Question

determine the empirical formula and molecular formula for chrysotile asbestos it has the following percentage composition 28.0 3% mg , 21.6 % Si 1.16% H ,49.21% O
molar molar mass 5 20.8 grams
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Answer 1

RELATIVE NO OF ATOMS:

(divide the % composition of the element with thier atomic mass)

% Mg

28.03/24 =1.16

% Si

21.6/28=0.77

%H

1.16/1=1.16

%O

49.21 /16=3.07

SIMPLE RATIO:

Mg

1.16/0.77=1.5

Si

0.77/0.77=1

H

1.16/0.77=1.5

O

3.07/0.77=4

Empirical formula = Mg 1.5 Si H 1.5 O 4

multiplying by 2 we get

Empirical formula = Mg 3 Si 2 H 3O 8

Empirical mass

= 3(24)+2(28)+3(1)+8(16)

=72+56+3+128

=259

Molar mass =520.8 g

molar mass= n(empirical mass)

n= molar mass / empirical mass

n=520.8/259

n=2

molecular formula = 2(Mg 3 Si 2 H 3O 8)

molecular formula = Mg 6 Si 4 H 6 O 16

Answer 2

% Mg

28.03/24 =1.16

% Si

21.6/28=0.77

%H

1.16/1=1.16

%O

49.21 /16=3.07

SIMPLE Ratio

Mg

1.16/0.77=1.5

Si

0.77/0.77=1

H

1.16/0.77=1.5

O

3.07/0.77=4

Empirical formula = Mg 1.5 Si H 1.5 O 4

multiplying by 2 we get

Empirical formula = Mg 3 Si 2 H 3O 8

Empirical mass

= 3(24)+2(28)+3(1)+8(16)

=72+56+3+128

=259

Molar mass =520.8 g

molar mass= n(empirical mass)

n= molar mass / empirical mass

n=520.8/259

n=2

molecular formula = 2(Mg 3 Si 2 H 3 O 8)

molecular formula = Mg 6 Si 2 H 6 O 16