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?
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: 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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