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

i think a, hope it .

explanation:

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.