Answer to Question #15642 in Other Chemistry for Abdullah
a gaseous organic compound X was burnt in the excess of oxygen.A 0.112 dm3 sample of X measured at s.t.p produced 0.88 grams of CO2 how many carbon atoms are present in one molecule of x ?
If organic compound X has formula C(x)H(y)O(z), the reaction is C(x)H(y)O(z) + k O2 = x CO2 + (y/2) H2O
IUPAC has defined standard reference conditions as being 0 °C and 100 kPa (1 bar) R = 8.314×10−2 L*bar*K(-1)*mol(-1) pV = nRT n = pV/(RT) n(X) = 1 bar * 0.112 L / (8.314×10−2 L*bar*K(-1)*mol(-1) * 273 K) = 0.005 mol n(CO2) = 0.88 g / 44 g/mol = 0.02 mol
x = n(CO2)/n(X) = 0.02 / 0.005 = 4 Four carbon atoms are present in one molecule of X.