Question

Menthol is a flavoring agent extracted from peppermint oil. It contains C, H, and O. In one combustion analysis, 10.00 mg of the substance yields 11.53 mg H2O and 28.16 mg CO2. What is the empirical formula of menthol? Add subscripts to complete the empirical formula

Answer 1

Answer:

C₁₀H₂₀O

Explanation:

The molecular formula must be $C_{x}H_yOz$. The combustion reaction will occur between the fuel and oxygen gas:

$C_{x}H_yOz$ + O₂ → CO₂ + H₂O

For Lavoisier law, the mass of the reactants must be equal to the mass of the products (mass conservation):

10.0 + mO₂ = 28.16 + 11.53

mO₂ = 29.69 mg

Supposing that all the oxygen and the menthol were consumed, let's calculate the number of moles of the compounds, knowing, for the Periodic Table, that:

MC = 12 g/mol, MO = 16 g/mol, MH = 1 g/mol

MCO₂ = 12 + 2x16 = 44 g/mol

MH₂O = 2x1 + 16 = 18 g/mol

MO₂ = 2x16 = 32 g/mol

n = mass (g)/molar mass

nCO₂ = 0.02816/44 = 6.4x10⁻⁴ mol

nH₂O = 0.01153/18 = 6.4x10⁻⁴ mol

nO₂ = 0.02969/32 = 9.3x10⁻⁴ mol

The molar number is proportional in the molecule, so, in CO₂, the number of C is 6.4x10⁻⁴ mol, and of O is 1.28x10⁻³. All the carbon of the methol will be in CO₂, and all the H will be in H₂O. The number of moles of O will be the difference of moles in H₂O and the O₂, then:

nC = 6.4x10⁻⁴ mol

nH = 2x6.4x10⁻⁴ = 1.28x10⁻³ mol

nO = (2x6.4x10⁻⁴ + 6.4x10⁻⁴) - (2x9.3x10⁻⁴) = 6.0x10⁻⁵

The empirical formula is the molecule formula with the small subscripts numbers, which represent the number of moles of the atoms in the molecule. So, let's divide all the number of moles for the small on 6.0x10⁻⁵.

nC = (6.4x10⁻⁴)/(6.0x10⁻⁵) = 10

nH = (1.28x10⁻³)/(6.0x10⁻⁵) = 20

nO = (6.0x10⁻⁵)/(6.0x10⁻⁵) = 1

So, the empirical formula of methol is C₁₀H₂₀O.