Respuestas
1. A y b. La fórmula empírica de un compuesto es la relación más pequeña (más sencilla), en número enteros, de los moles de átomos de cada elemento en el compuesto. Se obtiene a partir de la composición centesimal, o en masa del compuesto y de las respectivas masas atómicas gramo de los elementos que la forman.
c. permite calcular la composición centesimal o porcentual de cada uno de los elementos que conforman la fórmula de un compuesto, en un 100 u o en 100 g del compuesto.
2. a. UREA: CO(NH2OH)
Mm C: 1 x 12 = 12 g/mol
O:2 x 16 = 32 g/mol
N: 1 x 14 = 14 g/mol
H: 3 x 1 = 3 g/mol
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Mm = 61 g/mol
H: 61 g ---- 100 %
3 g ---- x
X = 4.918 %
b. C2H5OH
Mm C: 1 x 12 = 12 g/mol
O:1 x 16 = 16 g/mol
H: 6 x 1 = 6 g/mol
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Mm = 46 g/mol
H: 46 g ---- 100 %
3 g ---- x
X = 13.04 %
c. C6H6
Mm C: 6 x 12 = 72 g/mol
H: 6 x 1 = 6 g/mol
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Mm = 78 g/mol
H: 78 g ---- 100 %
3 g ---- x
X = 3.846 %
d. NH3
Mm N: 1 x 14 = 14 g/mol
H: 3 x 1 = 3 g/mol
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Mm = 17 g/mol
H: 17 g ---- 100 %
3 g ---- x
X = 17.65 %
e. SiH4
Mm Si: 1 x 28 = 28 g/mol
H: 4 x 1 = 4 g/mol
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Mm = 32 g/mol
H: 32 g ---- 100 %
4 g ---- x
X = 12.5 %
EL AMONIACO TIENE MAYOR PORCENTAJE DE HIDROGENO CON 17,65%
3.) CnH2n+2 C8H(2.8+2) = C8H18
Mm C8H18 = 114 g/mol
4. C6H15O3N
Mm C: 6 x 12 = 72 g/mol
H:15 x 1 = 15 g/mol
O:3 x 16 = 48 g/mol
N: 1 x 14 = 14 g/mol
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Mm = 149 g/mol
C: 149 g ---- 100 %
72 g ----- x
X= 48.32%
H: 149 g ---- 100 %
15 g ----- x
X= 10.06%
O: 149 g ---- 100 %
48 g ----- x
X= 32.21%
N: 149 g ---- 100 %
14 g ----- x
X= 9.39%
5. C: 40.1 g/ 12 g/mol = 3.342 mol
H: 6.6 g/ 1 g/mol = 6.6 mol
O: 53.3 g/16 g/mol = 3.332 mol
Dividir cada elemento entre el menor de los resultados
C: 3.342 mol = 3.332 mol = 1
H: 6.6 mol / 3.332 mol = 2
O: 3.332 mol / 3.332 mol = 1
FE= CH2O
7. C: 35.51 g/ 12 g/mol = 2.926 mol
H: 4.77 g/ 1 g/mol = 4.77 mol
N: 8.29 g/ 14 g/mol = 0.592 mol
O: 37.85 g/16 g/mol = 2.366 mol
Na: 13.06 g/23 g/mol = 0.568 mol
Dividir cada elemento entre el menor de los resultados
C: 2.926 mol / 0.56 mol = 5
H: 4.77 mol / 0.568 mol = 8
N: 0.592 mol / 0.568 mol = 1
O: 2.366 mol / 0.568 mol = 4
Na: 0.568 mol / 0.568 mol =1
FE: C5H8NO4Na
Mm FE
C: 5 x 12 = 60 g/mol
H: 8 x 1 = 8 g/mol
N: 1 x 14 = 14 g/mol
O: 4 x 16 = 64 g/mol
Na: 1 x 23 = 23 g/mol
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Mm FE = 169 g/mol
n= 169 g/mol / 169 g/mol
n = 1
Fmolecular = (C5H8O4Na)1 = C5H8NO4Na