Quantitive chemistry

• Law of conservation of mass:

  • No atoms are lost or made during a chemical reaction
  • Mass of the products equals the mass of the reactants
• Balanced equation of magnesium reacting with hydrochloric acid:

  Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g)
• Define relative atomic mass (RAM) and relative formula mass (RFM):

  • RAM: Average mass of atoms in an element considering masses and abundance of isotopes, relative to 12C
  • RFM: Sum of RAMs of all atoms in the formula
• Relative formula mass of:

  1. CaF2 - 78
  2. C6H12O6 - 180
• Explanation for carbon dioxide and water escaping from the test tube:

  • They are both gases
• Mean mass of magnesium produced:

  (3.3 + 3.5 + 3.2) / 3 = 3.3

  To increase precision:

  • Measure to more decimal places
  • Use a more sensitive balance/apparatus
• Avogadro's constant:

  • Number of atoms, molecules, or ions in a mole of a substance
  • Value is 6.02 x 10^23
• Formula linking mass, molecular mass, and moles:

  Mass = Mr x Moles
• Mass of 20 moles of calcium carbonate (CaCO3):

  Mass = Mr x Moles

  Mr = 100

  100 x 20 = 2000 g
• Amount of carbon dioxide in moles in 0.32 g:

  Moles = Mass / Mr

  0.32 / 44 = 0.007
• Mass of nitrogen needed to form 6.8 tonnes of ammonia:

  Step 1: Moles of ammonia = 400000

  Step 2: Moles of nitrogen = 200000

  Step 3: Mass of nitrogen = 5.6 tonnes
• Limiting reactant in a chemical reaction:

  • Involves two reactants
  • One reactant is used in excess to ensure complete reaction of the other
  • The reactant that is completely used up is the limiting reactant
• Grams of oxygen gas given off from 40.8 g of hydrogen peroxide:

  Step 1: Balanced equation: 2 H2O2 (l) → 2 H2O + O2 (g)

  Mr of H2O2 = 34

  Step 2: Calculate moles and mass of oxygen gas
• Oxygen will be given off from 40.8 g of hydrogen peroxide
• Balanced equation: 2 H2O2(l) → 2 H2O + O2(g)
• Number of moles in 40.8 g: 40.8/34 = 1.2 moles
• Ratio in the balanced equation of H2O2 : O2 = 2:1
• Therefore, number of moles of O2 = 0.6 moles
• Mass of oxygen = 0.6 x 32 (Mr of O2) = 19.2
• Two formulae that link concentration, mass, and volume together:
• Concentration (g per dm3) = Mass (g) / Volume (dm3)
• Concentration (mol per dm3) = nr of moles / volume (dm3)
• 31.0 cm3 of potassium hydroxide solution neutralised 25.0 cm3 of 2.0 moldm−3 nitric acid
• Calculate the concentration of the potassium hydroxide solution in moldm−3
• Calculate the moles of HNO3 used = Concentration x volume; 2 x 0.025 dm3 = 0.05 moles
• Calculate the moles of KOH; Ratio is 1:1, therefore number of moles of KOH = 0.05
• Calculate the concentration of KOH; Volume = Moles / concentration; 0.05 / 0.031 = 1.61
• Molar volume of a gas at room temperature and pressure:
• 1 mole of a gas at room temperature and pressure occupies 24 dm3
• Titration:
• A technique for finding the concentration of a solution by reacting a known volume of this solution with a solution of known concentration
• How to conduct a titration:
• Rinse the pipette with a solution of unknown concentration and measure out the known volume of this solution
• Add an indicator (a substance that changes color at the end of titration)
• Rinse the burette with a solution of known concentration, discard the liquid, and gradually add the solution of a known concentration
• When the indicator changes color (at the end point), record the volume added
• Perform suitable calculations to find the concentration
• Why is it not always possible to obtain the theoretical amount of product in a chemical reaction?
• The reaction may not go to completion because it is reversible
• Some of the product may be lost when it is separated from the reaction mixture