Chemistry calculations

A quick look at the calculating side of TITRATION, MOLES and PERCENTAGE YIELD 

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  • Chemistry calculations
    • Moles
    • Titration
      • Reacting an acid with a base to reach a point of neutralisation
      • Using the relationship between concentration and volume
  • Moles
  • Titration
    • Reacting an acid with a base to reach a point of neutralisation
    • Using the relationship between concentration and volume
  • Base unit = mole (n)
    • "The amount of substance that contains as many particles as there are atoms in 12g of carbon-12"
      • n (number of moles) = mass (in grams) divided by Mr (relative atomic mass)
        • Base unit = mole (n)
        • Requires a balanced chemical equation
          • Same number of each molecule on either side of the equation
        • Same number of each molecule on either side of the equation
        • Percentage yield
          • Chemistry calculations
            • = actual yield divided by theoretical yield x 100%
              • Example: 9.3g (actual) divided by 15.4 (theoretical) x 100 = 60.4%
              • x 100 to make it into a percentage
            • Actual yield = the mass of the product at the end of the experiment
          • Theorectical yield = the predicted yield, calculated using reactants and an equation
            • Actual yield = the mass of the product at the end of the experiment
          • Example: n = 15.4g of CO2 divided by 44 (Mr of CO2) = 0.35 mol
            • n (number of moles) = mass (in grams) divided by Mr (relative atomic mass)
            • Example: If I have ONE mole of oxygen I have 16g ( 1 mole x 16.0 Mr)
              • Concentration x volume = concentration x volume
                • Example: 10cm3 of 1M HCl + 10cm3 of ?M NaOH = NaCl + H2O
                  • Concentration x volume = concentration x volume
                    • 10 x 1 = 10 x ?
                  • Concentration of NaOH = 1M
                    • 10 x 1 = 10 x ?

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