The Atom and Relative Mass

• This topic focuses on counting particles that are too small to see, like atoms and molecules
• Measurements like mass and volume are used to count microscopic particles
• Reactions occur in specific ratios to form products, a concept known as stoichiometry

The Atom

• Atoms are the smallest part of an element
• 10^15 atoms are required to cover the head of a pin, according to JGR Briggs
• Masses of atoms are minuscule and measured in grams (g)
• Mass of the most abundant 'H: 1.66 x 10^-24 g
• Mass of the most abundant 12C: 1.99 x 10^-23 g
• Relative masses are used in calculations instead of actual masses due to the small size of atoms
• 1 atomic mass unit is assigned to the most abundant hydrogen atom mass of the most abundant carbon atom / mass of the most abundant hydrogen atom = 12.0

Key Definitions in Relative Mass

• One atom of carbon-12 has 12 atomic mass units
• All other atomic masses are computed by comparing them to 1/12 the mass of one atom of carbon-12
• The relative molecular mass (Mr) is the ratio of the average mass of one molecule of a substance to 1/12 the mass of a carbon-12 isotope, expressed on the carbon-12 scale.
• The relative formula mass (Mr) is the ratio of the average mass of one formula unit of the compound to 1/12 the mass of a carbon-12 isotope, expressed on the carbon-12 scale
• The relative isotopic mass (Ar) is the ratio of the mass of one atom of the isotope to 1/12 the mass of a carbon-12 isotope, expressed on the carbon-12 scale
• The relative atomic mass (Ar) is the ratio of the average mass of one atom of the element to 1/12 the mass of a carbon-12 isotope, expressed on the carbon-12 scale
• Isotopes of the same element have the same number of protons but differ in the number of neutrons
• The average mass is the weighted average of the relative isotopic masses of the isotopes according to their abundance

Calculating Relative Atomic Masses

• Use the percentages of the isotopes multiplied by the mass, divided by 100
• For Chlorine, Relative atomic mass of Cl = (35 x 3/4) + (37 x 1/4) = 35.5

Molecular and Empirical Formulae

• The molecular formula gives the definitive amount of atoms of each element in one molecule of a compound
• The empirical formula shows the atoms of elements in the simplest ratio
• H2O is the molecular formula and empirical formula
• Ethanoic acid (CH3CO2H / C2H4O2) is the molecular formula and the empirical formula is CH2O
• Glucose (C6H12O6) is the molecular formula and the empirical formula is CH2O
• To find the molecular formula, extra information such as the Mr is required

Calculating Empirical Formula from Experimental Data

• Finding the ratio of elements present allows deducing the empirical formula
• The mass is taken as a percentage, assuming the mass of compound is 100 g
• The mass is divided by the Ar of the element
• This gives the mole ratio
• Divide through by the smallest number to achieve the simplest whole number that represents the empirical formula

Moles and the Avogadro Constant

• Particles are too small and too many to count efficiently, so chemists use moles
• One mole of substance contains 6.02 x 10^23 (Avogadro constant) particles
• A mole is equal to 6.02 × 10^23 items, equivalent to a dozen equalling 12 items

One mole contains exactly 6.02 x 10^23 (Avogadro constant) elementary entities, broken down as:

• Na has a No. of atoms = 6.02 x 10^23
• S8 has a No. of molecules = 6.02 × 10^23 and No. of S atoms = 8 x 6.02 × 10^23
• MgCl2 has a No. of Mg2+ ions = 6.02 × 10^23 and No. of Cl- ions = 2 × 6.02 × 10^23

Molar Mass

• Molar mass (M) is the mass of one mole of substance with a unit of g mol-1
• Molar mass of element = Relative atomic mass in g mol-1
• Molar mass of compound = Relative molecular mass or Relative formula mass in g mol-1
• Molar mass, M, has a unit of g mol-1, while Mr and Ar have no units
• Molar mass of SO2 = 32.1 + 2 (16.0) = 64.1 g mol-1, Mr of SO2 = 64.1
• Molar mass of Na2CO3 = 2(23.0) + 12.0 + 3(16.0) = 106.0 g mol¯¹, Mr of Na2CO3 = 106.0

Molar Volume

• Equal volumes of all gases, under the same conditions of temperature and pressure, contain the same number of particles
• This is irrespective of the identity of the gas
• 1 mol of any gas occupies 22.7 dm3 at standard temperature pressure (s.t.p.) or 24.0 dm3 at room temperature and pressure (r.t.p.).
• This is known as the molar volume of gas
• 1 cm3 = 1 x 10–3 dm3 = 1 x 10–6 m3

Definitions: - r.t.p (room temperature and pressure) has 293 K (20 °C) and 1 atm, with a Molar volume of gas of 24.0 dm3 - s.t.p (standard temperature and pressure) has 273 K (0 °C) and 1 bar, with a Molar volume of gas of 22.7 dm3

Key Molar Relationships

If X is dissolved in an aqueous solution:

• concentration (in mol dm−3)  volume (in dm3) If X is a gas:
• Has a relationship between the volume of gas (in dm3,) and the number of moles
• Volume of gas (in dm3) at r.t.p. (20°C and 1 atm) is 24.0 dm3
• Volume of gas (in dm3) at s.t.p. (0°C and 1 bar) is 22.7 dm3

Chemical Reactions

Key points for chemical equations:

• Do NOT alter formula of compound/ions
• Balance only by changing the numbers (coefficient) in front of compounds/ions/elements
• Total atoms of each element on both side of the equation must be the same, for conservation of mass

Examples:

• Thermal Decomposition: CaCO3(s) → CaO(s) + CO2(g)
• Combustion: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
• Acid – base reaction: H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
• Acid – carbonate reaction: 2HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l)
• Ammonium salt − base reaction: NH4Cl(aq) + NaOH(aq) → NaCl(aq) + NH3(g) + H2O(l)
• Precipitation of insoluble salts: Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq)
• *Acid – metal reaction: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
• Displacement reaction: Mg(s) + Zn2+(aq) → Mg2+(aq) + Zn(s), and Cl2(g) + 2I−(aq) →2Cl −(aq) + I2(s)
• Redox reactions: MnO4−(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

Limiting Reagents

• In a chemical reaction, reagents can be in excess, in which case they aren't completely used up
• The amount of product that forms is determined by the limiting reagent which is completely used up, which stops the reaction

Calculations: no. of moles of O2 required = 0.2 / 4 × 5 = 0.25,

• Since no. of moles of O2 available is more than required, O2 is in excess and NH3 is the limiting reagent
• No. of moles of O2 left unreacted = 0.30 – 0.25 = 0.05 mol

Percentage Yield

actual mass of product / theoretical mass

Counting invisible particles such as atoms and molecules using mass and volume measurements. Stoichiometry explains reactions forming products in specific ratios. Relative masses are used in calculations instead of actual masses due to the small size of atoms.