What’s the empirical formula of a molecule containing 18.7% lithium, 16.3% carbon, and 65.0% oxygen?
Answers
Hi there,
☆ find no of mols of each element by divid over atomic mass
● n (Li) = 18.7 ÷ 6.94 = 2.69
● n (C) = 16.3 ÷ 12 = 1.35
● n (O) = 65.0 ÷ 16 = 4.06
☆ find the mole ratio by dividing over smallest mole value which is 1.35
○ Li = 2
○ C = 1
○ O = 3
♤ so, the empirical formula is
Li2CO3
♡ hope the answer is helpful
♡ Plz mark as brainliest
Answer:
Empirical formula is Li₂CO₃.
Explanation:
Percentage of oxygen= 65.0%
Percentage of lithium = 18.7%
Percentage of carbon= 16.3%
Empirical formula = ?
Solution:
Number of gram atoms of C = 16.3/12 = 1.4
Number of gram atoms of Li = 18.7/6.94 = 2.7
Number of gram atoms of O = 65.0/ 16 = 4.1
Atomic ratio:
Li : C : O
2.7/1.4 : 1.4/1.4 : 4.1/1.4
2 : 1 : 3
Li : C : O = 2 : 1 : 3
Empirical formula is Li₂CO₃.