CHO 연소분석. 0.5438 g sample 1.039 g CO2 0.6369 g H2O
0.5438 g CHO sample 1.039 g CO2 0.6369 g H2O
A 0.5438 g sample of a liquid
consisting of only C, H, and O was burned in pure oxygen
and 1.039 g of CO2 and 0.6369 g of H2O were obtained.
What is the empirical formula of the compound?
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▶ 참고: 원소분석 [ https://ywpop.tistory.com/64 ]
▶ 참고: CH 연소분석 [ https://ywpop.tistory.com/9606 ]
▶ 참고: CHO 연소분석 [ https://ywpop.tistory.com/2854 ]
각 성분의 질량을 계산하면,
1.039 × (12/44) = 0.28336 g C
0.6369 × (2/18) = 0.070767 g H
0.5438 – (0.28336 + 0.070767) = 0.18967 g O
각 성분의 몰수를 계산하면,
0.28336 / 12 = 0.023613 mol C
0.070767 / 1 = 0.070767 mol H
0.18967 / 16 = 0.011854 mol O
몰수의 가장 작은 정수비를 계산하면,
C : H : O = 0.023613 : 0.070767 : 0.011854
= 0.023613/0.011854 : 0.070767/0.011854 : 0.011854/0.011854
= 1.992 : 5.9699 : 1
≒ 2 : 6 : 1
---> 실험식 = C2H6O
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