Question

In the combustion of 50 ml of the hydrocarbon CHₓHy in excess oxygen gas, which results in 200 ml of carbon dioxide and 250 ml of water vapor at STP (Standard Temperature and Pressure), what is the molecular formula of this hydrocarbon?

:
A) C₃H₈
B) C₄H₁₀
C) C₅H₁₂
D) C6H₁₄

Answer 1

The molecular formula of the hydrocarbon in question is C4H10 (butane), determined by the volume of carbon dioxide and water produced during combustion at STP.

Step-by-step explanation:

The molecular formula of the hydrocarbon can be determined by analyzing the stoichiometry of combustion. Combustion of hydrocarbons results in carbon dioxide and water. Using the given information, we have 50 ml of the hydrocarbon yields 200 ml of carbon dioxide and 250 ml of water at STP. This means that we have a 1:4 ratio of hydrocarbon to carbon dioxide, and since volume ratios at STP can be translated to mole ratios for gases, we know that for each carbon atom in the hydrocarbon, one carbon dioxide molecule is produced. Therefore, the hydrocarbon has 4 carbon atoms. Moreover, for each 2 moles of water produced, 2x1=2 moles of hydrogen atoms must have been in the hydrocarbon. Hence, the molecular formula of the hydrocarbon is C4H10, which is option B, butane.