Lecture 8 Limiting reactants & Combustion analysis

Lecture 8 Limiting reactants & Combustion analysis...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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;
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 5

Lecture 8 Limiting reactants & Combustion analysis...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online