ByMpRoduct
With how many water molecules does uranium dioxide (UO2) crystallize when it loses 34.8% of its mass upon heating? M(U) = 238 g/mol
I have no problem calculating if the masses are already given. Unfortunately, I don't know how to do it with the percentage.
I would be very grateful if someone could calculate it for me.
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You know that the total mass of the crystallized uranium dioxide is 100%. If the water evaporates, it remains 100% – 34.8% =x%.
These x% of the total mass have the molar mass of uranium dioxide.
This can now be calculated with a three-set or a ratio calculation, which molar mass of water is 34.8% by weight.
And this should be a multiple of the molar mass of water molecules.
Would 28 water molecules possibly come out as a result?
Not with me;)
28 water molecules per uranium dioxide molecule would mean that you have 504 g/mol of water on 270 g/mol of uranium dioxide. While it can be clearly seen from the task that the water is about half as large as that of the uranium dioxide. So it can’t be right.
What’s your way of calculating? I have such a guess.
Do I have to multiply the 28 by 0.348 again so I have 10 water molecules?
I don’t understand how you think. If the 270 g/mol are only 65.2%, the 100% rough the heel must be more…
Of course (270 g/mol * 100/65.2) I come to 414 g/mol. Of which 34.8% are water, i.e. 144 g/mol. Water has 18 g/mol, so it is 8 mol of water to 1 mol of UO2.
All right. Thank you for your help. Of course, the 100% are the 270 g/mol MIT the water. Thus we have 288 g/mol and an unknown. I then calculated these unknowns with mathematical aids and then came to 8 water molecules. Is that right?
If 34,8 % water is released, the 100% yes UO2 and Be water. The UO2 with 270 g/mol then corresponds to 65.2%.
I thought if 270 g/mol are 100%, then 65.2% are 176.04 g/mol. That’s what I did with the water. Then I divided the uranium through the water and then came to about 28.