Concept explainers
A compound is 82.7% carbon and 17.3% hydrogen, and has a molar mass of approximately 60 g/mol. When 1.000 g of this compound burns in excess oxygen, the enthalpy change is −49.53 kJ.
- (a) What is the empirical formula of this compound?
- (b) What is the molecular formula of this compound?
- (c) What is the standard enthalpy of formation of this compound?
- (d) Two compounds that have this molecular formula appear in Appendix G. Which one was used in this exercise?
(a)
Interpretation:
Empirical formula of the compound gas has to be determined.
Answer to Problem 5.106QE
Empirical formula of the compound is
Explanation of Solution
Compound is 82.7% carbon and 17.3% hydrogen. Hence,
Number of moles of Carbon and Hydrogen is determined as follows,
Empirical formula of the compound is determined as follows,
Therefore, empirical formula of the compound is
(b)
Interpretation:
Molecular formula of the compound gas has to be determined.
Answer to Problem 5.106QE
Molecular formula of the compound is
Explanation of Solution
Empirical formula mass is determined as follows,
Molecular formula of the compound is
Therefore, molecular formula of the compound is
(c)
Interpretation:
Standard enthalpy of formation of the compound has to be determined.
Answer to Problem 5.106QE
Standard enthalpy of formation of the given compound is
Explanation of Solution
Number of moles in
Enthalpy change when
The balanced thermochemical equation for the combustion of cyclopropane is given as:
Standard enthalpy of formation values is given below,
Standard enthalpy of formation of the given compound can be calculated by the equation:
Substitute the values as follows,
Therefore, standard enthalpy of formation of the given compound is
(d)
Interpretation:
Compound used for the exercise has to be determined.
Answer to Problem 5.106QE
The compound used must be
Explanation of Solution
For
For
Standard enthalpy of formation of the given compound is
Standard enthalpy of formation of the given compound is nearly equal to the standard enthalpy of formation of normal butane. Hence, the compound used must be
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Chapter 5 Solutions
Chemistry
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