Chemical Calculations: Moles, Molarity, and Dilutions

Chemical Equations

Definition and Balancing of Chemical Equations

Chemical equations are symbolic representations of chemical reactions, showing the reactants and products involved. A balanced chemical equation reflects the law of conservation of mass, ensuring that the number of atoms of each element is the same on both sides of the equation.

• Reactants: Substances that undergo change during a reaction.

• Products: Substances formed as a result of the reaction.

• Balancing: Adjust coefficients to ensure equal numbers of each atom on both sides.

Example:

$ext{CH}_4 + 2 ext{O}_2 ightarrow ext{CO}_2 + 2 ext{H}_2 ext{O}$

• Number of C: 1 = 1

• Number of H: 4 = 4

• Number of O: 4 = 4

Writing Balanced Chemical Equations

To write balanced equations, use the valencies of elements to determine the correct formulas.

• Li + F2: $2 ext{Li} + ext{F}_2 ightarrow 2 ext{LiF}$

• Mg + Cl2: $ext{Mg} + ext{Cl}_2 ightarrow ext{MgCl}_2$

• Li + S: $2 ext{Li} + ext{S} ightarrow ext{Li}_2 ext{S}$

• Al + O2: $4 ext{Al} + 3 ext{O}_2 ightarrow 2 ext{Al}_2 ext{O}_3$

Reversible Reactions and Equilibrium

Some chemical reactions are reversible, indicated by a double arrow ($ightleftharpoons$). The system can reach a dynamic equilibrium where the rates of the forward and reverse reactions are equal.

Example: $ext{H}_2 ext{O} + ext{CO}_2 ightleftharpoons ext{H}_2 ext{CO}_3$

Le Châtelier's Principle

If a dynamic equilibrium is disturbed by changing the conditions (concentration, pressure, temperature), the position of equilibrium shifts to partially counteract the change.

• Increasing CO2: Shifts equilibrium to the right (more H2CO3).

• Pressure/Volume Effects: For $3 ext{H}_2(g) + ext{N}_2(g) ightleftharpoons 2 ext{NH}_3(l)$, increasing pressure or decreasing volume shifts equilibrium toward ammonia.

International System (SI) of Metric Units

SI Prefixes and Their Factors

The SI system uses prefixes to denote multiples or fractions of units such as meters, grams, moles, and seconds.

Prefix	Symbol	Factor	Decimal
Tera	T	1012	1,000,000,000,000
Giga	G	109	1,000,000,000
Mega	M	106	1,000,000
Kilo	k	103	1,000
Milli	m	10-3	0.001
Micro	μ	10-6	0.000001
Nano	n	10-9	0.000000001
Pico	p	10-12	0.000000000001
Femto	f	10-15	0.000000000000001

Moles and Molarity

Definition of a Mole

A mole is a unit representing $6.022 imes 10^{23}$ particles (Avogadro's number), whether atoms, molecules, ions, or electrons. It is analogous to the term "dozen" but for a much larger quantity.

• 1 mole of atoms: $6.022 imes 10^{23}$ atoms

• 1 mole of molecules: $6.022 imes 10^{23}$ molecules

Molar Mass

The molar mass of a substance is the mass of one mole of that substance, expressed in grams. It can be found on the periodic table and is numerically equal to the atomic or molecular mass in atomic mass units (amu).

• Example: Carbon-12 has a molar mass of 12 g/mol.

• Practice: 1 mol F = 19 g; 0.1 mol F = 1.9 g; 38 g F = 2 mol

Relative Molecular Mass (Mr)

The sum of the relative atomic masses of the constituent atoms in a compound.

• Water (H2O): $(2 imes 1) + 16 = 18$

• Urea (NH2CONH2): $14 + (2 imes 1) + 12 + 16 + 14 + (2 imes 1) = 60$

Calculating Moles

Use the following equation to calculate the number of moles:

$ext{Number of Moles (mol)} = rac{ ext{mass (g)}}{M_r ( ext{g})}$

• Example: 0.09 g water, $M_r = 18$ $ext{Number of Moles} = rac{0.09}{18} = 0.005$ mol or 5 mmol

• Example: 1 mg of substance with $M_r = 250$ $ext{Number of Moles} = rac{0.001}{250} = 4 imes 10^{-6}$ mol or 4 μmol

Stoichiometry in Reactions

Stoichiometry allows calculation of reactant and product quantities in chemical reactions.

