Unit 5: Chemical calculations

Unit 5: Chemical calculations

1. Do simple chemical calculations involving reacting masses

   • we can use simple proportion to calculate the mass of a product formed if we have enough information
     • 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?
       • 9.6g magnesium gives 48g magnesium sulfate so 1.2g magnesium gives 1.2 / 9.6 * 48 = 6 answer: 6g magnesium sulfate
     • 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?
       • 6.4g copper is formed from 2.4g magnesium so 32g copper gives 32 / 6.4 * 2.4 = 12 answer: 12g magnesium

2. Define the Avogadro constant and the mole

   • Avogadro constant
     • the number of defined particles (ions, atoms, molecules) in one mole of those particles
   • the mole
     • the relative formula mass of a substance in grams

3. Understand how to use the mole concept to calculate reacting masses

   • You can find the number of moles of atoms, molcules or ions using the formula:
     • number of moles = mass of substance taken (g) / mass of 1 mole of the substance (g)
     • e.g. How many moles of water are there in 4.5g of water?
       • water = H2O
       • H = 1 O = 16
         • 2*1 = 2g
         • 1/16 = 16g
         • 2 + 16 = 18
         • 1 mole of water = 18g
       • 4.5 / 18 = 0.25
         • there are 0.25 moles of water in 4.5g of water

4. Calculate stoichiometric reacting masses

   • stoichiometry = the ratios of the reactants and the products shown in a balanced chemical equation
   • e.g. Calculate the mass of oxygen needed to react with 12g of magnesium
     • 2Mg + O2 -----> 2MgO 2*24 + 2*16 -----> ?
     • 48g magnesium reacts with 32g of oxygen 12g magnesium will react with 12 / 48 * 32 = 8g of oxygen

5. Understand how to do calculations involving the concept of limiting reactants

   • the limiting reactant is the one that is NOT in excess
   • ex. 1.2g of magnesium is reacted with a solution containing 2.74g of hydrochloric acid. Which is the limiting reactant?
     • Mg + 2HCl -----> MgCl2 + H2
       • 1*24 = 24g Mg
       • 35.5 + 1 = 36.5g HCl
     • number of moles of magnesium = 1.2 / 24 = 0.05 moles
     • number of moles of hydrochloric acid = 2.74 / 36.5 = 0.075 moles
     • from the equation: 1 mole of magnesium reacts with 2 moles of hydrochloric acid
     • 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.

6. Understand how to do calculations involving the molar gas volume

   • molar gas volume = the volume occupied by 1 mole of any gas - 24dm cubed at room temperature
   • volume of gas (dm cubed) = number of moles * 24
   • ex. Calculate the volume of 0.2 moles of carbon dioxide at room temperature and pressure
     • vol. of gas = 0.2 * 24 = 4.8dm cubed

7. Calculate percentage yield and percentage purity

   • percentage yield
     • actual yield / predicted yield * 100
       • actual yield = the amount of product we get in a reaction
       • predicted yield = is found using relative atomic masses and the stoichiometry of the equation
     • 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.
       • the actual yield is 35.6g
       • 2Al + 3Cl2 -----> 2AlCl3
         • 1 mole of aluminium produces 1 mole of aluminium chloride
         • 27 (Al) + (3*35.5)(Cl) = 133.5g = AlCl3
         • 9g Al produces 9*133.5 / 27 = 44.5g AlCl3
         • % yield = actual yield / predicted yield * 100
           • 35.6 / 44.5 * 100 = 80%
   • percentage purity
     • mass of pure product / mass of impure product * 100
     • 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.
       • % purity = mass of pure product / mass of impure product * 100
         • 58.5 / 60 * 100 = 97.5 pure

8. Calculate the empirical formula and molecular formula of a compound

   • empirical formula = a chemical formula that shows the simplest ratio of atoms in a compound
     • 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.
       • Relative atomic masses: Sn = 119 Cl = 35.5
       • Step 1: Note the masses of each element
         • Sn = 29.75g Cl = 35.5g
       • Step 2: Divide by the relative atomic masses
         • 29.75 / 119 = 0.25 moles 35.5 / 35.5 = 1.0 mole
       • Step 3: Divide each number of moles by the lowest number of moles
         • 0.25 / 0.25 = 1 1.0 / 0.25 = 4
       • Step 4: Write the formula
         • SnCl4
   • molecular formula = a formula showing the type and number of each element present in a molecule
     • ex. A compound has the empirical formula CH2. Its relative formula mass is 84. Calculate its molecular formula.
       • Relative atomic masses: C = 12 H = 1
       • Step 1: Find the empirical formula mass
         • 12 + (2*1) = 14
       • Step 2: Divide relative formula mass by empirical formula mass
         • 84 / 14 = 6
       • Step 3: Multiply the empirical fomula by the number calculated in Step 2
         • 6 * CH2 = C6H12

9. Understand how to do calculations involving solution concentration in mol/dm3 (cubed)

   • The concentration of a solution is the amount of solute dissolved in 1dm3 of solution. The units of concentration are therefore mol/dm3
   • concentration (mol/dm3) = number of moles of solute / volume of solution (dm3)
   • ex. Calculate the concentration in mol/dm3 of a solution of sodium hydroxide, NaOH, containing 4g of sodium hydroxide in 50cm3 of solution.
     • Relatoive atomic masses: Na = 23 O = 16 H = 1
     • Step 1: change grams to moles
       • 4 / (23 + 16 + 1) = 0.1 mole NaOH
     • Step 2: change cm3 to dm3
       • 50 / 1000 = 0.05dm3
     • Step 3: substitute into equation
       • 0.1 / 0.05 = 2 mol/dm3 NaOH

10. Define relative atomic mass, relative molecular mass and relative formula mass

    • relative atomic mass
      • the mass of an atom on a scale where an atom of carbon-12 weighs exactly 12 units
    • relative molecular mass
      • the sum of all the relative atomic masses of all the atoms in a molecule
    • relative formula mass
      • the sum of the relative atomic masses of all the atoms or ions in a compound