Chemistry, asked by viveksathyan7826, 11 months ago

What’s the empirical formula of a molecule containing 18.7% lithium, 16.3% carbon, and 65.0% oxygen?

Answers

Answered by abuahmed2200a
41

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

Answered by RainbowSparkl1es
3

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.

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