.86 g, instead of 2.806 g. If you redo the calculation with that figure, you'll get something much more reasonable - U_n O_n*3.
As it turns out, UO_3 is one of the oxides of uranium. Others are UO_2, U_3O_8, and UO_4. If you have a mix of those oxides, it's reasonable to get 2.806 g. But you can't solve that problem with the information given (at least, not easily). I hope that helps!
How many grams of oxygen are added to the original sample? We just subtract the masses.
Omass = UOmass - Umass
Omass = 2.806 g - 2.38 g
Omass = 0.426 g
You give the atomic masses of uranium and oxygen. Those numbers tell you the number of grams that a 1 mole (6.02 x 10^23 atom) sample weighs - the atomic mass of uranium is 238 grams / mole. So, you can figure out how many uranium atoms are in the original sample, and how many oxygen atoms are added.
Umole = Umass / Uatomicmass
Umole = 2.38 g / (238 g/mol)
Umole = 0.01 mol
Omole = Omass / Oatomicmass
Omole = 0.426 g / (16 g/mol)
Omole = 0.026625 mol
So, the ratio of oxygen atoms to uranium atoms in the oxide is O:U = 0.026625 : 0.01. Now, there has to be a whole number of atoms in each oxide molecule, so we need to find a way to express that ratio that uses whole numbers.
0.026625 : 0.01 = 26625 : 10000
0.026625 : 0.01 = 5325 : 2000
0.026625 : 0.01 = 1065 : 400
0.026625 : 0.01 = 213 : 80
That's the simplest form. So, using the numbers you gave, a uranium oxide molecule has 80 atoms of uranium for every 213 atoms of oxygen; the chemical formula is U_n*80 O_n*213 (the integer constant n means that we only know the ratio of atoms, not the exact number, in the oxide).
Now, 293 atoms (at least!) is a pretty big molecule, and that tips me off that there's something wrong with our calculation. Maybe you made a typo - I think you meant to write that the resulting oxide weighed 2.86 g, instead of 2.806 g. If you redo the calculation with that figure, you'll get something much more reasonable - U_n O_n*3.
As it turns out, UO_3 is one of the oxides of uranium. Others are UO_2, U_3O_8, and UO_4. If you have a mix of those oxides, it's reasonable to get 2.806 g. But you can't solve that problem with the information given (at least, not easily). I hope that helps!
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you are right answer is U3O8...Can you clarify this again?
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