# Chemistry calculations

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

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- Created by: Joseph101 - Team GR
- Created on: 21-10-13 11:26

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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)

- 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

- Chemistry calculations
- 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)

- 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 x volume = concentration x volume
- Concentration of NaOH = 1M
- 10 x 1 = 10 x ?

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