Question #15642
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 ?
Expert's answer
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.
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.
