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Friday, November 13, 2015

Empirical Formulas...

Empirical Formulas
What is % Composition of C10H14O?
Calculate the mass of H in 25.4 g of C10H14O.
C= 10 x 12.01 = 120.1
H= 14 x 1.01 = 14.14
O= 1 x 16 = 16.00

120.1 + 14.14 + 16.00 = 150.24 g/mol

%C= (120.1/150.24)(100)= 79.94% C
%H= (14.14/150.24)(100)= 9.412% H
%O= (16.00/150.24)(100)= 10.65% O

(9.412%)(25.4 g)= 2.39 g H

MOLECULAR FORMULAS
-whole number multiple of empirical formula

  • ratio of masses
-Can be the same as the empirical formula

EX: P4O10

EMPIRICAL FORMULAS
-lowest whole number ratio of elements in a compound
-start with % composition and work to moles


2 comments:

  1. You did a great job summing up what we learned in class about empirical formula. The link and your examples really helped make it clear for me!

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  2. I really like that you clarified the formulas in such an easy way to understand.

    ReplyDelete