JEE Chemistry Some Basic Concepts Of Chemistry Notes Set 04

Dalton's Atomic Hypothesis and Its Modern Form

Dalton explained the laws of chemical combination by assuming that every element is made of extremely small particles, atoms, that take part in chemical change. All atoms of one element are alike and differ from those of other elements, in mass among other properties. Atoms are neither created nor destroyed in a chemical reaction. Atoms of different elements combine to form compounds, and in a given compound the kinds and relative numbers of atoms are fixed.
Modern knowledge changes several points: atoms are divisible (protons, neutrons and electrons); atoms of the same element can differ in mass (isotopes such as \(^{16}\)O, \(^{17}\)O, \(^{18}\)O); atoms of different elements can have the same mass (isobars such as \(^{40}\)Ca and \(^{40}\)Ar); atoms can be changed in nuclear reactions; and atoms need not combine in simple whole-number ratios (sucrose, C\(_{12}\)H\(_{22}\)O\(_{11}\)).

Atomic Mass, Molecular Mass and the Mole

The atomic mass unit is one twelfth of the mass of a carbon-12 atom: 1 amu = \( 1.66 \times 10^{-24} \) g = \( 1.66 \times 10^{-27} \) kg. The atomic mass of an element is the average mass of its atoms measured in this unit, so it has no dimension. The molecular mass is the sum of the atomic masses of all atoms in the molecule.
If an element has isotopes of masses \( a_1, a_2, \ldots \) with percentages \( x_1, x_2, \ldots \), then \[ \text{average atomic mass} = \frac{\sum a_i x_i}{\sum x_i}. \] The same weighting gives the average molar mass of a gas mixture: total mass divided by total moles.
One mole is the amount containing \( 6.022 \times 10^{23} \) particles (N\(_A\)). Its mass in grams equals the atomic or molecular mass, called the gram atomic mass (GAM) or gram molecular mass (GMM).

Properties and Measurement of Gases

Gases have very low density, expand without limit, are highly compressible and diffuse quickly. They fill their container, exert pressure by molecular collisions on the walls, and can be liquefied by cooling and compression. A gas is described by four variables: volume V, pressure P, temperature T and amount n.
Volume: 1 L = 1 dm\(^3\) = 1000 mL = 1000 cm\(^3\) = \( 10^{-3} \) m\(^3\).
Temperature: K = \(^\circ\)C + 273, and \( \dfrac{C}{5} = \dfrac{F - 32}{9} \). Both scales read the same at \(-\)40.
Pressure: force per unit area, measured with a manometer or barometer. 1 Pa = 1 N m\(^{-2}\); 1 bar = \( 10^5 \) Pa; 1 atm = 101325 Pa = 760 mm Hg = 760 torr; 1 torr = 133.3 Pa.

Ideal Gas Equation

\[ PV = nRT \] R depends only on the units: 0.0821 L atm K\(^{-1}\) mol\(^{-1}\) = 8.314 J K\(^{-1}\) mol\(^{-1}\) = 1.987 cal K\(^{-1}\) mol\(^{-1}\) (about 2 cal). Use the value of R that matches the units of P and V, and always convert temperature to kelvin.
At STP (0 \(^\circ\)C, 1 atm) one mole of an ideal gas occupies 22.4 L. Equal volumes of different gases under the same conditions have equal numbers of molecules (Avogadro's law), so for such gases the ratio of masses equals the ratio of molar masses. The equation also holds for a mixture, with n as the total moles.

Prefixes in the SI System

Worked Examples

Example 1. How many atoms are there in 4.4 g of CO\(_2\)?
Moles = 4.4/44 = 0.1. Each molecule has 3 atoms, so atoms = 0.1 \( \times \) 3 \( \times 6.022 \times 10^{23} = 1.81 \times 10^{23} \).

Example 2. A gas has density 1.43 g/L at STP. Find its molar mass.
Mass of 22.4 L = 1.43 \( \times \) 22.4 = 32 g, so the molar mass is 32 g/mol (oxygen).

Example 3. A mixture contains 0.25 mol N\(_2\) and 0.75 mol O\(_2\). Find the average molar mass.
Mass = 0.25 \( \times \) 28 + 0.75 \( \times \) 32 = 7 + 24 = 31 g for 1 mol, so the average molar mass is 31 g/mol.