Question

Complete combustion of 7.60 g of a hydrocarbon produced 23.0 g of CO2 and 11.8 g of H2O. What is the empirical formula for the hydrocarbon? I got C4H10 but was incorrect.

Answer 1

first calculate gm of carbon and hydrogen

molar mass of CO₂ = 44.01gm/mole

molar mass of carbon = 12.0107 gm/mole

that mean 44.01 gm CO₂ contain 12.0107 gm carbon we have to calculate then 23 gm CO₂ contain how many gram of carbon calculation are given below

gram of carbon = 12.010723/44 = 6.279 gm of carbon

molar mass of ​H₂O = 18.01528gm/mole

molar mass of hydrogen = 1.00794 gm/mole then molar mass of H₂ = 2.01588 gm/mol

that mean 18.01528 gm ​H₂O contain 2.01588 gm hydrogen we have to calculate then 11.8 gm ​H₂Ocontain how many gram of hydrogen calculation are given below

gram of hydrogenn = 2.0158811.8/18.01528 = 1.321gm of hydrogen

calculate % composition

7.60 = 100% then % of carbon = 1006.279/7.60 = 82.62%

% of hydrogen = 100- 82.62 = 17.38%

calculate atomic retio by dividing percentage composition by atomic mass

atomic retio of carbon = 82.62/12.01528 = 6.87

atomic retio of hydrogen = 17.38/1.0094 = 17.21

convert atomic retio to simplest retio by dividing atomic retio by smallest retio

simplest retio of carbon = 6.87/6.87 = 1

simplest retio of hydrogen = 17.21/6.87 = 2.50

convert fractional number to whole number by multiplying by two

for carbon = 12 = 2

for hydrogen = 2.52 = 5

empirical formula of compound is C₂H₅