Question

Esta formula es empirica 40,1% de C, 6,6% de H y 53,3% de O

Answer 1

1. calcular moles

C. 40.1 g / 12 g/mol = 3.34 mol
H:   6.6 g/ 1 g/mol = 6 mol
O: 53.3 g/16 g/mol = 3.33 mol

2. Dividir entre el menor de los resultados

C: 3.34 mol / 3.33 mol = 1
H:   6 mol / 3.33 mol  = 2
O: 3.33 mol / 3.33 mol = 1

3. Fórmula empirica: CH₂O

Answer 2

Respuesta:

$c = 40.1 + h = 6.6 + o = 53.3 = 100 \\ c = \frac{40.1}{12} = 3.34 \\ h = \frac{6.6}{1} = 6.6 \\ o = \frac{53.3}{16} = 3.33 \\ \\ \\ c = \frac{3.34}{3.33} = 1 \\ h = \frac{6.6}{3.33} = 1.98 = 2 \\ o = \frac{3.33}{3.33} = 1 \\ \\ ch2o$