Combustion of a 0.9835-g sample of a compound containing only carbon,hydrogen,and oxygen produced 1.900 g of CO2 and 1.070 g of H2O.a.What isthe empirical formula of the compound?
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Combustion of a 0.9835-g sample of a compound containing only carbon,hydrogen,and oxygen produced 1.900 g of CO2 and 1.070 g of H2O.a.What isthe empirical formula of the compound?
Combustion of a 0.9835-g sample of a compound containing only carbon,hydrogen,and oxygen produced 1.900 g of CO2 and 1.070 g of H2O.
a.What isthe empirical formula of the compound?
Combustion of a 0.9835-g sample of a compound containing only carbon,hydrogen,and oxygen produced 1.900 g of CO2 and 1.070 g of H2O.a.What isthe empirical formula of the compound?
化合物中含有C的质量:1.900*(12/44)=0.5182g
化合物中含有H的质量:1.070*(2/18)=0.1189g
化合物中含有O的质量:0.9835-0.5182-0.1189=0.3464g
则:分子式中系数之比:
0.5182/12 :0.1189/1 :0.3464/16
=0.04318 :0.1189 :0.02165
=4:11:2
所以,经验化学式为:C4H11O2或者[C4H11O2]n
混合物里只有H2,O2和C。生成物中的CO2中的C元素完全来自于混合物中的C。所以
C ~ CO2,我们知道C摩尔质量为12g,CO2摩尔质量为44g。根据比例计算有C质量为1.9/44*12= 0.5182g。相同方法可以计算H2的质量为1.070/18*2=0.1189g。所以O2质量为0.9835g-0.5182g- 0.1189g= 0.3464g所以这题的答案是?化合物的化学式是...
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混合物里只有H2,O2和C。生成物中的CO2中的C元素完全来自于混合物中的C。所以
C ~ CO2,我们知道C摩尔质量为12g,CO2摩尔质量为44g。根据比例计算有C质量为1.9/44*12= 0.5182g。相同方法可以计算H2的质量为1.070/18*2=0.1189g。所以O2质量为0.9835g-0.5182g- 0.1189g= 0.3464g
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