JEE Chemistry Some Basic Concepts Of Chemistry Notes Set 05

Atomic Mass, Molecular Mass and the Mole

Dalton's atomic theory treats matter as made of indivisible atoms. Atoms of one element are alike, atoms of different elements differ, and a chemical change only rearranges atoms without creating or destroying any. Since an atom is far too light to weigh directly, masses are compared with a standard. The modern standard is one-twelfth of the mass of a carbon-12 atom, called the unified atomic mass unit (u).

\( 1\,\text{u} = \dfrac{1}{12}\times \text{mass of one C-12 atom} = 1.66\times 10^{-24}\,\text{g} \)

Relative atomic mass = (mass of one atom of the element) / (1 u). It is a pure number and, for a single isotope, is close to the number of nucleons. Molecular mass is the sum of the atomic masses of all atoms in the formula.

A mole is the amount of substance that contains as many entities as there are atoms in exactly 12 g of carbon-12. That number is the Avogadro constant, \( N_A = 6.022\times 10^{23}\,\text{mol}^{-1} \). The mass of one mole of a substance in grams equals its atomic or molecular mass in u, and is called its molar mass.

Average atomic mass of an element with isotopes of masses \( a_1, a_2, \ldots \) and percentage abundances \( x_1, x_2, \ldots \):

\( \bar{A} = \dfrac{a_1x_1 + a_2x_2 + \cdots}{100} \)

Mean molar mass of a mixture of gases with mole numbers \( n_i \) and molar masses \( M_i \): \( \bar{M} = \dfrac{\sum n_iM_i}{\sum n_i} \).

Gases and Molar Volume

Gay-Lussac found that gases react in simple whole-number ratios of volume when measured at the same temperature and pressure. Avogadro explained this: equal volumes of all gases under the same conditions hold equal numbers of molecules (not atoms). Hence, counting molecules of a gas is the same as measuring its volume.

At STP (273 K and 1 atm) one mole of any ideal gas occupies 22.4 L. The conversions between mass, moles, number of particles and volume at STP are:

\( n = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{N}{N_A} = \dfrac{V_{\text{STP}}}{22.4\,\text{L}} \)

Worked example 1. How many moles of each element are present in 294 g of \( \ce{H2SO4} \)? Molar mass = 98 g mol^-1, so 294 g is 3 mol of the acid. Each formula unit has 2 H, 1 S and 4 O atoms, so the sample contains 6 mol H, 3 mol S and 12 mol O.

Worked example 2. Chlorine is 75% \( ^{35}\text{Cl} \) and 25% \( ^{37}\text{Cl} \). The average atomic mass is \( (75\times 35 + 25\times 37)/100 = 35.5 \) u.

Worked example 3. A gas mixture contains 16% \( \ce{O2} \), 80% \( \ce{N2} \), 3% \( \ce{CO2} \) and 1% \( \ce{SO2} \) by moles. Mean molar mass = \( (16\times 32 + 80\times 28 + 3\times 44 + 1\times 64)/100 = 29.48 \) g mol^-1.

Percentage Composition, Empirical and Molecular Formulae

The mass percentage of an element in a compound is the mass of that element in one mole of the compound divided by the molar mass, times 100. By the law of definite proportions it is the same for every sample of the compound.

The empirical formula gives the simplest whole-number ratio of atoms; the molecular formula gives the actual number of atoms in a molecule. The molecular formula is an integral multiple of the empirical formula, \( n = \dfrac{\text{molecular mass}}{\text{empirical formula mass}} \).

To find the empirical formula: (1) take 100 g of the compound so that percentages become grams, (2) divide each mass by the atomic mass to get moles, (3) divide all by the smallest value, (4) multiply by a small integer if any ratio is not whole (for example 1.5 becomes 3 after multiplying by 2).

Worked example 4. A compound has 40.0% C, 6.7% H and 53.3% O and a molar mass of 180 g mol^-1. Moles per 100 g: C 3.33, H 6.65, O 3.33, giving the ratio 1 : 2 : 1 and the empirical formula \( \ce{CH2O} \) (30 g mol^-1). Since 180/30 = 6, the molecular formula is \( \ce{C6H12O6} \).

If only the percentage of one element and the number of its atoms per molecule are known, the minimum molar mass is (mass of those atoms) / (mass fraction). For example, a compound with 8% sulphur has a least molar mass of 32/0.08 = 400 g mol^-1.

Stoichiometry of Chemical Equations

A balanced equation conserves atoms of every element (and charge, for ionic equations). Its coefficients give the mole ratio in which substances react or form, so for \( a\ce{A} + b\ce{B} \rightarrow c\ce{C} + d\ce{D} \):

\( \dfrac{n_A\,\text{reacted}}{a} = \dfrac{n_B\,\text{reacted}}{b} = \dfrac{n_C\,\text{formed}}{c} = \dfrac{n_D\,\text{formed}}{d} \)

Calculations can be mass-mass, mass-volume, mole-mole or volume-volume (gases); in each case first convert to moles, apply the coefficient ratio, then convert back.

