дано
m(CxHyOz) = 3.42 g
m(CO2) = 5.28 g
m(H2O) = 1.98g
M(CxHyOz) = M(H2O) * 19
CxHyOz-?
M(CO2) = 44 g/mol
n(CO2) = m(CO2) / M(CO2) = 5.28 / 44 = 0.12 mol
n(C) = n(CO2) = 0.12 mol
M(C) = 12 g/mol
m(C) = n(C) * M(C) = 0.12 *12 = 1.44 g
M(H2O) = 18 g/mol
n(H2O) = m(H2O) / M(H2O) = 1.98 / 18 = 0.11 mol
n(H) = 2*n(H2O) = 2 * 0.11 = 0.22 mol
M(H) = 1 g/mol
m(H) = n(H) * M(H) = 0.22 * 1 = 0.22 g
m(O) = m(CxHyOz) - (m(C) + m(H)) = 3.42 - ( 1.44 + 0.22) = 1.76 g
M(O) = 16 g/mol
n(O) = m(O) / M(O) = 1.76 / 16 = 0.11 mol
n(C) : n(O) : n(H) = 0.12 : 0.11: 0.22 = 12:11 : 22
C12O11H22
M(CxHyOz) = M(H2O) * 19 = 18 * 19 = 342 g/mol
M(C12H22O11) = 342 g/mol
ответ : САХАРОЗА
Объяснение:
В наверное ведь она содержит R-COH