Fundamentals of Chemical Reactions | General Chemistry 2

Chemical reactions:

A chemical reaction is a process in which one or more substances (the reactants) are converted to one or more different substances (the products). According to Dalton's atomic theory, chemical reactions cause the rearrangement of atoms, but do not cause either the creation or the destruction of atoms.

Chemical equations:

A chemical reaction is represented by a chemical equation. Each species to the left of the arrow is a reactant, and each species to the right is a product. The structure of a chemical equation is as follows:

• The chemical formulas of the reactants are on the left-hand side.
• The chemical formulas of the products are on the right-hand side.
• Reactants and products are separated by an arrow.
• The chemical formula of each substance is separated from the others by a '+' sign.
• Physical states can be specified in parentheses after the chemical formula of each substance as (s), (l), (g), or (aq) for solid, liquid, gas, and aqueous (dissolved in water), respectively.

Mg (s) + Cl₂ (g) → MgCl₂ (s)

Interpreting coefficients:

Coefficients in chemical equations can be interpreted as the number of molecules or the number of moles of a substance produced or consumed during the reaction.

3 H₂ + N₂ → 2 NH₃

Molecular interpretation: 3 molecules of H₂ react with 1 molecule of N₂ to form 2 molecules of NH₃.
Molar interpretation: 3 moles of H₂ react with 1 mole of N₂ to form 2 moles of NH₃.

Law of conservation of mass:

The Law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. This principle, first formulated by Antoine Lavoisier in 1789, implies that the total mass of the reactants is equal to the total mass of the products.
 

Microscopic and Macroscopic Views:

• Microscopic level: On the atomic level, atoms are simply rearranged to form new molecules, and no atoms are lost or gained.
• Macroscopic level: On a larger scale, the mass measured before and after the reaction in a closed system remains constant.

 If 10 grams of reactant A reacts with 15 grams of reactant B, the total mass of the products will be 25 grams.

Balancing chemical equations:

Balancing chemical equations involves making sure that the number of atoms for each element is the same on both sides of the equation, thereby satisfying the Law of conservation of mass. It can be done by changing the coefficients of the reactants and/or products, but never by changing their formulas.

Balance the following chemical equation: Li + Br₂ → LiBr

2 Li + Br₂  → 2 LiBr   (balanced)
2 Li + Br₂  → LiBr₂    (incorrect: the nature of the product is different ⇒ LiBr₂ ≠ LiBr)

How to balance chemical equations:

1. Write the unbalanced chemical equation.

2. Count the atoms of each element on both sides of the equation.

3. Change the coefficients (numbers in front of molecules) to balance the atoms for each element. Do not change the subscripts (numbers within molecules).

4. Balance the oxygen or hydrogen atoms last.

5. Simplify the coefficients if possible to make sure they are the smallest set of whole numbers.

6. Double-check the balance: Recount the atoms of each element to confirm the equation is balanced. 

Combination Reactions:

• Two or more reactants combine to form a single product.
• General form: A + B $\to$ AB
   

2 Na (s) + Cl₂ (g) → 2 NaCl (s)
CO₂ (g) + H₂O (l) → H₂CO₃ (aq)

Decomposition reactions:

• A single reactant breaks down into two or more products.
• General form: AB  →  A + B
   

CaCO₃ (s) → CaO (s) + CO₂ (g)
MgSO₄ (s) → MgO (s) + SO₃ (g)

Combustion reactions:

• A substance reacts with oxygen, releasing energy in the form of light and heat, and producing carbon dioxide and water if the substance is organic.
• General form: Fuel + O₂ → CO₂ + H₂O
   

CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O (g)
2 C₂H₆ (g) + 7 O₂ (g) → 4 CO₂ (g) + 6 H₂O (g)

Single displacement reactions:

• One element displaces another element in a compound, forming a new compound and releasing the displaced element.
• General form: A + BC → AC + B
   

Br₂ (l) + CaI₂ (aq) → CaBr₂ (aq) + I₂ (s)
Fe (s) + H₂SO₄ (aq) → FeSO₄ (aq) + H₂ (g)

Double displacement reactions:

• The ions of two compounds exchange places in an aqueous solution to form two new compounds.
• General form: AB + CD → AD + CB
   

NaCl (aq) + AgNO₃ (aq) → NaNO₃ (aq) + AgCl (s)
BaCl₂ (aq) + Na₂SO₄ (aq) → 2 NaCl (aq) + BaSO₄ (aq)

Empirical and molecular formula: 

• Empirical formula: The empirical formula of a compound gives the simplest whole-number ratio of the atoms of each element in the compound.
• Molecular formula: The molecular formula of a compound gives the actual number of atoms of each element in a molecule of the compound.
   

