Answer to Question #22045 in Inorganic Chemistry for Tiarra Hickman
Using atomic masses from the periodic table, we will find the quantity of moles of each element in the compound:
n(Li)= 3.478/6.939=0.501 mol;
n(H)= 0.505/1.008=0.501 mol;
n(O)= 8.017 /15.999=0.501 mol.
It is lithium hydroxide (LiOH).
The molar percentage each element of this compound is 33.3%.
The mass percentage each element of this compound is the following:
ω (Li)= (3.478/12.00)×100 = 28.98%;
ω(H)= (0.505/12.00)×100 = 4.21%;
ω(O)= (8.017/12.00)×100 = 66.81%.
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