Lecture 8 Limiting reactants &amp; Combustion analysis

# Lecture 8 Limiting reactants &amp; Combustion analysis...

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Lec-8: Limiting reactants & Combustion analysis Roy A. Lacey, Stony Brook University; Che 131, Spring 2011 1 Combustion Analysis C a H b + excess O 2 ---> a CO 2 (g) + b/2 H 2 O The percent of carbon and hydrogen in C a H b can be determined from the mass of H 2 O and CO 2 produced. Roy A. Lacey, Stony Brook University; Che 131, Spring 2011 2

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Molecular Formula The molecular formula is equal to the empirical formula or a simple multiple of it. Find the molecular formula for a compound whose composition is: 24 3% C 4 1% H and 71 6% C and composition is: 24.3% C, 4.1% H and 71.6% Cl and has a molar mass of 99.0 g Assume 100 g of compound 24.3 g C 41 gH 4.1 g H 71.6 g Cl 24.3 2.0 C 12. / g mol gm o l = Empirical formula 4.1 4.0 H 1.0 / g mol g mol = C H Cl 242 = 12 1 = = CH 2 Cl 71.6 2.0 Cl 35.5 / g mol o l = 1 2 222 Molecular formula Æ (CH 2 Cl) n Roy A. Lacey, Stony Brook University; Che 131, Spring 2011 3 n = (molar mass)/(mass empirical formula) C 2 H 4 Cl 2 Writing Balanced Chemical Equations Fe C + FeC Fe Cl 2 FeCl 3 Fe C FeC 3 2 Fe Cl 2 FeCl 3 Balance Cl 2 Fe Cl 2 FeCl 3 3 2 Balance Fe Roy A. Lacey, Stony Brook University;
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Lecture 8 Limiting reactants &amp; Combustion analysis...

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