Molecule of glucose : 

Empirical formula = CH₂O
Molecular formula = C₆H₁₂O₆

How to determine the empirical formula:

1. Obtain the mass of each element present in the compound (usually given in grams).
2. Convert the mass of each element to moles using the molar mass of the element.
3. Divide the moles of each element by the smallest number of moles to obtain the simplest whole-number ratio.
    

A compound is found to contain 40.00% carbon, 6.72% hydrogen, and 53.28% oxygen by mass.

Convert to moles:

• Carbon: $\frac{40.00 \mathrm{g}}{12.01 \mathrm{g}.{\mathrm{mol}}^{-1}}$ = 3.33 moles
• Hydrogen: $\frac{6.72 \mathrm{g}}{1.01 \mathrm{g}.{\mathrm{mol}}^{-1}}$ = 6.65 moles
• Oxygen: $\frac{53.28 \mathrm{g}}{16.00 \mathrm{g}.{\mathrm{mol}}^{-1}}$ = 3.33 moles

Simplest ratio:

• Carbon: $\frac{3.33}{3.33}$ ​= 1
• Hydrogen: $\frac{6.65}{3.33}$ = 2
• Oxygen: $\frac{3.33}{3.33}$ ​= 1

Empirical formula: CH₂O

How to determine the molecular formula:

1. Determine the empirical formula.
2. Calculate the empirical formula mass (EFM), which is the sum of the atomic masses of all atoms in the empirical formula.
3. Divide the molar mass of the compound by the empirical formula mass to obtain the multiplication factor (n).
4. Multiply the subscripts in the empirical formula by the factor n to obtain the molecular formula.
    

A compound has an empirical formula CH₂O and a molar mass of 180.16 g/mol.

• EFM: 12.01 + (2 x 1.01) + 16.00 = 30.02 g/mol
• Factor n: $\frac{180.16 \mathrm{g}.{\mathrm{mol}}^{-1}}{30.02 \mathrm{g}.{\mathrm{mol}}^{-1}}$ ​= 6
• Molecular formula: C₆H₁₂O₆

Combustion analysis:

The combustion analysis is an analysis used to determine the empirical formula of an organic compounds by burning the compound in excess oxygen and analyzing the products of the combustion.
 

Steps in combustion analysis:

1. The organic compound is burned in an excess of oxygen, ensuring complete combustion. The primary products of combustion for hydrocarbons and organic compounds are carbon dioxide (CO₂) and water (H₂O).
2. The CO₂ and H₂O produced are collected and measured. The masses of CO₂ and H₂O are determined, often using gravimetric or volumetric methods.
3. The amount of CO₂ produced is used to determine the moles of carbon in the original compound. The amount of H₂O produced is used to determine the moles of hydrogen in the original compound.
4. If the compound contains oxygen, the moles of oxygen can be determined by subtracting the mass of carbon and hydrogen from the total mass of the compound. 
5. The moles of each element are converted to the simplest whole-number ratio. This ratio provides the empirical formula of the compound.

• Mass of original compound = 1.00 g; Mass of CO₂ produced = 2.20 g; Mass of H₂O produced = 0.90 g
• Moles of C = $\frac{2.20 \mathrm{g}}{44.01 \mathrm{g}.{\mathrm{mol}}^{-1}}$ = 0.050 moles
• Moles of H = 2 x $\frac{0.90 \mathrm{g}}{18.02 \mathrm{g}.{\mathrm{mol}}^{-1}}$ = 0.100 moles
• Mass of C = 0.050 moles x 12.01 g/mol = 0.600 g
• Mass of H = 0.100 moles x 1.01 g/mol = 0.101 g
• Mass of O = 1.00 g - (0.600 g + 0.101 g) = 0.299 g
• Moles of O = $\frac{0.299 \mathrm{g}}{16.00 \mathrm{g}.{\mathrm{mol}}^{-1}}$ ​= 0.0187 moles
• Ratio C : H : O = $\frac{0.050}{0.0187}$ : $\frac{0.100}{0.0187}$ : $\frac{0.0187}{0.0187}$ = 2 : 6 : 1
• Empirical formula: C₂H₆O

How to determine empirical formula through combustion analysis:

1. Obtain mass of original compound and mass of CO₂ and H₂O produced.

2. Convert masses of CO₂ and H₂O to moles of C (= mole of CO₂) and H (= 2 x mole of H₂O).

3. Determine mass of O (= mass of original compound - mass of C - mass of H)

4. Convert mass of O to mole of O.

5. Calculate the simplest whole-number ratio of atoms of each element.

6. Write the empirical formula