Question

A partir de la siguiente formula empírica, establezca la respectiva formula empírica

A) C=64% , H=4,44%, CL=31,56%

B) C=24,74%, H= 4,12% O=32,98% N=14,44% Na=23,72%

Answer 1

A) C=64% , H=4,44%, CL=31,56%

C: 64 g/ 12 g/mol = 5.333 mol
H: 4.44 g/ 1 g/mol = 4.44 mol
Cl: 31.56 g/ 35.45 g/mol = 0.892 mol
dividir

C: 5.333 mol / 0.892 mol = 6
H: 4.44 mol / 0.892 mol = 5
Cl: 0.892 mol / 0.892 mol = 1

FE: C6H5Cl

B) C=24,74%, H= 4,12% O=32,98% N=14,44% Na=23,72%

C: 24.74 g / 12 g/mol = 2.061 mol
H: 4.12 g7 1 g/mol = 4.12 mol
O: 32.98 g / 16 g/mol = 2.061 mol
N: 14.44 g/ 14 g/mol = 1.031 mol
Na: 23.72 g/ 23 g/mol = 1.031 mo

DIVIDIR
C: 2.061 mol / 1.031 mol = 2
H: 4.12 mol / 1.031 mol = 4
O: 2.061 mol / 1.031 mol = 2
N: 1.031 mol / 1.031 mol = 1
Na: 1.031 mol / 1.031 mol = 1

FE: C2H4O2NNa