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發問:
on complete combustion, a 3.41 g of compound (CxHyOz) produced 8.6 g of carbon dioxide and 1.51 g of water.WHAT IS The empirical formula of the compound . HOW TO DO??????
最佳解答:
Molar mass of C = 12 g mol-1 Molar mass of H = 1 g mol-1 Molar mass of O = 16 g mol-1 Molar mass of CO2 = 12 + 16x2 = 44 g mol-1 Molar mass of H2O = 1x2 + 16 = 18 g mol-1 Fraction by mass of C in CO2 = 12/44 Fraction by mass of H in H2O = 2/18 ===== Unbalance equation : CxHyOz + O2 → CO2 + H2O In 3.41 g of the compound : Mass of C = Mass of C in CO2 formed = 8.6 x (12/44) = 2.345 g Mass of H = Mass of H in H2O formed = 1.51 x (2/18) = 0.168 g Mass of O = 3.41 - (2.345 + 0.168) = 0.897 g ===== In the compound, mole ratio C : H : O = (2.345/12) : (0.168/1) : (0.897/16) = 0.195 : 0.168 : 0.056 = 3.5 : 3 : 1 = 7 : 6 : 2 Hence, empirical formula = C7H6O2
其他解答:
The mass of C = 8.6 x 12/(12+16x2) = 2.35g The mass of H = 1.51 x 2/(16+2) = 0.168g The mass of O = 3.41 - 2.35 - 0.168 = 0.892g The number of mol of C = 2.35/12 = 0.196 The number of mol of H = 0.168/1 = 0.168 The number of mol of O = 0.892/16 = 0.0556 The ratio of mol C:H:O = 0.196/0.0556 : 0.168/0.0556 : 0.0556/0.0556 C:H:O = 3.53 : 3.02 : 1 = 7:6:2 so the empirical formula of the compound is C7H6O2
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f.4 chemistry發問:
on complete combustion, a 3.41 g of compound (CxHyOz) produced 8.6 g of carbon dioxide and 1.51 g of water.WHAT IS The empirical formula of the compound . HOW TO DO??????
最佳解答:
Molar mass of C = 12 g mol-1 Molar mass of H = 1 g mol-1 Molar mass of O = 16 g mol-1 Molar mass of CO2 = 12 + 16x2 = 44 g mol-1 Molar mass of H2O = 1x2 + 16 = 18 g mol-1 Fraction by mass of C in CO2 = 12/44 Fraction by mass of H in H2O = 2/18 ===== Unbalance equation : CxHyOz + O2 → CO2 + H2O In 3.41 g of the compound : Mass of C = Mass of C in CO2 formed = 8.6 x (12/44) = 2.345 g Mass of H = Mass of H in H2O formed = 1.51 x (2/18) = 0.168 g Mass of O = 3.41 - (2.345 + 0.168) = 0.897 g ===== In the compound, mole ratio C : H : O = (2.345/12) : (0.168/1) : (0.897/16) = 0.195 : 0.168 : 0.056 = 3.5 : 3 : 1 = 7 : 6 : 2 Hence, empirical formula = C7H6O2
其他解答:
The mass of C = 8.6 x 12/(12+16x2) = 2.35g The mass of H = 1.51 x 2/(16+2) = 0.168g The mass of O = 3.41 - 2.35 - 0.168 = 0.892g The number of mol of C = 2.35/12 = 0.196 The number of mol of H = 0.168/1 = 0.168 The number of mol of O = 0.892/16 = 0.0556 The ratio of mol C:H:O = 0.196/0.0556 : 0.168/0.0556 : 0.0556/0.0556 C:H:O = 3.53 : 3.02 : 1 = 7:6:2 so the empirical formula of the compound is C7H6O2
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