Question

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Answer 1

The empirical formula would be  $C_2H_3O_2$ and the molecular formula would be  $C_4 H_6 O_4$

Empirical and molecular formula

From the illustration:

C = 40.6 g                   H = 5.1 g                      O = 54.2 g

Divide each element by its molar weight:

C = 40.6/12 = 3.4                    H = 5.1/1 = 5.1                  O = 54.2/16 = 3.4

Divide by the smallest:

C = 3.4/3.4 = 1                        H = 5.1/3.4 = 1.5                 O = 3.4/3.4 = 1

Thus, the empirical formula is $C_2H_3O_2$

To find the molecular formula:

[ $C_2H_3O_2$]n = molecular formula

where n = molar mass/empirical formula mass

Empirical formula mass = 24 + 3 + 32 = 59

n = 118.084/59 = 2

Thus, the molecular formula is $C_4 H_6 O_4$ which is more or less the same as the empirical formula.

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Answer 2

The Empirical formula will be C₂H₃O₂  and the Molecular formula would be  C₄H₆O₄

What is Empirical and Molecular Formula ?

Empirical formula is a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.

Molecular formula is a chemical formula that gives the total number of atoms of each element in each molecule of a substance.

Given ;

• C = 40.6 g
• H = 5.1 g
• O = 54.2 g

Divide each element by its molar weight:

• C =  40.6 / 12 = 3.4
• H = 5.1 / 1 = 5.1                
• O = 54.2 / 16 = 3.4

Divide by the smallest Value:

• C = 3.4/3.4 = 1
• H = 5.1/3.4 = 1.5                
• O = 3.4/3.4 = 1

Thus, the Empirical formula is C₂H₃O₂

Now, to find the molecular formula ;

[C₂H₃O₂ ]n = molecular formula

where n = molar mass / empirical formula mass

Empirical formula mass = 24 + 3 + 32 = 59

n = 118.084 / 59 = 2

[C₂H₃O₂ ]n = molecular formula

[C₂H₃O₂ ]2 = C₄H₆O₄

Thus, the Molecular formula is C₂H₃O₂

Learn More about Empirical formula here ; brainly.com/question/14044066

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