What are the Mass and Empirical Formula of the Combusted Hydrocarbon?

Click For Summary
SUMMARY

The combustion of a hydrocarbon sample produced 21.83 g of CO2 and 4.47 g of H2O, resulting in a total mass of 6.47 g for the hydrocarbon. The empirical formula derived from the molar ratios of carbon and hydrogen is C1H1. Additionally, the heat of combustion per empirical formula unit is calculated to be approximately 624.5 kJ. The discussion also touches on the equilibrium constant for a reaction involving species A and B, with specific rates provided for the forward and backward reactions.

PREREQUISITES
  • Understanding of stoichiometry and molar mass calculations
  • Knowledge of combustion reactions and empirical formulas
  • Familiarity with equilibrium concepts and rate laws
  • Basic thermodynamics principles, including heat of combustion
NEXT STEPS
  • Study the calculation of empirical formulas from combustion data
  • Learn about the relationship between reaction rates and equilibrium constants
  • Explore the kinetic interpretation of chemical equilibrium
  • Investigate the concept of heat of formation (ΔHfº) in thermodynamics
USEFUL FOR

Chemistry students, educators, and professionals involved in chemical analysis, combustion studies, and thermodynamics will benefit from this discussion.

fabrave21
Messages
2
Reaction score
0
2 question I don't get ... =\ any help would be appreciated.
I don't even know where to start ...

Homework Statement



2. A sample of a hydrocarbon is combusted completely in O2 to produce 21.83 g CO2, 4.47 g H2O and 311 kJ of heat. A) What is the mass of the hydrocarbon sample that was combusted? B) What is the empirical formula of the hydrocarbon? C) Calculate the value of ΔHfº per empirical-formula unit of the hydrocarbon.7. Suppose you have some reaction: 2A ---> B and B ---> 2A The rate for the formation of B from A is 3.8 x 102 s-1 and the rate for the formation of A from B is 8.6 x 10-1 s-1. A) What is the equilibrium constant for the equilibrium reaction, 2A <----> B ?
B) Suppose that you started with 2 moles of A and at some later time you measure .9 moles of B in a total volume of 1L, is the reaction at equilibrium? Explain.

Homework Equations


The Attempt at a Solution



2. mols C = 21.83/44 =.496

mols H = 4.47/18.0 =,248 x2 =.496

mols = g/Mwt then mols x Mwt = g

The mass of C is .496 x12.0 = .5.97 g C
The mass of H = .496 x 1.00 = ,496 g H
Total mass is 6.47 g for the hydrocarbon

The empirical formula is the ratio of the mols of C and H which makes it C1H1

6.47/ 13 = .496 mols
heat of combustion per formula weight is 311KJ/ .498 mols

= 624.497992 kJ

7. ?
 
Last edited:
Physics news on Phys.org
figured out number 2. Still don't get number 7
 
Have you heard about kinetic intepretation of equilibrium?

Write expressions for
- reaction equilibrium
- forward reaction speed
- backward reaction speed

Compare denominator with forward reaction speed and denominator with backward reaction speed. Do you think you can combine rate coefficients to get equilibrium constnats?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
14K
  • · Replies 4 ·
Replies
4
Views
3K