1. Ethylene glycol, commonly used in antifreezes, contains only carbon, hydrogen, and oxygen. When a sample of it is combusted in air, a 23.46 g sample produces 20.42 g of water and 33.27 g of carbon dioxide. Determine its empirical formula. If it has a molar mass of 62.0 g, what is its molecular formula

Empirical formula = CH3O

Molecular formula = C2H6O2

Explanation:

Step 1: Data given

Mass of the sample = 23.46 grams

Mass of H2O = 20.42 grams

Molar mass of H2O = 18.02 g/mol

Mass of CO2 = 33.27 grams

Molar mass of CO2 = 44.01 G:mol

Atomic mass of C = 12.01 g/mol

Atomic mass of O = 16.0 g/mol

Atomic mass of H = 1.01 g/mol

Molar mass of the compound = 62.0 g/mol

Step 2: Calculate moles of H2O

Moles H2O = 20.42 grams / 18.02 g/mol

Moles H2O = 1.133 moles

Step 3: Calculate moles H

For 1 mol H2O we have 2 moles H  

For 1.133 moles H2O we have 2* 1.133 = 2.266 moles H

Step 4: Calculate mass H

Mass H = 2.266 moles * 1.01 g/mol

Mass H = 2.29 grams

Step 5: Calculate moles CO2

Moles CO2 = 33.27 grams / 44.01 g/mol

Moles CO2 = 0.7560 moles

Step 6: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.7560 moles CO2 we have 0.7560 moles C

Step 7: Calculate mass C

Mass C = 0.7560 moles * 12.01 g/mol

Mass C = 9.08 grams

Step 8: Calculate mass O

Mass O = 23.46 grams - 9.08 grams - 2.29 grams

Mass O =12.09 grams

Step 9: Calculate moles O

Moles O = 12.09 grams / 16.0 g/moles

Moles O = 0.7556

Step 10: Calculate mol ratio

We divide by the smallest amount of moles  

C: 0.7560 moles / 0.7556 moles =1  

H: 2.266 moles / 0.7556 moles =3

O; 0.7556 / 0.7560 moles = 1

This means for 1 mol C we have 3 moles H and 1 mol O

The empirical formula is CH3O

Step 11: Calculate the molecular formula

The molar mass of the empirical formula is 31 g/mol

Step 11: Calculate molecular formula

We have to multiply the empirical formula by n

n = 62.0 g/mol / 31g/mol = 2

Molecular formula = 2*(CH3O)

Molecular formula = C2H6O2

conduction is the transfer of thermal energy (heat in transfer) due to collisions between the molecules in the object. collisions between adjacent atoms and molecules transfer kinetic energy from the warmer to the cooler object. the objects must be in physical contact.