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화학/원소(연소)분석

실험식. hydrate 20.3% Cu 8.95% Si 36.3% F 34.5% H2O

by 영원파란 2022. 11. 9.

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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.

 

 

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