The empirical formula of hydrocarbon is C₃H₈
The molecular formula of hydrocarbon is C₆H₁₆
Empirical formula calculation
Hydrocarbon contain carbon and hydrogen
Step 1: find the mass carbon (C) in carbon dioxide (CO₂) and hydrogen (H ) in water
mass of of element = molar mass of element/ molar mass molecule x total mass of molecule
From periodic table the molar mass of C =12, for CO₂ = 12+( 16 x2) =44 g/mol, for H = 1.00 g/mol, for H₂O = (2 x1)+16 = 18 g/mol
mass of C = 12/44 x 2.641 =0.7203 g
since there are 2 atom of H in H₂O the molar mass of H = 1 x2 = 2 g/mol
mass of H is therefore = 2/18 x 1.442 =0.1602 g
Step 2: find the moles of C and H
moles = mass÷ molar mass
moles of C = 0.7203 g÷ 12 g/mol = 0.060 moles
moles of H = 0.1602÷ 1 g/mol = 0.1602 moles
Step 3: find the mole ratio of C and H by dividing each mole by smallest mole ( 0.06)
for C = 0.06/0.06 =1
For H = 0.1602/0.06 =2.67
multiply by 3 to remove the decimal
For C = 1 x3 =3
For H = 2.67 x3 =8
therefore the empirical formula = C₃H₈
The molecular formula calculation
[C₃H₈]n = 88.1 g/mol
[12 x 3)+( 1 x8)]n =88.1 g/mol
44 n = 88.1
divide both side by 44
n=2
therefore [C₃H₈]₂ = C₆H₁₆
1) Answer is: the empirical formula of the hydrocarbon is C₃H₈.
Chemical reaction: CₓHₐ + O₂ → xC + a/2H₂O.
m(CO₂) = 2.641 g.; mass of carbon dioxide.
n(CO₂) = m(CO₂) ÷ M(CO₂).
n(CO₂) = 2.641 g ÷ 44.01 g/mol.
n(CO₂) = n(C) = 0.06 mol; amount of carbon.
m(H₂O) = 1.442 g.
n(H₂O) = 1.442 g ÷ 18 g/mol.
n(H₂O) = 0.08 mol.
n(H) = 2 · n(H₂O) = 0.16 mol; amount of hydrogen.
n(C) : n(H) = 0.06 mol : 0.16 mol /0.06 mol.
n(C) : n(H) = 1 : 2. 67 /×3.
n(C) : n(H) = 3 : 8.
2) Answer is: the molecular formula of hydrocarbon is C₆H₁₆.
M(C₃H₈) = 44.05 g/mol; molar mass of empirical formula.
M(CₓHₐ) = 88.1 g/mol; molar mass of hydrocarbon.
M(CₓHₐ) ÷ M(C₃H₈) = 88.1 g/mol ÷ 44.05 g/mol.
M(CₓHₐ) ÷ M(C₃H₈) = 2.
The molar mass of hydrocarbon is two times higher than molar mass of empirical formula, so number of carbon atoms is six and number of hydrogen atoms sixteen.
