# 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 ?