# 10.00mg of a substance yields 11.53mg H2O and 28.16mg CO2. What is the empirical formula of menthol? This is so confusing, help would be much appreciated!?

Nov 19, 2015

Empirical formula: ${C}_{\text{10"H_"20}} O$. It corresponds to the empirical and molecular formula of menthol.

#### Explanation:

1) calculate mmoles of $C {O}_{2}$ and water produced:

${n}_{\text{CO2" = "28.16 mg"/"44.01 mg/mmol}} = 0.63985 m m o l$

these are also the mmoles of carbon in ten milligrams of substance.

${n}_{\text{H2O" = "11.53 mg"/"18.02 mg/mmol}} = 0.63985 m m o l$

We deduce that ten milligrams of substance contain:

$0.63985 \cdot 2 = 1.2797 m m o l$ of hydrogen (double of water mmoles)

We can deduce soon that the molecule contains a number of hydrogen atoms that is the double of carbon atoms; so the empirical formula should be (provisionally) ${C}_{x} {H}_{\text{2x}} {O}_{y}$.

2) calculate moles of dioxygen (${O}_{2}$) necessary to the complete combustion, applying the law of mass conservation

$\text{products' mass} = 28.16 m g + 11.53 m g = 39.69 m g$

then:

$\text{reactants' mass} = 39.69 m g$

Therefore the ${O}_{2}$ mass needed for combustion must be equal to

${m}_{\text{O2}} = 39.69 m g - 10.00 m g = 29.69 m g$

then we can calculate oxygen moles:

${n}_{O} = \text{29.69 mg"/"16.00 mg/mmol} = 1.85563 m m o l$

3) calculate number of moles of oxygen in the ten milligrams of substance, applying the conservation law to oxygen alone

The number of mmoles of oxygen atoms in the products is the double of $C {O}_{2}$ mmoles plus the moles of water:

${n}_{\text{O(products)}} = 0.63985 \cdot 2 + 0.63985 = 1.91956 m m o l$ (O total in the products or reactants")

The mmole number of oxygen atoms in the substance is obtained by subtracting the oxygen number of mmoles used to combust the substance from this total number of atoms in mmoles:

${n}_{\text{O (substance)}} = 1.91956 - 1.85563 = 0 , 06393 m m o l$ (O combined in 10 mg substance).

4) Calculate the empirical formula by taking the moles ratios and getting the minimum integer indexes of C, H and O.

We see that carbon mmoles are approximately ten times oxygen mmoles:

$r a t i o \frac{C}{O} = {n}_{C} / {n}_{O} = \frac{0.63985}{0.06393} = 10 , 00$

Therefore we conclude that putting at least one oxygen atom in the empirical formula, we must put ten times this number in the carbon index an the double in hydrogen index, so we get:

${C}_{\text{10"H_"20}} {O}_{1}$, that is: ${C}_{\text{10"H_"20}} O$.

This is also the molecular formula of menthol .