Question

Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g of H2O. Menthol has a molar mass of 156.27 g/mol. What is the molecular formula of menthol?

Answer 1

Answer:

$\boxed{\text{C _{10} H _{20} O}}$

Explanation:

In a combustion experiment, all the carbon ends up as CO₂, and all the hydrogen ends up as water.

Data:

Mass of menthol = 0.1005 g

     Mass of CO₂ = 0.2829 g

     Mass of H₂O = 0.1159   g

Calculations:

(a) Mass of each element

$\text{Mass of C} = \text{0.2829 g CO _{2} } \times \dfrac{\text{12.01 g C}}{\text{44.01 g CO _{2} }} = \text{0.077 20 g C}\\\\\text{Mass of H} = \text{0.1159 g H _{2} O} \times \dfrac{\text{2.016 g H}}{\text{18.02 g H _{2} O}} = \text{0.012 97 g H}\\\\\text{Mass of O} = \text{mass of menthol - mass of C - mass of H}\\= \text{0.1005 - 0.077 20 - 0.01297}= \text{0.010 33 g O}$

(b) Moles of each element

$\text{Moles of C} = \text{0.077 20 g C} \times \dfrac{\text{1 mol C}}{\text{12.01 g C}} = 6.428 \times 10^{-3}\text{ mol C}\\\\\text{Moles of H} = \text{0.012 97 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{ 0.012 86 mol H}\\\\\text{Moles of O} = \text{0.010 33 g O} \times \dfrac{\text{1 mol O }}{\text{16.00 g O}} = 6.458 \times 10^{-4}\text{ mol O}$

(c) Molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{6.428 \times 10^{-3}}{6.458 \times 10^{-4}} = 9.954\\\\\text{H: } \dfrac{0.012 86}{6.458 \times 10^{-4}} = 19.92\\\\\text{O: } \dfrac{6.458 \times 10^{-4}}{6.458 \times 10^{-4}} = 1$

(d) Round the ratios to the nearest integer

C:H:O = 10:20:1

(e) Write the empirical formula

The empirical formula is C₁₀H₂₀O.

(f) Calculate the empirical formula mass

 10 × C = 10 × 12.01  = 120.1    u

20 × H = 20 × 1.008 =  20.16  u

  1 × O = 1 × 16.00    =   16.00 u

                EF  mass =  156.3    u

(g) Divide the molecular mass by the empirical formula mass.  

$n = \dfrac{\text{MM}}{\text{EFM}} = \dfrac{156.27}{156.3} = 0.9998 \approx 1$

(h) Determine the molecular formula

$\text{MF} = \text{(EF)}_{n} = \rm (C_{10}H_{20}O)_{1} = \textbf{C _{10} H _{20} O}\\\text{The molecular formula of menthol is } \boxed{\textbf{C _{10} H _{20} O}}$

Answer 2

Answer: The molecular formula for the menthol is $C_{10}H_{20}O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=0.2829g$

Mass of $H_2O=0.1159g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

• For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.2829 g of carbon dioxide, $\frac{12}{44}\times 0.2829=0.077g$ of carbon will be contained.

• For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.1159 g of water, $\frac{2}{18}\times 0.1159=0.0129g$ of hydrogen will be contained.

• Mass of oxygen in the compound = (0.1005) - (0.077 + 0.0129) = 0.0106 g

To formulate the empirical formula, we need to follow some steps:

• Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0129g}{1g/mole}=0.0129moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0106g}{16g/mole}=0.00066moles$

• Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00066 moles.

For Carbon = $\frac{0.0064}{0.00066}=9.69\approx 10$

For Hydrogen  = $\frac{0.0129}{0.00064}=19.54\approx 20$

For Oxygen  = $\frac{0.00066}{0.00066}=1$

• Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

Hence, the empirical formula for the given compound is $C_{10}H_{20}O_1=C_{10}H_{20}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 156.27 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

$n=\frac{156.27g/mol}{156g/mol}=1$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O$

Thus, the molecular formula for the menthol is $C_{10}H_{20}O$