# Answer to Question #22045 in Inorganic Chemistry for Tiarra Hickman

Question #22045
A 12.00-g sample of a compound is analyzed and found to be composed of 3.478 g lithium, 0.505 g of hydrogen, and 0.817 g oxygen. What is the percentage composition of this compound?
1
2013-01-17T10:21:57-0500
Begin by determining the molar mass of each compound involved in the reaction. &
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:
&omega; (Li)= (3.478/12.00)&times;100 = 28.98%;
&omega;(H)= (0.505/12.00)&times;100 = 4.21%;
&omega;(O)= (8.017/12.00)&times;100 = 66.81%.

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