CHEMISTRYCAFE

1.1 Relative Masses of Atoms and Molecules

• Isotopes are atoms of the same element that have the same number of protons but different number of neutrons.

• Relative atomic mass is the ratio of the average mass of one atom of the element to 1/12 the mass of an atom of Carbon-12 isotope.

• Relative isotopic mass is the ratio of the mass of an isotope to 1/12 the mass of an atom of Carbon-12 isotope.

• Relative molecular mass is the ratio of the mass of a molecule to 1/12 the mass of an atom of Carbon-12 isotope.

• Relative formula mass is the ratio of the average mass of one formula unit of an ionic compound to 1/12 the mass of a Carbon-12 isotope.

• Calculating relative atomic mass is based on the proportion of each isotope of an element in the sample.

• Ar = ∑ (Isotope Mass x Percentage Abundance)/100

 

 

1.2 Moles and Avogadro’s Constant

• One mole, n, of a substance contains as many particles as the number of atoms of carbon in 0.012 kg of carbon-12 isotope. The number is known as Avogadro’s constant, L and has a value of 6.02 x 10^23. Unit for moles: mol.

• Number of particles = nL

 

 

1.3 Formulas

• Molar mass of an atom/molecule is equal to its relative atomic mass or molecular mass.

• Number of Moles of an Element, n = (Mass of an element(g))/ (Molar Mass of Atom) (gmol^-1)

• Number of moles of a Molecule, n = (Mass of a molecule(g))/ (Molar Mass of Molecule) (gmol^-1)

• Number of moles, n = concentration x volume, V (Check units!)

• At r.t.p (293K and 1 atm), Number of mole, n = Volume of gas(dm^3)/24(dm^3)

• At s.t.p (273K and 1 bar), Number of mole, n = Volume of gas(dm^3)/22.7(dm^3)

​

 

1.3.1 Empirical and Molecular Formula

• Empirical Formula shows the ratio of the atoms of the different elements in the compound.

1. Change percentage composition to mass, m(g)

2. Divide the mass by its molar mass of atom or its Ar to get the number of moles.

3. Divide the number of moles f each element by the smallest number of moles to obtain the ratio.

• Assuming 100g of an organic compound,

​

​

​

​

​

​

​

​

​

​

​

​

The empirical formula of the compound is CH2O.

• Molecular formula shows the actual number of atoms of each element present in one molecular of a compound.

• Molecular formula = n (Empirical Formula) where n = 1,2,3.

• n= Relative Molecular Mass/ Relative Mass in the empirical formula

 

 

1.4 Combustion Analysis

• The molecular formula of hydrocarbons can be determined by complete combustion of oxygen.

• CxHy + (x + y/4) O2 = xCO2 + y/2 H2O

• An alkali is used to remove CO2. The number of moles of CO2 formed is equal to the number of moles of C in the hydrocarbons.

• H2O exists as liquid under r.t.p. The moles of H2O are equal to half of the moles of H in the hydrocarbon.

​

 

1.5 Reacting Masses and Volumes

• Reacting Masses

1. Limiting Reagent is the reagent that is completely used up at the end of the reaction and determines the yield of the reaction.

2. Percentage Yield = Actual Yield/ Theoretical Yield * 100%

3. Percentage Purity = Mass of Pure Substance/ Mass of Sample * 100%

• Volume of Gases (Repeated Earlier)

1. At r.t.p, Volume of gas(dm^3) = n * 24dm^3

2. At s.t.p, Volume of gas(dm^3) = n * 22.7 dm^3

• Concentrations of Solutions (Dilution)

1. Concentration of solution = n(mol)/V(dm^3)

2. In the dilution of a solution, number of moles of substance remains the same.

3. c1v1(before dilution) = c2v2(after dilution)

​

1.6 Volumetric Analysis

​

• Acid-Base Reactions

1. Acid-base titration is used for the quantitative analysis of acid-base reactions.

2. Equivalence point of a titration is reached when the reactants have just neutralized each other according to the stoichiometric ratio given by the balanced equation of the reaction.

3. End-point of a titration is reached when the indicator in the titration has just changed color.

​

​

​

​

​

​

​

​

​

​

​

• Sampling involves collection a portion from a given solution however concentration does not change but moles do.

• Back Titration

1.  Back titration is used when direct titration cannot be done

2. Sample X is reacted with known excess of reagent Y. The excess unreacted reagent Y can be titrated against reagent Z to determine the amount of unreacted Y. The amount of X can then be determined using stoichiometry.

• Double-Indicator Titration

1. Double-indicator titration is used when there are two stages of reaction and two indicators can be used to detect each stage of neutralization.

2. For example, a carbonate reacts with acid in two stages.

   1. Na2CO3 + HCl = NaHCO3 + NaCl

   2. NaHCO3 + HCl = NaCl + CO2 + H2O

   3. Overall: Na2CO3 + 2HCl = 2NaCl + CO2 + H2O

   4. Phenolphthalein and Methyl Orange are suitable indicators for stages 1 and 2 respectively.

​

1.7 Redox Reactions

​

​

​

​

​

​

​

​

​

​

​

​

​

​

• Oxidation state is the oxidation form that an atom is in.

• Oxidising agents are species that oxidise other species while being reduced.

• Reducing agents are species that reduce other species while being oxidised.

• Redox reactions are where both oxidation and reduction occur.

• The electrons given out during oxidation must equal to the number of electrons during reduction by conservation of charges.

​

1.7.1 Oxidation Number

​

​

​

​

​

​

​

​

​

​

​

​

​

​

​

​

​

​

​

​

​

1.8 Balancing Redox Reactions

​

​

​

​

​

​

​

​

​

​

​

​

​

​

​

• Disproportionation is a reaction where an element in a substance undergoes both oxidation and reduction to give two products.

• Comproportionation is the opposite with two reactants of the same element with two different oxidation states undergoes oxidation and reduction to produce one product with the same oxidation state.

​

© 2018 Shanmugam Udhaya All Rights Reserved