Question

Complete combustion of a compound containing hydrogen and carbon produced 2.641 g of carbon dioxide and 1.442 grams of water as the only products. the molar mass of the hydrocarbon is 88.1 g/mol. what are the empirical and molecular formulas?

Answer 1

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.

Answer 2

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