CBSE Class 11 Chemistry Chapter 1: Some Basic Concepts of Chemistry NCERT Solutions

The molecular mass of a compound is the sum of the atomic masses of all the atoms in one molecule of the compound. We use the atomic masses of the constituent elements to calculate it.

1. H₂O (Water):

   The molecule contains 2 hydrogen atoms and 1 oxygen atom.

   Molecular mass of H₂O = (2 × Atomic mass of Hydrogen) + (1 × Atomic mass of Oxygen)

   $$=\; [2\; \backslash times\; (1.008\; \backslash text\{\; u\})]\; +\; [1\; \backslash times\; (16.00\; \backslash text\{\; u\})]$$

   $$=\; 2.016\; \backslash text\{\; u\}\; +\; 16.00\; \backslash text\{\; u\}$$

   $$=\; 18.016\; \backslash text\{\; u\}$$

   Rounding to two decimal places, the molecular mass of water is 18.02 u.

2. CO₂ (Carbon dioxide):

   The molecule contains 1 carbon atom and 2 oxygen atoms.

   Molecular mass of CO₂ = (1 × Atomic mass of Carbon) + (2 × Atomic mass of Oxygen)

   $$=\; [1\; \backslash times\; (12.011\; \backslash text\{\; u\})]\; +\; [2\; \backslash times\; (16.00\; \backslash text\{\; u\})]$$

   $$=\; 12.011\; \backslash text\{\; u\}\; +\; 32.00\; \backslash text\{\; u\}$$

   $$=\; 44.011\; \backslash text\{\; u\}$$

   Rounding to two decimal places, the molecular mass of carbon dioxide is 44.01 u.

3. CH₄ (Methane):

   The molecule contains 1 carbon atom and 4 hydrogen atoms.

   Molecular mass of CH₄ = (1 × Atomic mass of Carbon) + (4 × Atomic mass of Hydrogen)

   $$=\; [1\; \backslash times\; (12.011\; \backslash text\{\; u\})]\; +\; [4\; \backslash times\; (1.008\; \backslash text\{\; u\})]$$

   $$=\; 12.011\; \backslash text\{\; u\}\; +\; 4.032\; \backslash text\{\; u\}$$

   $$=\; 16.043\; \backslash text\{\; u\}$$

   The molecular mass of methane is 16.043 u.

The molecular formula of sodium sulphate is $Na\_2SO\_4$. To calculate the mass percent of each element, we first need to find the molar mass of the compound.

The atomic masses are approximately: Na = 23.0 u, S = 32.066 u, O = 16.00 u.

Molar mass of $Na\_2SO\_4$ = (2 × Atomic mass of Na) + (1 × Atomic mass of S) + (4 × Atomic mass of O)

$$=\; [2\; \backslash times\; (23.0\; \backslash text\{\; u\})]\; +\; [1\; \backslash times\; (32.066\; \backslash text\{\; u\})]\; +\; [4\; \backslash times\; (16.00\; \backslash text\{\; u\})]$$

$$=\; 46.0\; \backslash text\{\; u\}\; +\; 32.066\; \backslash text\{\; u\}\; +\; 64.0\; \backslash text\{\; u\}$$

$$=\; 142.066\; \backslash text\{\; u\}$$

The formula for mass percent of an element in a compound is:

$$\backslash text\{Mass\; percent\; of\; an\; element\}\; =\; \backslash frac\{\backslash text\{Total\; mass\; of\; that\; element\; in\; the\; compound\}\}\{\backslash text\{Molar\; mass\; of\; the\; compound\}\}\; \backslash times\; 100$$

Now, we calculate the mass percent for each element:

Mass percent of Sodium (Na):

$$=\; \backslash frac\{(2\; \backslash times\; 23.0\; \backslash text\{\; u\})\}\{142.066\; \backslash text\{\; u\}\}\; \backslash times\; 100\; =\; \backslash frac\{46.0\}\{142.066\}\; \backslash times\; 100\; \backslash approx\; 32.379\backslash \%$$

Rounded to one decimal place, the mass percent of sodium is 32.4%.

Mass percent of Sulphur (S):

$$=\; \backslash frac\{(1\; \backslash times\; 32.066\; \backslash text\{\; u\})\}\{142.066\; \backslash text\{\; u\}\}\; \backslash times\; 100\; =\; \backslash frac\{32.066\}\{142.066\}\; \backslash times\; 100\; \backslash approx\; 22.57\backslash \%$$

Rounded to one decimal place, the mass percent of sulphur is 22.6%.

Mass percent of Oxygen (O):

$$=\; \backslash frac\{(4\; \backslash times\; 16.00\; \backslash text\{\; u\})\}\{142.066\; \backslash text\{\; u\}\}\; \backslash times\; 100\; =\; \backslash frac\{64.0\}\{142.066\}\; \backslash times\; 100\; \backslash approx\; 45.049\backslash \%$$

Rounded to one decimal place, the mass percent of oxygen is 45.0%. (Note: The sum of percentages should ideally be 100%. 32.4 + 22.6 + 45.0 = 100.0%).

We are given the percentage composition by mass of an iron oxide: 69.9% iron and 30.1% oxygen.

To find the empirical formula, we first convert these percentages into moles by dividing by the respective atomic masses.

Atomic mass of Iron (Fe) = 55.85 u

Atomic mass of Oxygen (O) = 16.00 u

Step 1: Calculate the relative moles of each element.

Relative moles of Iron (Fe) = $\backslash frac\{\backslash text\{\%\; of\; Iron\}\}\{\backslash text\{Atomic\; mass\; of\; Fe\}\}\; =\; \backslash frac\{69.9\}\{55.85\}\; \backslash approx\; 1.25$

Relative moles of Oxygen (O) = $\backslash frac\{\backslash text\{\%\; of\; Oxygen\}\}\{\backslash text\{Atomic\; mass\; of\; O\}\}\; =\; \backslash frac\{30.1\}\{16.00\}\; \backslash approx\; 1.88$

Step 2: Find the simplest whole-number ratio of moles.

To find the simplest ratio, we divide the number of moles of each element by the smallest number of moles calculated.

Smallest number of moles = 1.25

Ratio of Fe = $\backslash frac\{1.25\}\{1.25\}\; =\; 1$

Ratio of O = $\backslash frac\{1.88\}\{1.25\}\; \backslash approx\; 1.504$

Since we have a ratio involving a half (1.5), we multiply both ratios by 2 to get whole numbers.

Whole number ratio of Fe = $1\; \backslash times\; 2\; =\; 2$

Whole number ratio of O = $1.504\; \backslash times\; 2\; \backslash approx\; 3$

Step 3: Write the empirical formula.

The simplest whole-number ratio of Fe to O is 2:3. Therefore, the empirical formula of the iron oxide is Fe₂O₃.