Respuesta :

C=52.11/12, H=13.14/1 O=34.75/16, C=4.3425/2.171875  H=13.14/2.171875  O=2.171875/2.171875, C=1.99 H=6 O=1, C=2 H=6 O=1 Empirical Formula=C2H6O
Oseni

The empirical formula of a compound with  52.11% carbon, 13.14% hydrogen, and 34.75% oxygen will be [tex]C_2H_6O[/tex]

Empirical formula

The composition of the compound is as follows:

C = 52.11                            H = 13.14                                   O = 34.75

Divide through by their respective molar masses

C = 52.11/12 = 4.3              H = 13.14/1 = 13.14                O = 34.75/16 = 2.17

Divide by the smallest

C = 4.3/2.17 = 2                   H = 13.14/2.17= 6               O = 2.17/2.17 = 1

Thus, the empirical formula of the compound is [tex]C_2H_6O[/tex]

More on empirical formulas can be found here: https://brainly.com/question/14044066

#SPJ2

Otras preguntas