Formulae, Equations, and Amounts of Substances
Writing Formulas
Writing chemical formulas can be done effectively using valencies or charges on ions. In the method using valencies, one should begin by writing the symbols of the elements and then place the valency below each symbol before crossing the valencies. For example, to write the formula for water, H₂O, we consider that hydrogen needs one electron and oxygen needs two.
Formula & Charges of Common Polyatomic Ions
Common polyatomic ions include Ammonium (NH4^+) with a valency of 1, Hydroxide (OH^-) also with a valency of 1, Nitrate (NO3^-) with a valency of 1, Hydrogencarbonate (HCO3^-) with a valency of 1, Carbonate (CO3^{2-}) with a valency of 2, Sulfate (SO4^{2-}) with a valency of 2, and Phosphate (PO4^{3-}) with a valency of 3.
Writing Balanced Chemical Equations
To write balanced chemical equations, one must first write the correct symbols and formulas, then balance the equation to ensure that the number of atoms of each element in the reactants equals the number of atoms in the products. Finally, it's important to add state symbols: (s), (l), (g), or (aq).
Examples of Balancing Equations
Examples of balanced equations include: Ca(NO3)2 (aq) + Na2SO4 (aq) \rightarrow CaSO4 (s) + 2NaNO3 (aq), C3H8 (g) + 5O2 (g) \rightarrow 3CO2 (g) + 4H2O (g), and N2 (g) + 3H2 (g) \rightarrow 2NH3 (g).
Ionic Equations
The process of writing ionic equations involves writing a balanced full equation, replacing ionic compounds with their separate ions, deleting identical ions on both sides, and finally writing the net ionic equation.
Types of Chemical Reactions & Observations
1- Displacement Reactions
Displacement reactions occur when one element replaces a less reactive element in a compound. An example of metal displacement in aqueous solutions is Mg (s) + CuSO4 (aq) \rightarrow MgSO4 (aq) + Cu (s), where the observation is that the blue color of the solution fades and magnesium turns brown. An example of a thermite reaction is 2Al (s) + Fe2O3 (s) \rightarrow Al2O3 (s) + 2Fe (s). Metal displacement with acids can be exemplified by Mg (s) + 2HCl (aq) \rightarrow MgCl2 (aq) + H2 (g), where bubbles of hydrogen gas are observed, and solid metal disappears. Halogen displacement can be illustrated by Cl2 (aq) + 2KBr (aq) \rightarrow 2KCl (aq) + Br2 (aq), where chlorine is reduced, bromide ions are oxidized, resulting in a color change from colorless to orange-brown.
2- Precipitation Reactions
Precipitation reactions occur when an insoluble solid is formed upon mixing substances in solutions. For instance, in the reaction CO2 (g) + Ca(OH)2 (aq) \rightarrow CaCO3 (s) + H2O (l), lime water turns milky or cloudy indicating a test for CO2. Another example is AgNO3 (aq) + NaCl (aq) \rightarrow AgCl (s) + NaNO3 (aq), used to test for halides by adding silver ions. Ba(NO3)2 (aq) + Na2SO4 (aq) \rightarrow BaSO4 (s) + 2NaNO_3 (aq) tests for sulfate ions by adding barium ions, resulting in a white precipitate.
3- Neutralization Reactions
Neutralization reactions are characterized by the reaction between acids and bases. For instance, KOH (aq) + HNO3 (aq) \rightarrow KNO3 (aq) + H2O (l) sees an increase in temperature in an exothermic reaction, while CuO (s) + H2SO4 (aq) \rightarrow CuSO4 (aq) + H2O (g) results in a formed solution. Furthermore, reactions of acids with metal carbonates or hydrogen carbonates, such as Li2CO3 + 2HCl \rightarrow 2LiCl + H2O + CO2 and NaHCO3 + HCl \rightarrow NaCl + H2O + CO2 produce bubbles of carbon dioxide gas; notably, these reactions are not redox reactions.
Relative Atomic Mass (Ar) & Relative Molecular Mass (Mr)
Carbon-12 serves as the standard atom, with a mass of 12 atomic mass units. The relative atomic mass (Ar) is the weighted average mass of atoms of an element compared to 1/12 the mass of an atom of Carbon-12 derived from the periodic table. The relative molecular mass (Mr) is the sum of relative atomic masses of all atoms in a covalent molecule. For giant structures, we reference the relative formula mass. For example, Cl2 has a Mr of 35.5 x 2 = 71, and CaCO3 has a Mr of 40 + 12 + (16 x 3) = 100. For hydrated compounds like CuSO4 . 5H2O, it calculates as 64 + 32 + (16 x 4) + (1 x 10) + (16 x 5) = 250, while Fe2O3 . 3H_2O calculates to (56 x 2) + (16 x 3) + (1 x 6) + (16 x 3) = 214.
Molar Mass (M)
Molar mass is defined as the mass per mole of a substance in grams per mole (g mol$^{-1}$) and is equal to Ar or Mr expressed in grams.
