C3 Quantitative Chemistry

• Law of conservation of mass


  No atoms are lost or made during a chemical reaction so the mass of the products equals the mass of the reactants
• Balanced equation of magnesium reacting with hydrochloric acid


  Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
• Relative atomic mass (RAM)


  Average mass of atoms in an element taking into account masses and abundance of its isotopes, relative to 12C
• Relative formula mass (RFM)


  Sum of RAM's of all atoms in the formula
• Avogadro's constant


  The number of atoms, molecules or ions in a mole of a given substance. The value of the constant is 6.02 x 10^23.
• Limiting reactant


  In a chemical reaction involving two reactants, the reactant that is completely used up is called the limiting reactant because it limits the amount of products
• Concentration (g per dm3)

  Mass (g)/Volume (dm3)
• Concentration (mol per dm3)


  Number of moles/Volume (dm3)
• Molar volume of a gas at room temperature and pressure

  1 mole of a gas occupies 24 dm3
• Titration


  A technique for finding the concentration of a solution by reacting a known volume of this solution with a solution of known concentration
• Conducting a titration


  1. Rinse the pipette and measure out the known volume

  2. Add an indicator

  3. Rinse the burette and gradually add the solution of known concentration

  4. Record the volume added when the indicator changes colour

  5. Obtain concordant volume results

  6. Perform suitable calculations
• Reasons why it is not always possible to obtain the theoretical amount of product in a chemical reaction

  • The reaction may not go to completion because it is reversible
  • Some of the product may be lost when it is separated from the reaction mixture
  • Some of the reactants may react in ways different to the expected reaction (side reactions may occur)
• Percentage yield of a product in a chemical reaction


  Actual mass of a product / Maximum theoretical mass of product x 100%
• Atom economy


  A measure of the amount of starting materials that end up as useful products. It is a ratio of the relative formula mass of desired product to the sum of relative formula masses of reactants.
• What is the relative formula mass of: A) CaF2 B) B) C6H12O6

  CaF2 - (Ar values: Ca = 40, F = 19) 40 + 19 + 19 = 78

  C2H12O6 - (Ar values: C = 12, H = 1, O = 16) (12 x 6) + (1 x 12) + (16 x 6) = 180
• The following reaction occurs in a test tube under a Bunsen Burner:

  4 MgO(s) + CH4 (g) → 4 Mg(s) + 2 H2O(g) + CO2 (g)

  The carbon dioxide and water escape from the test tube.

  Use the equation to explain why.

  They are both gases
• Calculate the mean of magnesium produced and suggest how you could increase the precision of the results

  (3.3 + 3.5 + 3.2) / 3 = 3.3 Measure to more decimal places or use a more sensitive balance / apparatus

  
• What is the formula that links mass, molecular mass and moles together

  Mass = Mr x Moles
• What is the mass of: 20 moles of calcium carbonate, CaCO3


  Mass = Mr x Moles (Mr = 100) 100 x 20 = 2000 g
• Calculate the amount of carbon dioxide in moles in 0.32 g of carbon dioxide. Relative atomic masses (A r ): carbon = 12, oxygen = 16

  Moles = Mass / Mr 0.32 / 44 = 0.007
• N2(g) + 3 H2(g) ⇌ 2 NH3(g)

  Calculate the mass of nitrogen needed to form 6.8 tonnes of ammonia. Relative atomic masses (A r ): H = 1; N = 14


  Step 1 - Work out the number of number of moles of ammonia (Mr of ammonia = 17) 6800000 / 17 = 400000 moles of ammonia

  Step 2 - Use the balanced equation and number of moles of ammonia to work out the number of moles of nitrogen The ratio of nitrogen to ammonia is 1:2 Therefore the number of moles of nitrogen is 400000/2 = 200000

  Step 3 - Work out the mass of nitrogen (Mr of N2 is 28) 200000 x 28 = 5600000 g = 5.6 tonnes.
• Hydrogen peroxide decomposes in water to form water and oxygen. How many grams of oxygen gas will be given off from 40.8 g of hydrogen peroxide?

  Step 1: Write the balanced equation 2 H2O2(l)→2 H2O + O2(g) Mr of H2O2= 34

  Step 2: Number of moles in 40.8 g : 40.8/34 = 1.2 moles Ratio in the balanced equation of H2O2 : O2 = 2:1

  Step 3 :Therefore number of moles of O2 = 0.6 moles Step 4: Mass of oxygen = 0.6 x 32 (Mr of O2 ) = 19.2
• 31.0 cm3 of potassium hydroxide solution neutralised 25.0 cm3 of 2.0 moldm−3 nitric acid.

  HNO3 + KOH → KNO3 + H2O

  Calculate the concentration of the potassium hydroxide solution in moldm−3


  Step 1: Calculate the moles of HNO3 used = Concentration x volume 2 x 0.025 dm3 (25/1000 to convert the units) = 0.05 moles

  Step 2 : Calculate the moles of KOH Ratio is 1:1 therefore number of moles of KOH = 0.05

  Step 3 : Calculate the concentration of KOH Volume = Moles/concentration; 0.05 / 0.031 = 1.61
• Understand how to calculate the percentage yield from the following data

  Understand how to calculate the percentage yield from the following data

  
• What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2 ?


  Step 1 - Write the balanced equation N2 + 3 H2 → 2 NH3

  Step 2 - Calculate the theoretical amount of NH3 .Moles NH3 (ratio of H2 to NH3 is 3:2); of 20/1.5 = 13.3 moles 13.3 X 17 (Mr of NH3 ) = 227

  Step 3 - Calculate percentage yield of NH3 40.5/227 x 100 = 17.8%
• Look at the equations for the two reactions that produce CuCI2

  Reaction I: CuCO3 (s) + 2 HCI (aq) -> CuCI2 (aq) + H2O (l) + CO2 (g)

  Reaction II: CuO (s) + 2 HCI (aq) -> CuCI2 (aq) + H2O (l)

  Reactive formula masses: CuO=79.5; HCI=36.5; CuCI2=134.5; H2O=18

  Which reaction has a better atom economy? 


  Reaction II (look at the reactants):

  Total formula mass of reactants = 152.5

  Formula mass of CuCI2 = 134.5

  (134.5/152.5)X 100% = 88.2%