Class 9

9.4.5 Chemical Formula of Covalent Compounds

This is the fifth lecture from Chapter 4: “Stoichiometry” in the new Class 9 Chemistry book (Punjab Board – PCTB). It discusses the chemical formula of covalent compounds. The lecture also includes a multiple-choice quiz, short question and long question notes.

MCQs Based Quiz

9.4.5 Chemical Formula of Covalent Compounds

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A compound has empirical formula CH₂ and molar mass 42 g/mol. Its molecular formula is: (Empirical formula mass = 14 g/mol)

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The empirical formula of a compound is CH2O. Its molar mass is 60 gmol-1. Determine its molecular formula.

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What is the empirical formula of hydrogen peroxide (H2O2)?

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What is the empirical formula of benzene (C6H6)

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If a compound has an empirical formula of CH₂O and molar mass of 180 g/mol, what is its molecular formula? (Empirical formula mass = 30 g/mol)

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Short Questions

What do you need to write the chemical formula of a covalent compound?

To write the chemical formula of a covalent compound we would need:

  1. Its empirical formula
  2. Its molar mass
  3. Its empirical formula mass

How can you determine the empirical formula of a compound?

Empirical formula of a compound can be determined by calculating the simplest whole number ratio of atoms present in that compound. This can be done by using experimental data on the mass percent composition of the compound.

Empirical formula of hydrogen peroxide is HO. Its molar mass is 34 gmol-1. Determine its molecular formula.

$$\begin{aligned}
\text{Molar mass of hydrogen peroxide} &= 34\ \text{g mol}^{-1} \\
\text{Empirical formula mass (HO)} &= 17\ \text{g mol}^{-1} \\
n &= \frac{34}{17} = 2 \\
\text{Molecular formula} &= \text{(HO)2} = \mathrm{H_2O_2}
\end{aligned}$$

Write down the name of three such compounds which have different empirical and molecular formulas.

  1. Benzene: Molecular formula = C6H6 and empirical formula = CH
  2. Acetylene: Molecular formula = C2H2 and empirical formula = CH
  3. Hydrogen peroxide: Molecular formula = H2O2 and empirical formula = HO

Empirical formula of a compound is CH. Its molecular mass is 78 gmol-1. Find out its molecular formula.

$$\begin{aligned}
\text{Empirical formula} &= \text{CH}\\
\text{Molecular mass} &= 78 gmol^{-1}\\
\text{Empirical formula mass} &= 13 gmol^{-1}\\
\text{Molecular formula} &= \text{n(Empirical formula)}\\
\text{n} &= \frac{\text{Molar mass}}{\text{empirical formula mass}}\\
&= \frac{78}{13} = 6\\
\text{Molecular formula} &= (CH) n\\
&= (CH)6 = {C_6 H_6} \\
\end{aligned}$$

The empirical formula of a compound is CH2O. Its molar mass is 180 gmol-1. Determine its molecular formula.

$$\begin{aligned}
\text{Empirical formula} &= CH_{2}O\\
\text{Molecular mass} &= 180 gmol^{-1}\\
\text{Empirical formula mass} &= 30 gmol^{-1}\\
\text{Molecular formula} &= \text{n(Empirical formula)}\\
\text{n} &= \frac{\text{Molar mass}}{\text{empirical formula mass}}\\
&= \frac{180}{30} = 6\\
\text{Molecular formula} &= (CH_{2}O)n\\
&= (CH_{2}O)6 = {C_6 H_{12} O_6} \\
\end{aligned}$$

The empirical formula of a compound is CH2O. Its molar mass is 60 gmol-1. Determine its molecular formula.

$$\begin{aligned}
\text{Empirical formula} &= CH_{2}O\\
\text{Molecular mass} &= 60 gmol^{-1}\\
\text{Empirical formula mass} &= 30 gmol^{-1}\\
\text{Molecular formula} &= \text{n(Empirical formula)}\\
\text{n} &= \frac{\text{Molar mass}}{\text{empirical formula mass}}\\
&= \frac{60}{30} = 2\\
\text{Molecular formula} &= (CH_{2}O)n\\
&= (CH_{2}O)2 = {C_2 H_4 O_2} \\
\end{aligned}$$