Chemistry, 14.12.2019 20:31, angie249

Match each molecular formula to the corresponding empirical formula.
a.)c4h8o2
b.)c4h12o4
c.)c2 h10 o2

1.) ch3o
2.) ch5o
3.) c2h4o

Answers: 1

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Answers

answered: josephvaldez518

Molecular Formula == Empirical Formula

A.)C₄H₈O₂ 3.) C₂H₄O

B.)C₄H₁₂O₄ 1.) CH₃O

C.)C₂H₁₀O₂ 2.) CH₅O

Explanation:

Data Given

Molecular Formula

A.)C₄H₈O₂

B.)C₄H₁₂O₄

C.)C₂H₁₀O₂

Empirical Formula

1.) CH₃O

2.) CH₅O

3.) C₂H₄O

Match each molecular formula to the corresponding empirical formula = ?

Solution:

As we know

Empirical formula shows the simplest ratio of atoms in the molecule but not whole numbers of atoms in a compound or molecule

So we look at the ratio of atoms in each given molecule and will find the simplest ratio

A.)C₄H₈O₂

Look at the ratio of atoms

C : H : O

4 : 8 : 2

Divide the all values on the smallest values to get simplest number ratio

C : H : O

4/2 : 8/2 : 2/2

2 : 4 : 1

So the simplest ratio of carbon, hydrogen and oxygen is 2, 4 and 1 respectively

C : H : O

2 : 4 : 1

so the empirical formula will be

3.) C₂H₄O

The match pair will be

Molecular formula == Empirical formula

C₄H₈O₂ C₂H₄O

B.)C₄H₁₂O₄

Look at the ratio of atoms

C : H : O

4 : 12 : 4

Divide the all values on the smallest values to get simplest number ratio

C : H : O

4/4 : 12/4 : 4/4

1 : 3 : 1

So the simplest ratio of carbon, hydrogen and oxygen is 1, 3 and 1 respectively

C : H : O

1 : 3 : 1

so the empirical formula will be

1.) CH₃O

The match pair will be

Molecular formula == Empirical formula

C₄H₁₂O₄ CH₃O

C.)C₂H₁₀O₂

Look at the ratio of atoms

C : H : O

2 : 10 : 2

Divide the all values on the smallest values to get simplest number ratio

C : H : O

2/2 : 10/2 : 2/2

1 : 5 : 1

So the simplest ratio of carbon, hydrogen and oxygen is 1, 5 and 1 respectively

C : H : O

1 : 5 : 1

so the empirical formula will be

2.) CH₅O

The match pair will be

Molecular formula == Empirical formula

C₂H₁₀O₂ CH₅O

answered: Guest

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in $h_{2}s$ , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between $h_{2}s$ molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent $h_{2}s$ molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.