a)
H: 1,587/1 = 1,587 :1,587 => 1 atom H
N: 22,22/14 = 1,587 :1,587 => 1 atom N
O: 77,19/16 = 4,82 :1,587 => 3 atomi O
Formula bruta (F.B) ⇒ (HNO₃)n
Formula moleculara (F.M) ⇒ (HNO₃)n = 63 ⇔ 1n+14n+38n = 63 ⇔ 63n = 63 ⇒ n = 1 ⇒ HNO₃ - acid azotic
b) H:N:O = 1:14:48
c) MHNO₃ = 63g/mol
63g HNO₃.......................3*16g O
99g HNO₃...........................x = 75,428g O
d) n = m/M = 25,2/63 = 0,4 moli HNO₃
1 mol HNO₃...............N.A molecule = 6,022*10²³
0,4 moli HNO₃.....................x = 2,4088*10²³ molecule
e) 1 mol = 1000 mmoli
16 mmoli = 0,016 moli
63g HNO₃..............3 moli O
xg HNO₃...............0,016 moli O
x = 0,336g HNO₃
