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Empirical & molecular formulae   Empirical Formula   - the simplest whole-number ratio of atoms in a compound.  Molecular Formula   - the actual numbers of atoms present in a molecule of the compound.  - it may be the same of the empirical formula or an integral multiple of it.  © my-chem-tutor.blogspot.com (1) using composition by mass  (2) using combustion data  Calculation of empirical  & molecular formulae
(I) using composition by mass Caffeine was found to contain: 49.48% carbon, 5.19% hydrogen, 16.48% oxygen and 28.85% nitrogen. It's Mr is 194.19 g/mol. Calculate its empirical and molecular formula?   For 100g of caffeine,  Empirical Formula: C 4 H 5 N 2 O [Mass = 97g = 0.5 x Mr]  Molecular Formula: C 8 H 10 N 4 O 2   © my-chem-tutor.blogspot.com QUESTION
(II) using combustion data 10 cm 3  of a gaseous hydrocarbon was mixed with 100cm 3  of oxygen and ignited. 80 cm 3  of gas remained which was reduced to 50 cm 3  on shaking with excess aq. KOH. Find the molecular formula of the hydrocarbon at r.t.p.  © my-chem-tutor.blogspot.com QUESTION Information from question: CO 2  reacts with aq. KOH   Vol of CO 2  formed = 80 - 50 = 30 cm 3 Vol of O 2  remained = 50 cm 3 Vol of O 2  reacted = = 100 - 50 = 50 cm 3 Always derive as much data from the question as possible! TIP
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Atoms, Molecules & Stoichometry (II)

  • 1. Empirical & molecular formulae Empirical Formula - the simplest whole-number ratio of atoms in a compound. Molecular Formula - the actual numbers of atoms present in a molecule of the compound. - it may be the same of the empirical formula or an integral multiple of it. © my-chem-tutor.blogspot.com (1) using composition by mass (2) using combustion data Calculation of empirical & molecular formulae
  • 2. (I) using composition by mass Caffeine was found to contain: 49.48% carbon, 5.19% hydrogen, 16.48% oxygen and 28.85% nitrogen. It's Mr is 194.19 g/mol. Calculate its empirical and molecular formula? For 100g of caffeine, Empirical Formula: C 4 H 5 N 2 O [Mass = 97g = 0.5 x Mr] Molecular Formula: C 8 H 10 N 4 O 2 © my-chem-tutor.blogspot.com QUESTION
  • 3. (II) using combustion data 10 cm 3 of a gaseous hydrocarbon was mixed with 100cm 3 of oxygen and ignited. 80 cm 3 of gas remained which was reduced to 50 cm 3 on shaking with excess aq. KOH. Find the molecular formula of the hydrocarbon at r.t.p. © my-chem-tutor.blogspot.com QUESTION Information from question: CO 2 reacts with aq. KOH  Vol of CO 2 formed = 80 - 50 = 30 cm 3 Vol of O 2 remained = 50 cm 3 Vol of O 2 reacted = = 100 - 50 = 50 cm 3 Always derive as much data from the question as possible! TIP
  • 4.