A hydrocarbon of unknown formula C_xH_ywas submitted to combustion analysis with the following results.What is the empirical formula of the hydrocarbon? A)C₅H₂ B)C₅H₄ C)C₁₀H₄ D)C₁₀H₈

Question

Quizplus · Accepted Answer

From the combustion analysis results: 
C = 66.67%
H = 11.11%
S = 22.22%

Assume a 100g sample of the hydrocarbon, then the mass of each element present can be calculated as:
C = 66.67g C
H = 11.11g H
S = 22.22g S

Next, convert each mass to moles:
moles C = (66.67g C) / (12.011 g/mol) = 5.5551 mol C
moles H = (11.11g H) / (1.008 g/mol) = 11.0342 mol H
moles S = (22.22g S) / (32.06 g/mol) = 0.6922 mol S

Divide each mole value by the smallest of the three to obtain the simplest mole ratio:
C:H:S = 5.5551/0.6922 = 8.0291 : 1 : 1.0011

Since we need integer values, we can round the ratio to the nearest whole number:
C:H:S = 8 : 1 : 1

Therefore, the empirical formula is CS8H and simplifying it further to the smallest whole number ratio gives the final empirical formula of CS1H. However, none of the answer choices have H1, but we can assume that it is meant to be H2 because hydrocarbons typically have at least two hydrogen atoms per carbon atom. Thus, choice B is the best option.

A Hydrocarbon of Unknown Formula CxHy was Submitted to Combustion

A Hydrocarbon of Unknown Formula C_xH_ywas Submitted to Combustion