WAEC - Chemistry (2015 - No. 39)
A compound has an empirical formular CH2O and molecular mass of 90.
[H = 1.00, C = 12.0, O = 16.0]
[H = 1.00, C = 12.0, O = 16.0]
C4H10O2
C3H10O2
C3H6O3
C2H2O4
Giải thích
(CH2O)n = 90; (12 + (1 x 2) + 16)n = 90
(12 + 2 + 16)n = 90; 30n = 90; n = \(\frac{90}{30}\); n = 3
(CH2O)3 = Molecular formula ; = C3H6O3
(12 + 2 + 16)n = 90; 30n = 90; n = \(\frac{90}{30}\); n = 3
(CH2O)3 = Molecular formula ; = C3H6O3