CHO 0.0956 g 0.1356 g CO2 0.0833 g H2O 62.1 g/mol

CHO 0.0956 g 0.1356 g CO2 0.0833 g H2O 62.1 g/mol

 

 

An unknown compound has the formula CHO.

You burn 0.0956 g of the compound

and isolate 0.1356 g of CO2 and 0.0833 g of H2O.

What is the empirical formula of the compound?

If the molar mass is 62.1 g/mol, what is the molecular formula?

 

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▶ 참고: 원소분석 [ https://ywpop.tistory.com/64 ]

▶ 참고: CHO 연소분석 [ https://ywpop.tistory.com/2854 ]

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각 성분의 질량을 계산하면,

C: 0.1356 g × (12/44) = 0.03698 g

H: 0.0833 g × (2/18) = 0.009256 g

O: 0.0956 – (0.03698 + 0.009256) = 0.049364 g

 

 

 

각 성분의 몰수를 계산하면,

C: 0.03698 g / (12 g/mol) = 0.0030817 mol

H: 0.009256 / 1 = 0.009256 mol

O: 0.049364 / 16 = 0.00308525 mol

 

 

 

몰수의 가장 작은 정수비를 계산하면,

C : H : O = 0.0030817 : 0.009256 : 0.00308525

= 0.0030817/0.0030817 : 0.009256/0.0030817 : 0.00308525/0.0030817

= 1 : 3 : 1

---> 실험식 = CH3O

---> 실험식량 = (12) + 3(1) + (16) = 31

 

 

 

분자량 / 실험식량 = 62.1 / 31 = 2 = n

 

 

 

분자식 = n(실험식) = 2(CH3O) = C2H6O2

 

 

 

답: 실험식 = CH3O, 분자식 = C2H6O2