General Chemistry
Stoichiometry, electronic structure, redox, thermochemistry, phase equilibria, pH and electrochemistry. Theory, examples, and solved exercises.
Complete Theory
4The atom is the smallest unit of an element that retains its chemical properties. It consists of a nucleus containing protons (positive charge) and neutrons (neutral), collectively called nucleons, surrounded by a cloud of electrons (negative charge) moving in orbitals.
The atomic number is the number of protons in the nucleus — it is the element's "ID number": all atoms with are carbon, with oxygen, etc. The mass number is the sum of protons and neutrons.
The atomic number is the number of protons in the nucleus — it is the element's "ID number": all atoms with are carbon, with oxygen, etc. The mass number is the sum of protons and neutrons.
- Nuclides and isotopes: nuclides of the same element (same ) with different neutron number are called isotopes. Example: (6p+6n), (6p+7n), (6p+8n). Isotopes have the same chemical properties but different masses, and some are radioactive.
- Radioactive decay: unstable nuclei (radioisotopes) emit radiation to become more stable: ( nucleus, 2p+2n), (a neutron becomes a proton + electron), (positron), (high-energy photon).
- Nuclear fission: a heavy nucleus (e.g., ) absorbs a neutron and splits into two lighter nuclei, releasing energy and neutrons that trigger a chain reaction (the basis of nuclear energy).
- Nuclear fusion: two light nuclei (e.g., and ) combine to form a heavier nucleus, releasing enormous amounts of energy — the process that powers the Sun and stars.
The relative atomic mass of an element is the weighted average of the masses of its naturally occurring isotopes, measured relative to the atomic mass unit u (1/12 the mass of , equal to ). Since isotopes have different natural abundances, the atomic mass listed on the periodic table is a weighted average. Example: natural chlorine is 75.78% (34.97 u) and 24.22% (36.97 u) → u.
The relative molecular mass is obtained by summing the atomic masses of all atoms in the chemical formula: where is the number of atoms of element in the formula.
The empirical (minimum) formula expresses the simplest whole-number ratio of atoms in the compound. The molecular formula gives the actual number of atoms of each element in a single molecule: it is an integer multiple of the empirical formula (). Example: glucose has empirical formula and molecular formula ().
Real-world applications: determining empirical and molecular formulas is fundamental for identifying new drugs and natural compounds through elemental analysis and mass spectrometry.
The relative molecular mass is obtained by summing the atomic masses of all atoms in the chemical formula: where is the number of atoms of element in the formula.
The empirical (minimum) formula expresses the simplest whole-number ratio of atoms in the compound. The molecular formula gives the actual number of atoms of each element in a single molecule: it is an integer multiple of the empirical formula (). Example: glucose has empirical formula and molecular formula ().
Real-world applications: determining empirical and molecular formulas is fundamental for identifying new drugs and natural compounds through elemental analysis and mass spectrometry.
The mole is the SI unit for amount of substance: 1 mole contains exactly elementary entities (atoms, molecules, ions, or particles), where is Avogadro's constant. It is the "chemist's dozen": just as a dozen always means 12 items, a mole always means particles, regardless of the substance.
The mass of one mole of a substance is the molar mass , expressed in g/mol, numerically equal to the relative atomic or molecular mass. For example: means 1 mole of water (i.e., molecules) has a mass of 18 grams.
The fundamental relation linking mass (), amount of substance (), and number of particles ():
Scale intuition: is an immense number. If you had a mole of 1 cm³ marbles, they would cover the entire Earth's surface with a layer over 10 km deep! This is why we use the mole to weigh substances on laboratory scales (grams) while implicitly counting an enormous number of atoms.
Real-world applications: the mole is essential in every chemistry lab for preparing solutions of known concentration, dosing reactants, and interpreting reactions in quantitative terms.
The mass of one mole of a substance is the molar mass , expressed in g/mol, numerically equal to the relative atomic or molecular mass. For example: means 1 mole of water (i.e., molecules) has a mass of 18 grams.
The fundamental relation linking mass (), amount of substance (), and number of particles ():
Scale intuition: is an immense number. If you had a mole of 1 cm³ marbles, they would cover the entire Earth's surface with a layer over 10 km deep! This is why we use the mole to weigh substances on laboratory scales (grams) while implicitly counting an enormous number of atoms.
Real-world applications: the mole is essential in every chemistry lab for preparing solutions of known concentration, dosing reactants, and interpreting reactions in quantitative terms.
A balanced chemical equation respects the law of conservation of mass: the number of atoms of each element is the same on both sides of the arrow. The stoichiometric coefficients indicate the molar ratios between reactants and products.
Balancing: proceed by trial, starting with atoms that appear in the fewest species and leaving H and O for last. For more complex reactions, use the algebraic method (system of linear equations) or the ionic-electronic method (for redox reactions).
