2.842 g sample BCl 0.153 L 0.664 g B Cl2 0.688 L

2.842 g sample BCl 0.153 L 0.664 g B Cl2 0.688 L

 

 

A 2.842 g sample of a gaseous compound of boron and chlorine

occupies 0.153 L at a particular temperature and pressure.

This sample is decomposed to give 0.664 g of solid boron and

enough Cl2 gas to occupy 0.688 L at the original temperature and pressure.

Determine the empirical formula and molecular formula of the compound.

 

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▶ 참고: 원소분석

[ https://ywpop.tistory.com/64 ]

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B의 몰질량 = 10.81 g/mol 이므로,

0.664 g / (10.81 g/mol) = 0.06142 mol B

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

Cl의 질량 = 2.842 – 0.664 = 2.178 g

 

Cl의 몰질량 = 35.45 g/mol 이므로,

2.178 g / (35.45 g/mol) = 0.06144 mol Cl

 

 

 

몰수의 가장 작은 정수비를 계산하면,

B : Cl = 0.06142 : 0.06144 ≒ 1 : 1

---> 실험식 = BCl

---> 실험식량 = 10.81 + 35.45 = 46.26

 

 

 

Cl2의 몰질량 = 70.91 g/mol 이므로,

2.178 g / (70.91 g/mol) = 0.030715 mol Cl2

 

 

 

“기체의 부피 ∝ 기체의 몰수” 이므로,

BxClx : Cl2 = 0.153 L : 0.688 L = ? mol : 0.030715 mol

? = 0.153 × 0.030715 / 0.688 = 0.0068305 mol BxClx

 

 

 

BxClx의 몰질량을 계산하면,

2.842 g / 0.0068305 mol = 416.07 g/mol

 

 

 

분자량 / 실험식량 = n

416.07 / 46.26 = 8.99 ≒ 9 = n

 

 

 

분자식 = n(실험식) = 9(BCl) = B9Cl9