ArmyForever972
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Sagot :

AyIin

Given :-

  • Mass % of carbon (C) = 42.8%
  • Mass % of oxygen (O) = 57.1%

To Find :-

  • Empirical formula.

Solution :-

◆ Atomic mass of carbon (C) = 12g

◆ Atomic mass of oxygen (O) = 16g

  • No of moles of carbon (C) = 42.8/12

= 3.57

  • No of moles of oxygen (O) = 57.1/16

= 3.57

Simplest ratio :-

= 3.57 : 3.57

= 1 : 1

Empirical Formula = CO

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