When 2.56 g of a compound containing only carbon, hydrogen, and oxygen is burned completely, 4.89 g of CO2 and 3.00 g of h2o are produced. What is the empirical formula of the compound

The empirical formula is CH3O2

Explanation:

From The question, number of moles of CO2 = 4.89/44=0.111mol

Number of moles of H2O=3/18 = 0.1667moles

Since 1mole of CO2 contains 1mole of carbon and 2 moles of O2,

Implies that there 0.111 moles of C and 0.222moles of O2

Similarly also 0.1667moles of water contains 2 moles of H2 hence = 0.333 moles of H2

Dividing by smallest number = 0.111mol

For C = 0.111/0.111 = 1

For H = 0.333 / 0.111 = 3

For O = 0.222/0.111 = 2

Hence the empirical formula = CH3O2

C2H5O

Explanation:

Mass of C = molar mass of C/molar mass of CO2 x mass of CO2

= 12/44 x 4.89 = 1.3336 g

Mass of H = molar mass of H2/molar mass of H2O x mass of H2O

= 2/18 x 3 = 0.33333 g

Mass of O = mass of compound - mass of C - mass of H

= 2.56 - 1.3336 - 0.33333 = 0.98307g

Since moles = mass/atomic mass

Moles of C : H : O = 1.3336/12 : 0.33333/1 : 0.98307/16

= 0.1111 : 0.33333 : 0.061

Dividing by the smallest

= 2 : 5 : 1

The empirical formula is C2H5O