Question

Complete combustion of 7.50 g of a hydrocarbon produced 23.1 g of CO2 and 10.6 g of H2O. What is the empirical formula for the hydrocarbon ?

Answer 1

Chemical reaction: CₓHₐ + O₂ → xCO₂ + a/2H₂O.
m(CₓHₐ) = 7,50 g.
m(CO₂) = 23,1 g.
m(H₂O) = 10,6 g.
n(CO₂) = m(CO₂) ÷ M(CO₂).
n(CO₂) = 23,1 g ÷ 44 g/mol.
n(CO₂) = n(C) = 0,525 mol.
n(H₂O) = 10,6 g ÷ 18 g/mol = 0,588.
n(H₂O) : n(H) = 1 : 2.
n(H) = 1,176 mol.
n(H) : n(C) = 1,176 mol : 0,525 mol.
n(H) : n(C) = 2,25 : 1 /×4.
n(H) : n(C) = 9 : 4.
C₄H₉.

Answer 2

A CH compound is combusted to produce CO2 and H2O
CnHm + O2 -----> CO2 + H2O
Mass of CO2 = 23.1g
Mass of H2O = 10.6g
Calculate by mass of the compounds
For Carbon C, divide by molecular weight of CO2 and multiply with Carbon
molecular weight. So C in grams = 23.1 x (12.01 / 44.01) = 6.3 g C
For Hydrogen H, divide by molecular weight of H2O and multiply with Hydrogen molecular weight. So H in grams = 10.6 x (2.01 / 18.01) = 0.53 g C
= 1.18 of H
Calculate the moles for C and H
6.3 grams of C x (1 mole/12.01 g C) = 0.524 moles of C
1.18 grams of H x (1 mole/1.008 g H) = 1.17 moles of H
Divides by both mole entities with smallest
C = 0.524 / 0.524 = 1 x 4 = 4
H = 1.17 / 0.524 = 2.23 x 4 = 10
The empirical formula is C4H10.