Moles
The term moles refers to the SI unit for the amount of substance (mol), which signifies an amount of substance that contains the same number of particles as 12 g of C-12. The mass of one mole corresponds to the relative atomic mass or molecular mass expressed in grams, with 1 mole constituting 6.02 imes 10^{23} particles (Avogadro's constant). The corresponding formula is n = \frac{m}{M}. Examples illustrate that 320 g of Calcium leads to n = 320 / 40 = 8 mol, and for 80 g of CuSO4, it leads to n = 80 / (64 + 32 + 4 x 16) = 0.5 mol, while the mass of 6 moles of CaCO3 computes as m = 6 x (40 + 12 + 16 x 3) = 600 g.
Avogadro's Constant
Avogadro's constant quantifies the number of particles in one mole, being 6.02 imes 10^{23}. In practical terms, 1 mole of He equals 6.02 imes 10^{23} He atoms, while 1 mole of O2 equals 6.02 imes 10^{23} oxygen molecules. Moreover, 1 mole of O2 contains 2 imes 6.02 imes 10^{23} oxygen atoms; similarly, 1 mole of NaCl comprises 6.02 imes 10^{23} sodium ions and 6.02 imes 10^{23} chloride ions, totaling up to 2 imes 6.02 imes 10^{23} ions.
Calculating Reacting Masses From Equations
When calculating reacting masses, begin by determining the moles of a known substance using n = \frac{m}{M}. Subsequently, the equation ratio is employed to find moles of the required substance, and finally, the mass of this substance can be calculated using the formula m = n \times M. For example, in the equation SO3 + H2O \rightarrow H2SO4, to form 75.0 g of H2SO4, calculate moles of H2SO4 as 75 / 98.1 = 0.765 mol, resulting in moles of SO3 as 0.765 mol (1:1 ratio). Mass of SO3 is thus computed as 0.765 x 80.1 = 61.3 g.
Working Out Equations From Reacting Masses
To derive equations from reacting masses, start by writing down the molar mass of each substance, calculate the moles of each substance with n = \frac{m}{M}, and then divide each mole quantity by the smallest number to establish a whole number ratio. For instance, if 1.2 g C reacts with 16.2 g ZnO to yield 4.4 g CO2 and 13 g Zn, you find moles are: C = 0.1, ZnO = 0.2, CO2 = 0.1, and Zn = 0.2, yielding the ratio 1 : 2 : 1 : 2, corresponding to the equation C + 2ZnO \rightarrow CO_2 + 2Zn.
Percentage Yield & Percentage Purity
Yield represents the amount of product created, with theoretical yield being the maximum possible mass (as calculated) and actual yield being the actual mass obtained (as measured). The percentage yield can be calculated with the formula \% \text{ Yield } = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100.
Atom Economy
Atom economy measures the percentage of atoms from the starting materials that end up in the desired product. Addition reactions achieve a 100% atom economy, while elimination and substitution reactions typically have lower atom economies. Multistep reactions may exhibit even lower atom economies. The formula for atom economy is \text{Atom economy} = \frac{\text{molar mass of the desired product}}{\text{Sum of molar masses of all products}} \times 100.
Molar Gas Volume
The molar gas volume refers to the volume occupied by one mole of any gas at room temperature and pressure, measuring 24 dm³ or 24,000 cm³. Thus, the molar volume (Vm) is noted as 24 dm³ mol$^{-1}$ at r.t.p. or 24,000 cm³ mol$^{-1}$ at r.t.p.
The Ideal Gas Equation
The Ideal Gas Equation is represented as pV = nRT, where p denotes pressure in pascals (Pa), V is the volume in cubic meters (m³), T represents the temperature in kelvin (K), n is the amount of substance in moles (mol), and R is the gas constant (8.31 Jmol$^{-1}$ K$^{-1}$).
Concentration of Solutions
Concentration can be expressed as mass concentration, which is the mass of solute per volume of solution (g dm$^{-3}$), or molar concentration, defined as the number of moles per volume (mol dm$^{-3}$). This can be calculated with n = c \times V.
Concentration in ppm
Concentration in parts per million (ppm) indicates the number of parts of one substance in one million parts of another. For solutions, it is computed with concentration in ppm = (mass of solute / mass of solvent) x 10$^{6}$, and for gases, it is concentration in ppm = (volume of gas / volume of air) x 10$^{6}$.
Finding Empirical & Molecular Formula
The molecular formula reveals the actual number of atoms of each element within a molecule, while the empirical formula presents the simplest whole number ratio of atoms for each element in a compound. The relationship between molecular and empirical formulas is defined by the equation: Molecular Formula = Empirical Formula x (Mr of the compound / Mr of the empirical formula).
Finding Empirical Formula using combustion analysis
To determine empirical formulas through combustion analysis, one must follow several steps: calculate the mass of carbon and hydrogen derived from the carbon dioxide and water produced, calculate the mass of oxygen, and subsequently compute the empirical formula. The molecular formula can then be identified as long as the Mr of the compound is known.