The limiting reagent is the one consumed first, determining the maximum amount of product obtainable. The other reactants are in excess and some will remain unreacted.
To identify the limiting reagent, compare the ratios (moles divided by stoichiometric coefficient) of each reactant: the smallest value corresponds to the limiting reagent. Think of it like a recipe: if a sandwich needs 2 slices of bread and 1 slice of cheese, and you have 10 slices of bread and 4 of cheese, the cheese is the limiting ingredient (you stop at 4 sandwiches).
The reaction yield measures efficiency: The actual yield is always lower than the theoretical yield due to side reactions, mechanical losses, and incomplete equilibria.
Real-world applications: industrially, identifying the limiting reagent optimizes costs (the cheaper reactant is used in excess). Yield calculation is crucial for evaluating process efficiency (e.g., pharmaceutical syntheses, the Haber-Bosch ammonia process).
Balancing: proceed by trial, starting with atoms that appear in the fewest species and leaving H and O for last. For more complex reactions, use the algebraic method (system of linear equations) or the ionic-electronic method (for redox reactions).
The limiting reagent is the one consumed first, determining the maximum amount of product obtainable. The other reactants are in excess and some will remain unreacted.
To identify the limiting reagent, compare the ratios (moles divided by stoichiometric coefficient) of each reactant: the smallest value corresponds to the limiting reagent. Think of it like a recipe: if a sandwich needs 2 slices of bread and 1 slice of cheese, and you have 10 slices of bread and 4 of cheese, the cheese is the limiting ingredient (you stop at 4 sandwiches).
The reaction yield measures efficiency: The actual yield is always lower than the theoretical yield due to side reactions, mechanical losses, and incomplete equilibria.
Real-world applications: industrially, identifying the limiting reagent optimizes costs (the cheaper reactant is used in excess). Yield calculation is crucial for evaluating process efficiency (e.g., pharmaceutical syntheses, the Haber-Bosch ammonia process).
Worked Examples
2Example 1Moles, molecules, and mass
Given
of water
Find
Number of moles
Number of molecules
Number of H atoms
Step-by-step solution
1To find the number of moles, divide the mass of water by its molar mass: . This means 36 g of water contain exactly 2 moles of H₂O molecules.
2The number of molecules is obtained by multiplying moles by Avogadro's constant: . This is an enormous number — roughly 2000 billion billion molecules!
3Since each water molecule contains 2 hydrogen atoms, multiply the number of molecules by 2: . Similarly, there are oxygen atoms (one per molecule).
✓ Final result: , molecules, H atoms
Example 2Limiting reagent
Given
Reaction:
,
Find
Limiting reagent
Moles of produced
Leftover excess reagent
Step-by-step solution
1To determine the limiting reagent, compare the ratio moles/stoichiometric coefficient () for each reactant. For nitrogen: . For hydrogen: . The smallest value indicates the reactant that will run out first → is the limiting reagent.
2Using the stoichiometric ratio , calculate the moles of ammonia producible from the available : .
3Calculate the moles of actually consumed: . Subtract from the initial amount to find the leftover: remaining unreacted.
✓ Final result: Limiting ; 3.33 mol produced; 0.33 mol left over
Exercises with Solutions
3Exercise 1Mole and molar massMedium
Problem to solve
How many moles and molecules are contained in of carbon dioxide ()? The molar mass is .
Given data
m = 88 gM = 44 g/mol
Step-by-step solution
1Calculate the number of moles by dividing mass by molar mass: .
2The number of molecules is obtained by multiplying moles by Avogadro's constant: molecules.
✓ Final answer: , molecules
Exercise 2Empirical formulaHard
Problem to solve
An unknown compound contains 40.0% C, 6.7% H, and 53.3% O by mass. Determine its empirical formula.
Given data
C 40.0%H 6.7%O 53.3%
Step-by-step solution
1Consider 100 g of compound: we have 40.0 g C, 6.7 g H, 53.3 g O. Calculate moles of each: C mol; H mol; O mol.
2Divide all values by the smallest (3.33) to get the integer ratio: C ; H ; O .
3The empirical formula is therefore . This is the formula of glucose (molecular , multiple ) and many other sugars.
✓ Final answer: Empirical formula (e.g. glucose: molecular )
Exercise 3Limiting reagent and yieldHard
Problem to solve
In the reaction , of hydrogen react with of oxygen. What is the theoretical mass of water? (, , g/mol)
Given data
m(H_2)=4 gm(O_2)=40 gM(H_2)=2, M(O_2)=32, M(H_2O)=18 g/mol
Step-by-step solution
1Calculate the initial moles: mol; mol.
2Determine the limiting reagent by comparing : : ; : . The smallest (1) indicates is limiting.
3From the stoichiometric ratio , the moles of water produced are mol. Convert to mass: of water.
✓ Final answer: limiting; theoretical water
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