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 elements as follows: Hydrogen (H) ≈ 1.008 u, Carbon (C) ≈ 12.011 u, Oxygen (O) ≈ 16.00 u.

1. H₂O (Water):

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

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

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

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

   Rounded to two decimal places, the molecular mass is 18.02 u.

2. CO₂ (Carbon Dioxide):

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

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

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

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

   Rounded to two decimal places, the molecular mass is 44.01 u.

3. CH₄ (Methane):

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

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

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

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

   Rounded to two decimal places, the molecular mass is 16.04 u.

The molecular formula of sodium sulphate is $Na\_2SO\_4$. First, we calculate the molar mass of sodium sulphate using the atomic masses: Sodium (Na) ≈ 23.0 g/mol, Sulphur (S) ≈ 32.066 g/mol, Oxygen (O) ≈ 16.00 g/mol.

Molar mass of $Na\_2SO\_4\; =\; (2\; \backslash times\; \backslash text\{Atomic\; mass\; of\; Na\})\; +\; (1\; \backslash times\; \backslash text\{Atomic\; mass\; of\; S\})\; +\; (4\; \backslash times\; \backslash text\{Atomic\; mass\; of\; O\})$

= $(2\; \backslash times\; 23.0\; \backslash text\{\; g/mol\})\; +\; (1\; \backslash times\; 32.066\; \backslash text\{\; g/mol\})\; +\; (4\; \backslash times\; 16.00\; \backslash text\{\; g/mol\})$

= $46.0\; \backslash text\{\; g/mol\}\; +\; 32.066\; \backslash text\{\; g/mol\}\; +\; 64.0\; \backslash text\{\; g/mol\}$

= $142.066\; \backslash text\{\; g/mol\}$

The mass percentage of an element in a compound is calculated using the formula:

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

Now, we calculate the mass percentage for each element:

Mass percent of Sodium (Na):

= $\backslash frac\{2\; \backslash times\; \backslash text\{Atomic\; mass\; of\; Na\}\}\{\backslash text\{Molar\; mass\; of\; \}\; Na\_2SO\_4\}\; \backslash times\; 100\backslash \%$

= $\backslash frac\{2\; \backslash times\; 23.0\; \backslash text\{\; g/mol\}\}\{142.066\; \backslash text\{\; g/mol\}\}\; \backslash times\; 100\backslash \%$

= $\backslash frac\{46.0\}\{142.066\}\; \backslash times\; 100\backslash \%$

= $0.32379\; \backslash times\; 100\backslash \%$

= 32.38%

Mass percent of Sulphur (S):

= $\backslash frac\{1\; \backslash times\; \backslash text\{Atomic\; mass\; of\; S\}\}\{\backslash text\{Molar\; mass\; of\; \}\; Na\_2SO\_4\}\; \backslash times\; 100\backslash \%$

= $\backslash frac\{32.066\; \backslash text\{\; g/mol\}\}\{142.066\; \backslash text\{\; g/mol\}\}\; \backslash times\; 100\backslash \%$

= $0.2257\; \backslash times\; 100\backslash \%$

= 22.57%

Mass percent of Oxygen (O):

= $\backslash frac\{4\; \backslash times\; \backslash text\{Atomic\; mass\; of\; O\}\}\{\backslash text\{Molar\; mass\; of\; \}\; Na\_2SO\_4\}\; \backslash times\; 100\backslash \%$

= $\backslash frac\{4\; \backslash times\; 16.00\; \backslash text\{\; g/mol\}\}\{142.066\; \backslash text\{\; g/mol\}\}\; \backslash times\; 100\backslash \%$

= $\backslash frac\{64.0\}\{142.066\}\; \backslash times\; 100\backslash \%$

= $0.45049\; \backslash times\; 100\backslash \%$

= 45.05%

The mass percentages are approximately Na: 32.38%, S: 22.57%, and O: 45.05%. The sum of these percentages is 32.38 + 22.57 + 45.05 = 100.00%.

We are given the percentage composition by mass of iron and oxygen in an iron oxide. We need to find the empirical formula, which represents the simplest whole-number ratio of atoms in the compound. We will use the atomic masses: Iron (Fe) ≈ 55.85 g/mol, Oxygen (O) ≈ 16.00 g/mol.

Given:

% of iron by mass = 69.9 %

% of oxygen by mass = 30.1 %

Step 1: Convert percentage composition to moles.\nAssume we have 100 g of the compound. Then, we have 69.9 g of iron and 30.1 g of oxygen.

Relative moles of iron = \frac{\text{Mass of iron}}{\text{Atomic mass of iron}} = \frac{69.9 \text{ g}}{55.85 \text{ g/mol}} \approx 1.25 \text{ mol}

Relative moles of oxygen = \frac{\text{Mass of oxygen}}{\text{Atomic mass of oxygen}} = \frac{30.1 \text{ g}}{16.00 \text{ g/mol}} \approx 1.88 \text{ mol}

Step 2: Find the simplest whole-number ratio of moles.\nTo 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 mol (from iron).

Ratio of iron = \frac{1.25 \text{ mol}}{1.25 \text{ mol}} = 1

Ratio of oxygen = \frac{1.88 \text{ mol}}{1.25 \text{ mol}} \approx 1.504

Step 3: Convert the ratio to whole numbers.\nThe ratio of iron to oxygen is approximately 1 : 1.5. To convert this to whole numbers, we multiply both numbers by 2.

Iron ratio (whole number) = 1 \times 2 = 2

Oxygen ratio (whole number) = 1.504 \times 2 \approx 3

The simplest whole-number ratio of iron to oxygen atoms is 2:3.

Therefore, the empirical formula of the iron oxide is Fe₂O₃.