Respuesta :
Answer:
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
Answer:
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