a) Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO2 and 1.37 mg of H2O. If the compound contains only carbon and hydrogen, what is its empirical formula?

b) 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. What is the empirical formula for menthol? If menthol has a molar mass of 156 g>mol, what is its molecular formula?

So, in [tex]5.86\times 10^{-3}g[/tex] of carbon dioxide, [tex]\frac{12}{44}\times 5.86\times 10^{-3}=1.60\times 10^{-3}g[/tex] of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in [tex]1.37\times 10^{-3}g[/tex] of water, [tex]\frac{2}{18}\times 1.37\times 10^{-3}=0.152\times 10^{-3}g[/tex] of hydrogen will be contained.

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

Step 1: Converting the given masses into moles.

Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.60\times 10^{-3}g}{12g/mole}=0.133\times 10^{-3}moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.152\times 10^{-3}g}{1g/mole}=0.152\times 10^{-3}moles[/tex]

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 [tex]0.133\times 10^{-3}[/tex] moles.

For Carbon = [tex]\frac{0.133\times 10^{-3}}{0.133\times 10^{-3}}=1[/tex]

For Hydrogen = [tex]\frac{0.152\times 10^{-3}}{0.133\times 10^{-3}}=1.14\approx 1[/tex]

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

Hence, the empirical formula for the given compound is [tex]CH[/tex]

For b:

The chemical equation for the combustion of menthol follows:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

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

We are given:

Mass of [tex]CO_2[/tex] = 0.2829 g

Mass of [tex]H_2O[/tex] = 0.1159 g

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, [tex]\frac{12}{44}\times 0.2829=0.077g[/tex] 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, [tex]\frac{2}{18}\times 0.1159=0.013g[/tex] of hydrogen will be contained.

Mass of oxygen in the compound = (0.1005) - (0.077 + 0.013) = 0.105 g

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

Step 1: Converting the given masses into moles.

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

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.013g}{1g/mole}=0.013moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0105g}{16g/mole}=0.00065moles[/tex]

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 0.00065 moles.

For Carbon = [tex]\frac{0.0064}{0.00065}=9.84\approx 10[/tex]

For Hydrogen = [tex]\frac{0.013}{0.00065}=20[/tex]

For Oxygen = [tex]\frac{0.00065}{0.00065}=1[/tex]

Step 3: Taking the mole ratio as their subscripts.

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

The empirical formula for the given compound is [tex]C_{10}H_{20}O[/tex]

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:

[tex]n=\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]

We are given:

Mass of molecular formula = 156 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

[tex]n=\frac{156g/mol}{156g/mol}=1[/tex]

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

[tex]C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O[/tex]

Hence, the empirical and molecular formula for the given organic compound are [tex]C_{10}H_{20}O[/tex]