Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.
Upcoming SlideShare
×

# Atoms, Molecules & Stoichometry (II)

576 views

Published on

Published in: Technology, Education
• Full Name
Comment goes here.

Are you sure you want to Yes No
Your message goes here
• Be the first to comment

• Be the first to like this

### Atoms, Molecules & Stoichometry (II)

1. 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. 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. 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. 4. © my-chem-tutor.blogspot.com <ul><li>Concept: </li></ul><ul><li>volume ratios = mole ratios [Avogadro’s law]  </li></ul><ul><li>1 mol of C x H y gives x moles of CO 2 </li></ul><ul><li>10 cm 3 of C x H y gives 30 cm 3 of CO 2 </li></ul><ul><li>x = 3 </li></ul><ul><li>1 mol of C x H y reacts with (x+y/4) moles of O 2 </li></ul><ul><li>10 cm 3 of C x H y reacts with 50 cm 3 of O 2 </li></ul><ul><li>(x+y/4) = 5 </li></ul><ul><li>y = 8 </li></ul><ul><li>Molecular formula: C 3 H 8 </li></ul>