실험식. hydrate 20.3% Cu 8.95% Si 36.3% F 34.5% H2O
실험식. hydrate 20.3% Cu 8.95% Si 36.3% F 34.5% H2O
A certain hydrate is found to have the composition:
20.3% Cu, 8.95% Si, 36.3% F and 34.5% H2O.
What is the empirical formula of this hydrate?
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▶ 참고: 원소분석
[ https://ywpop.tistory.com/64 ]
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시료의 질량 = 100 g 이라 가정하면,
> 20.3 g Cu
> 8.95 g Si
> 36.3 g F
> 34.5 g H2O
각 성분의 몰수 계산
> 20.3 g / (63.55 g/mol) = 0.319 mol Cu
( 참고 https://ywpop.tistory.com/7738 )
> 8.95 / 28.09 = 0.319 mol Si
> 36.3 / 19.00 = 1.91 mol F
> 34.5 / 18.02 = 1.91 mol H2O
몰수의 가장 작은 정수비 계산
Cu : Si : F : H2O = 0.319 : 0.319 : 1.91 : 1.91
= 0.319/0.319 : 0.319/0.319 : 1.91/0.319 : 1.91/0.319
= 1 : 1 : 5.99 : 5.99
≒ 1 : 1 : 6 : 6
---> 실험식 = CuSiF6*6H2O
답: CuSiF6*6H2O
[키워드] A certain hydrate is found to have the following percent composition by mass: 20.3% Cu, 8.95% Si, 36.3% F, and 34.5% H2O.