Determine la masa molecular de los siguientes compuestos :Fe2 O3 (oxido férrico)Na OH (hidróxido de sodio)H N O3 (ACIDO NITRICO)C55 H72 Mg N4 O5 (CLOROFILA)K Mn O4 (PERGAMANGATO DE POTASIO)Cu Fe Po4 (?

Fe2 O3 (oxido férrico)

Fe : 2 x 55.

84 = 111.

68 g / mol

O : 3 x 16 = 48 g / mol

```````````````````````````````` Mm = 159.

68 g / mol

Na OH (hidróxido de sodio)

Na : 1 x 23 = 23 g / mol

O : 1 x 16 = 16 g / mol H : 1 x 1 = 1 g / mol

`````````````````````````````` Mm = 40 g / mol

H N O3 (ACIDO NITRICO)

H : 1 x 1 = 1 g / mol

N : 1 x 14 = 14 g / mol

O : 3 x 16 = 48 g / mol

`````````````````````````` Mm = 63 g / mol

C55 H72 Mg N4 O5 (CLOROFILA)

C : 55 x 12 = 660 g / mol

H : 72 x 1 = 72 g / mol

Mg : 1 x 24 = 24 g / mol

N : 4 x 14 = 56 g / mol

O : 5 x 16 = 80 g / mol

`````````````````````````````` Mm = 892 g / mol

K Mn O4 (PERGAMANGATO DE POTASIO)

K : 1 x 39 = 39 g / mol

Mm : 1 x 54.

9 = 54.

9 g / mol

O : 4 x 16 = 64 g / mol

```````````````````````````````` Mm = 157.

9 g / mol

Cu Fe Po4 (FOSFATO CUPROSO FERROSO)

Cu : 1 x 63.

54 = 63.

54 g / mol

Fe : 1 x 55.

84 = 55.

84 g / mol

P : 1 x 30.

97 = 30.

97 g / mol

O : 4 x 16 = 64 g / mol

```````````````````````````````````` Mm = 214.

35 g / mol

Na H Co3 (BICARBONATO DE SODIO)

Na : 1 x 23 = 23 g / mol

H : 1 x 1 = 1 g / mol

C : 1 x 12 = 12 g / mol

O : 3 x 16 = 48 g / mol

`````````````````````````````` Mm = 84 g / mol.