Question

Ethylene glycol is used in automobile antifreeze and in the production of polyester. The name glycol stems from the sweet taste of this poisonous compound. Combustion of 6.38 g of this compound gives 9.06 g of CO2 and 5.58 g of H2O. The compound only contains C, H, and O. What are the mass percentages of the elements in ethylene glycol? Using these perecentage, calculate the empirical formula for this compound. If the molecular mass of this compound is 62 amu, what is the molecular formula?

Answer 1

`\begin{gathered} Empirical\text{ formula: }CH_3O \\ Molecular\text{ formula: }C_2H_6O_2 \end{gathered}`

Explanations:

Given the following parameters

Mass of CO2 = 9.06grams

Mass of H2O =5.58grams

Since Carbon contains 12g/mol of carbon out of 44g/mol of CO2, hence the required mass of carbon required is given as:

`Mass\text{ of C}=(12)/(44)*9.06=2.47grams`

Similarly, hydrogen contains 2g/mol of hydrogen out of 18g/mol of H2O, hence the required mass of hydrogen required is given as;

`\begin{gathered} Mass\text{ of H}_2=(2)/(18)*5.58 \\ Mass\text{ of H}_2=0.62grams \end{gathered}`

Determine the mass of the compound

Mass of oxygen = 6.38 - (2.47 + 0.62)

Mass of oxygen = 3.29grams

Determine the moles of the elements

Mole of Carbon = 2.47/12 = 0.206moles

Mole of Hydrogen = 0.62/1 = 0.62moles

Mole of Oxygen = 3.29/16 = 0.206moles

Divide by the lowest number of moles

For Carbon: 0.206/0.206 = 1

For Hydrogen: 0.62/0.206 = 3

For Oxygen: 0.206/0.206 = 1

Determine the empirical formula

Empirical formula = CH3O

Determine the molecular formula

If the molecular mass of this compound is 62 amu, hence;

`\begin{gathered} (CH_3O)n=62 \\ (12+3(1)+16)n=62 \\ 31n=62 \\ n=2 \\ (CH_3O)_2=C_2H_6O_2 \end{gathered}`

• Hence the ,molecular formula ,of the compound is ,C2H6O2