• Example: $ext{C} + ext{O}_2 ightarrow ext{CO}_2$ 1.2 g C = 0.1 mol; 0.1 mol O2 required (3.2 g); 0.1 mol CO2 produced (4.4 g)

Solutions and Concentration

Definitions

• Solution: Homogeneous mixture of two or more substances.

• Solvent: Substance in which the solute is dissolved; present in greatest amount.

• Solute: Substance dissolved in the solvent; present in lesser amount.

Concentration of a Solution

Concentration is the amount of solute dissolved in a given volume of solvent.

$ext{Concentration} = rac{ ext{Mass (g)}}{ ext{Volume (l)}}$

• Units: g/L, mg/ml, mg/ml-1

Molarity (M)

Molarity is the number of moles of solute per liter of solution.

• 1 M solution: 1 mole in 1 liter

• Key equations:

$ext{Molarity (M)} = rac{ ext{Number of moles}}{ ext{Volume (l)}}$

$ext{Molarity (M)} = rac{ ext{mass (g)}}{M_r ( ext{g}) imes ext{volume (l)}}$

$ext{Molarity (M)} = rac{ ext{mass (g)} imes 1000}{M_r ( ext{g}) imes ext{volume (ml)}}$

• Example 1: 500 ml of 0.2 M NaOH ($M_r = 40$) $0.2 imes 40 imes 0.5 = 4$ g NaOH required

• Example 2: 100 ml of 2 M KCl ($M_r = 74.5$) $2 imes 74.5 imes 0.1 = 14.9$ g KCl

• Example 3: 2 g NaH2PO4 in 1 l ($M_r = 120$) $2/120 = 0.0167$ M or 16.7 mM

Percentage Solutions

• % mass/volume (% m/v): Grams of substance per 100 ml

• % volume/volume (% v/v): Volume (ml) of substance per 100 ml

Example: 1 M NaCl solution contains 58.5 g in 1 l % concentration = $rac{58.5}{1000} imes 100 ext{%} = 5.85 ext{%}$

Normal Saline Solution

Normal saline is the closest approximation to the concentration of NaCl in blood plasma: 0.9% w/v NaCl.

Dilutions and Serial Dilutions

Dilutions

Dilution is the process of reducing the concentration of a solute in solution, usually by adding more solvent. The amount of solute remains the same, but the volume increases, lowering the concentration.

Producing Solutions of Specific Concentrations

To prepare a solution of desired concentration from a stock solution, use the dilution equation:

$V_2 imes C_2 = V_1 imes C_1$

• $V_1$ = Initial volume

• $C_1$ = Initial concentration

• $V_2$ = Final volume

• $C_2$ = Final concentration

Example: 250 ml of 0.1 M solution diluted to 1 l $C_2 = rac{250 imes 0.1}{1000} = 0.025$ M

Serial Dilution

Serial dilution involves stepwise dilution of a solution, often used in experimental sciences to achieve very low concentrations.

• Dilution factor for each tube: $rac{ ext{volume of sample}}{ ext{volume of sample} + ext{volume of diluent}}$

• Ten-fold dilution: 1 ml sample + 9 ml diluent = $rac{1}{10} = 10^{-1}$

• Total dilution factor: Multiply dilution factors of each step. For two ten-fold dilutions: $10^{-1} imes 10^{-1} = 10^{-2}$

Applications of Serial Dilution

• Microbiology: Estimating concentration or number of cells/organisms.

• Biochemistry: Obtaining desired reagent concentrations.

• Pharmaceutical science: Achieving necessary chemical concentrations efficiently.

Summary of Learning Outcomes

• Understand how to balance chemical equations.

• Determine molar mass from atomic mass.

• Calculate and use molar concentrations.

• Prepare and understand serial dilutions.

Additional info: These notes cover foundational concepts in chemical calculations, including chemical equations, mole concept, molarity, concentration, and dilution techniques, all essential for General Chemistry students.