calcula la formula molecular de un compuesto sabiedo que tiene los sgt porsentajes P=30% H=20% O=56% su masa de 720 UMA
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Mm P = 30.9 g/mol
H = 1 g/mol
O = 16 g/mol
calcular moles de cada elemento
P: 31 g /30.9 g/mol = 1 mol
H: 20 g / 1 g/mol = 20 mol
O: 48 g / 16 g/mol = 3 mol
dividir entre el menor de los resultados
P: 1 mol / 1 mol = 1
H: 20 mol / 1 mol = 20
O: 3 mol / 1 mol = 3
FE = PH₂₀O₃
MASA MOLECULAR DE LA FE
P: 1 x 30.9 = 30.9 g/mol
H: 20 x 1 g = 20 g/mol
O: 3 x 16 g = 48 g/mol
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Mm = 98.9 g/mol ≈ 99 g/mol
n = 720 g/mol / 99 g/mol
n = 7
FM = (FE)n
FM = (PH₂₀O₃)₇
FM = P₇H₁₄₀O₂₁
H = 1 g/mol
O = 16 g/mol
calcular moles de cada elemento
P: 31 g /30.9 g/mol = 1 mol
H: 20 g / 1 g/mol = 20 mol
O: 48 g / 16 g/mol = 3 mol
dividir entre el menor de los resultados
P: 1 mol / 1 mol = 1
H: 20 mol / 1 mol = 20
O: 3 mol / 1 mol = 3
FE = PH₂₀O₃
MASA MOLECULAR DE LA FE
P: 1 x 30.9 = 30.9 g/mol
H: 20 x 1 g = 20 g/mol
O: 3 x 16 g = 48 g/mol
``````````````````````````````````````````
Mm = 98.9 g/mol ≈ 99 g/mol
n = 720 g/mol / 99 g/mol
n = 7
FM = (FE)n
FM = (PH₂₀O₃)₇
FM = P₇H₁₄₀O₂₁
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