Worked example 5. 245 g of \( \ce{KClO3} \) is heated: \( 2\ce{KClO3} \rightarrow 2\ce{KCl} + 3\ce{O2} \). Moles of \( \ce{KClO3} \) = 245/122.5 = 2. Oxygen formed = 3 mol, which is 67.2 L at STP. KCl formed = 2 mol = 149 g.

Limiting Reagent, Percentage Yield and Excess

The reactant that is used up first is the limiting reagent; it fixes the maximum amount of product. To find it, divide the moles of each reactant by its coefficient: the smallest quotient identifies the limiting reagent. The other reactants are in excess, and the excess left over = moles taken - moles used.

\( \%\,\text{yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}}\times 100 \). When a reaction has several steps with yields \( y_1, y_2, \ldots \), the overall fraction is the product of the step yields.

Worked example 6. 5 g of \( \ce{H2} \) is mixed with 32 g of \( \ce{O2} \) for \( 2\ce{H2} + \ce{O2} \rightarrow 2\ce{H2O} \). Moles: \( \ce{H2} \) = 2.5 and \( \ce{O2} \) = 1. Quotients: 2.5/2 = 1.25 and 1/1 = 1, so \( \ce{O2} \) limits. Water formed = 2 mol (36 g), and \( \ce{H2} \) left = 2.5 - 2 = 0.5 mol.

Principle of Atom Conservation (POAC) and Purity

When a problem has a chain of reactions or parallel reactions, write an atom balance for one element instead of balancing every equation: the moles of that element in all reactants equal its moles in all products. This avoids intermediate calculations. For a sequence \( \ce{M} \rightarrow \ce{MX2} \rightarrow \ce{M3X8} \rightarrow \ldots \) the metal atoms are conserved, so 1 mol of \( \ce{M3X8} \) must come from 3 mol of M.

For an impure sample, only the pure part reacts: \( \%\,\text{purity} = \dfrac{\text{mass of pure compound}}{\text{mass of sample}}\times 100 \). If a 10 g sample of chalk gives 2.24 L of \( \ce{CO2} \) at STP, then 0.1 mol of \( \ce{CaCO3} \) (10 g) reacted, so the sample is 100% pure.

Concentration Terms

Molarity \( M = \dfrac{n_{\text{solute}}}{V_{\text{solution}}\,(\text{L})} \) depends on temperature because the volume does. Molality \( m = \dfrac{n_{\text{solute}}}{\text{mass of solvent (kg)}} \), the mole fraction \( x_A = \dfrac{n_A}{n_A + n_B} \), and mass percent are independent of temperature.

% w/w = (mass of solute / mass of solution) × 100; % w/v = (mass of solute in g / volume of solution in mL) × 100; % v/v = (volume of solute / volume of solution) × 100; \( \text{ppm} = \dfrac{\text{mass of solute}}{\text{mass of solution}}\times 10^6 \).

Interconversion uses the density \( d \) (g mL^-1) of the solution. For a solution of w% by mass and solute molar mass \( M_2 \):

\( M = \dfrac{10\,d\,w}{M_2}, \qquad m = \dfrac{1000\,M}{1000\,d - M M_2} \)

Worked example 7. Concentrated sulphuric acid is 98% by mass with density 1.84 g mL^-1. Molarity = \( 10\times 1.84\times 98/98 = 18.4 \) M.

Worked example 8. An ethanol-water solution is 2 M with density 1.025 g mL^-1. In 1 L: mass = 1025 g, ethanol = 92 g, water = 933 g, so molality = 2/0.933 = 2.14 m.

Dilution and Mixing

On dilution the moles of solute do not change: \( M_1V_1 = M_2V_2 \). When solutions of the same solute are mixed, add the moles and add the volumes (assuming no volume change): \( M_{\text{final}} = \dfrac{M_1V_1 + M_2V_2}{V_1 + V_2} \). For mixing by mass, add the masses of solute and the masses of solution.

Worked example 9. Diluting 50 mL of 12 M HCl to 0.5 M needs a final volume of \( 50\times 12/0.5 = 1200 \) mL, so 1150 mL of water is added. Mixing 200 mL of 0.5 M and 300 mL of 1.5 M solutions gives \( (0.1 + 0.45)/0.5 = 1.1 \) M.

Worked example 10. For a salt such as \( \ce{CaCl2} \), count ions: 0.1 M \( \ce{CaCl2} \) gives 0.1 M \( \ce{Ca^{2+}} \) and 0.2 M \( \ce{Cl^-} \). When two solutions are mixed, add the moles of the same ion before dividing by the total volume.

Quick Reminders for the Exam

Convert everything to moles first. Balance the equation before using ratios. Check the limiting reagent whenever two reactant quantities are given. Use 22.4 L mol^-1 only at STP. Molality needs the mass of solvent, not of solution, and molarity needs the volume of solution, not of solvent.