Question

Help ........,,,,.........

Answer 1

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