例四这道化学题,有大神帮忙解答下详细过程?
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解析:
n(C有) = n(CO2) = 2.64 g / 44g/mol = 0.06 mol
m(C有) = 0.06 mol × 12 g/mol = 0.72 g
n(H有) = n(H2O)× 2 = (0.54 g / 18 g/mol)× 2 = 0.06 mol
m(H有) = 0.06 mol × 1 g/mol = 0.06 g
所以m(O有)= 0.94 g - 0.72 g - 0.06 g = 0.16 g
n(O有) = 0.16 g / 16 g/mol = 0.01 mol
n(C有):n(H有):n(O有)= 0.06 mol:0.06 mol:0.01 mol = 6:6:1
所以此有机物分子式为:C6H6O
即:C6H5-OH(苯酚)
n(C有) = n(CO2) = 2.64 g / 44g/mol = 0.06 mol
m(C有) = 0.06 mol × 12 g/mol = 0.72 g
n(H有) = n(H2O)× 2 = (0.54 g / 18 g/mol)× 2 = 0.06 mol
m(H有) = 0.06 mol × 1 g/mol = 0.06 g
所以m(O有)= 0.94 g - 0.72 g - 0.06 g = 0.16 g
n(O有) = 0.16 g / 16 g/mol = 0.01 mol
n(C有):n(H有):n(O有)= 0.06 mol:0.06 mol:0.01 mol = 6:6:1
所以此有机物分子式为:C6H6O
即:C6H5-OH(苯酚)
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