Unit 5: Chemical calculations

Do simple chemical calculations involving reacting masses
1. we can use simple proportion to calculate the mass of a product formed if we have enough information
  1. ex. 1: A student obtains 48g of magnesium sulfate from 9.6g of magnesium. What mass of magnesium sulfate can the student get from 1.2g of magnesium?
    1. 9.6g magnesium gives 48g magnesium sulfate so 1.2g magnesium gives 1.2 / 9.6 * 48 = 6 answer: 6g magnesium sulfate
  2. ex. 2: In the reaction Mg + CuSO4 -----> MgSO4 + Cu 6.4g of copper are formed from 2.4g of magnesium. What mass of magnesium is needed to get 32g of copper?
    1. 6.4g copper is formed from 2.4g magnesium so 32g copper gives 32 / 6.4 * 2.4 = 12 answer: 12g magnesium
Define the Avogadro constant and the mole
1. Avogadro constant
  1. the number of defined particles (ions, atoms, molecules) in one mole of those particles
2. the mole
  1. the relative formula mass of a substance in grams
Understand how to use the mole concept to calculate reacting masses
1. You can find the number of moles of atoms, molcules or ions using the formula:
  1. number of moles = mass of substance taken (g) / mass of 1 mole of the substance (g)
  2. e.g. How many moles of water are there in 4.5g of water?
    1. water = H2O
    2. H = 1 O = 16
      1. 2*1 = 2g
      2. 1/16 = 16g
      3. 2 + 16 = 18
      4. 1 mole of water = 18g
    3. 4.5 / 18 = 0.25
      1. there are 0.25 moles of water in 4.5g of water
Calculate stoichiometric reacting masses
1. stoichiometry = the ratios of the reactants and the products shown in a balanced chemical equation
2. e.g. Calculate the mass of oxygen needed to react with 12g of magnesium
  1. 2Mg + O2 -----> 2MgO 2*24 + 2*16 -----> ?
  2. 48g magnesium reacts with 32g of oxygen 12g magnesium will react with 12 / 48 * 32 = 8g of oxygen
Understand how to do calculations involving the concept of limiting reactants
1. the limiting reactant is the one that is NOT in excess
2. ex. 1.2g of magnesium is reacted with a solution containing 2.74g of hydrochloric acid. Which is the limiting reactant?
  1. Mg + 2HCl -----> MgCl2 + H2
    1. 1*24 = 24g Mg
    2. 35.5 + 1 = 36.5g HCl
  2. number of moles of magnesium = 1.2 / 24 = 0.05 moles
  3. number of moles of hydrochloric acid = 2.74 / 36.5 = 0.075 moles
  4. from the equation: 1 mole of magnesium reacts with 2 moles of hydrochloric acid
  5. to react completely, 0.05 moles Mg will need to react with 2*0.05 = 0.1 mole of HCl. But we only have 0.075 moles of HCl, so the HCl is the limiting reactant.
Understand how to do calculations involving the molar gas volume
1. molar gas volume = the volume occupied by 1 mole of any gas - 24dm cubed at room temperature
2. volume of gas (dm cubed) = number of moles * 24
3. ex. Calculate the volume of 0.2 moles of carbon dioxide at room temperature and pressure
  1. vol. of gas = 0.2 * 24 = 4.8dm cubed
Calculate percentage yield and percentage purity
1. percentage yield
  1. actual yield / predicted yield * 100
    1. actual yield = the amount of product we get in a reaction
    2. predicted yield = is found using relative atomic masses and the stoichiometry of the equation
  2. e.g. A student reacts 9g of aluminium powder with excess chlorine. The mass of aluminium chloride produced is 35.6g. Calculate the percentage yield.
    1. the actual yield is 35.6g
    2. 2Al + 3Cl2 -----> 2AlCl3
      1. 1 mole of aluminium produces 1 mole of aluminium chloride
      2. 27 (Al) + (3*35.5)(Cl) = 133.5g = AlCl3
      3. 9g Al produces 9*133.5 / 27 = 44.5g AlCl3
      4. % yield = actual yield / predicted yield * 100
        35.6 / 44.5 * 100 = 80%
2. percentage purity
  1. mass of pure product / mass of impure product * 100
  2. e.g. A chemist made 60g of aspirin. Chemical analysis showed that this sample contained 58.5g of pure aspirin and 1.5g of impurities. Calculate the percentage purity of this aspirin sample.
    1. % purity = mass of pure product / mass of impure product * 100
      1. 58.5 / 60 * 100 = 97.5 pure
Calculate the empirical formula and molecular formula of a compound
1. empirical formula = a chemical formula that shows the simplest ratio of atoms in a compound
  1. ex. Analysis of a compound of tin (Sn) and chlorine (Cl) showed that the tin chloride contained 29.75g of tin an d35.5g of chlorine. Calculate the empirical formula of tin chloride.
    1. Relative atomic masses: Sn = 119 Cl = 35.5
    2. Step 1: Note the masses of each element
      1. Sn = 29.75g Cl = 35.5g
    3. Step 2: Divide by the relative atomic masses
      1. 29.75 / 119 = 0.25 moles 35.5 / 35.5 = 1.0 mole
    4. Step 3: Divide each number of moles by the lowest number of moles
      1. 0.25 / 0.25 = 1 1.0 / 0.25 = 4
    5. Step 4: Write the formula
      1. SnCl4
2. molecular formula = a formula showing the type and number of each element present in a molecule
  1. ex. A compound has the empirical formula CH2. Its relative formula mass is 84. Calculate its molecular formula.
    1. Relative atomic masses: C = 12 H = 1
    2. Step 1: Find the empirical formula mass
      1. 12 + (2*1) = 14
    3. Step 2: Divide relative formula mass by empirical formula mass
      1. 84 / 14 = 6
    4. Step 3: Multiply the empirical fomula by the number calculated in Step 2
      1. 6 * CH2 = C6H12
Understand how to do calculations involving solution concentration in mol/dm3 (cubed)
1. The concentration of a solution is the amount of solute dissolved in 1dm3 of solution. The units of concentration are therefore mol/dm3
2. concentration (mol/dm3) = number of moles of solute / volume of solution (dm3)
3. ex. Calculate the concentration in mol/dm3 of a solution of sodium hydroxide, NaOH, containing 4g of sodium hydroxide in 50cm3 of solution.
  1. Relatoive atomic masses: Na = 23 O = 16 H = 1
  2. Step 1: change grams to moles
    1. 4 / (23 + 16 + 1) = 0.1 mole NaOH
  3. Step 2: change cm3 to dm3
    1. 50 / 1000 = 0.05dm3
  4. Step 3: substitute into equation
    1. 0.1 / 0.05 = 2 mol/dm3 NaOH
Define relative atomic mass, relative molecular mass and relative formula mass
1. relative atomic mass
  1. the mass of an atom on a scale where an atom of carbon-12 weighs exactly 12 units
2. relative molecular mass
  1. the sum of all the relative atomic masses of all the atoms in a molecule
3. relative formula mass
  1. the sum of the relative atomic masses of all the atoms or ions in a compound

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Unit 5: Chemical calculations

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