Menthol the substance we can smell in mentholated cough drops is composed of carbon, hydrogen and oxygen. A 0.1005 grams sample of menthol is combusted, producing 0.2829 grams of carbon dioxide and 0.1159 grams of water. What is the empirical formula for menthol? If the compound has a molar mass of 156 g/mol, what is its molecular formula?

Empirical and molecular formula are both C₁₀H₂₀O

Explanation:

The formula of menthol is CₐHₓOₙ. In the problem we need to find a, x and n.

The combustion of menthol occurs as follows:

CₐHₓOₙ + O₂ → aCO₂ + 1/2 H₂O

That means moles of carbon dioxide are equal to moles of carbon in menthol and moles of hydrogen are twice moles of water.

Moles CO₂ and mass C:

0.2829g CO₂ * (1mol / 44g) = 6.4295x10⁻³ moles CO₂ = Moles C * 12.01 = 0.0772g C

0.1159g H₂O * (1mol / 18g) = 6.4389x10⁻³ moles H₂O * 2 = 0.01288 moles H * 1.0g/mol = 0.01288g H

That means mass of O in the sample is:

Mass O = 0.1005g - 0.0772g C - 0.01288g H = 0.01042g O * (1mol / 16g) = 6.51x10⁻⁴ moles O

Empirical formula is the simplest whole number ratio of moles of each element present in a compound. Dividing each number of moles in moles O (The smaller number of moles):

H = 0.01288 moles H / 6.51x10⁻⁴ moles O = 19.8 ≈ 20

C =  6.4295x10⁻³ moles C / 6.51x10⁻⁴ moles O = 9.9 ≈ 10

O = 6.64x10⁻⁴ moles O / 6.51x10⁻⁴ moles O = 1

That means empirical formula is:

As molar mass of this empirical formula is:

10*12g/mol + 20*1g/mol + 1*16g/mol = 156g/mol = Molar mass of the compound,

Molecular formula is also C₁₀H₂₀O

nuclear fission produces elements that are heavier than helium.

the elements that are used in nuclear fission and their products are much heavier than helium.