Question

A common fuel additive that is composed of C, H, and O enhanced the performance of gasoline began being phased out in 1999 because of its contamination of drinking water. When 12.1 g of the compound are burned, 30.2 g of CO2and 14.8 g of H2O are formed. What is the empirical formula of the compound?

Answer 1

C₅ H₁₂ O

Step-by-step explanation:

44 g of CO₂ contains 12 g of C

30.2 g of CO₂ will contain 12 x 30.2 / 44 = 8.236 g of C .

18 g of H₂O contains 2 g of hydrogen

14.8 g of H₂0 will contain 1.644 g of H .

total compound = 12.1 out of which 8.236 g is C and 1.644 g is H , rest will be O

gram of O = 2.22

moles of C, O, H in the given compound = 8.236 / 12 , 2.22 / 16 , 1.644 / 1

= .6863 , .13875 , 1.644

ratio of their moles = 4.946 : 1 : 11.84

rounding off to digits

ratio = 5 : 1 : 12

empirical formula = C₅ H₁₂ O