Answer to Question #107471 in General Chemistry for Casey D

Question

what is the empirical formula of a compound that is 63.2% Mn and 36.8% O?

Answer 1

Examle 1

For example, the mass of a substance = 100g, then m(Mn)=63.2g and m(O)=36.8g.

n(Mn)=m(Mn)/M(Mn)=63.2/55=1.15mol

n(O)=m(O)/M(O)=36.8/16=2.3mol

n(Mn)/n(O)=1.15/2.3=1/2

Empirical formula MnO₂

Examle 2

w(Mn)=63.2% M(Mn)=55

63.2/55=1.15

w(O)=36.8% M(O)=16

36.8/16=2.3

1.15/1.15=1

2.3/1.15=2

Empirical formula MnO2