0.7152 g sample excess H5IO6 0.3408 g Hg2Cl2 percent

0.7152 g sample excess H5IO6 0.3408 g Hg2Cl2 percent

 

 

The mercury in a 0.7152 g sample was precipitated

with an excess of paraperiodic acid, H5IO6:

5Hg^2+ + 2H5IO6 → Hg5(IO6)2 + 10H^+

The precipitate was filtered, washed free of precipitating agents,

dried and weighed, 0.3408 g being recovered.

Calculate the percentage of Hg2Cl2 in the sample.

 

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5Hg^2+(aq) + 2H5IO6(aq) → Hg5(IO6)2(s) + 10H^+(aq)

 

 

 

Hg5(IO6)2의 몰질량 = 1448.75 g/mol 이므로,

0.3408 g / (1448.75 g/mol) = 0.000235237 mol Hg5(IO6)2

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

Hg5(IO6)2 : Hg = 1 : 5 계수비(= 몰수비) 이므로,

0.000235237 mol Hg5(IO6)2 × (5 mol Hg^2+ / 1 mol Hg5(IO6)2)

= 0.000235237 × 5 = 0.001176 mol Hg^2+

 

 

 

Hg2Cl2 : Hg = 1 : 2 = ? mol : 0.001176 mol

 

? = 0.001176 / 2 = 0.0005880 mol Hg2Cl2

 

 

 

Hg2Cl2의 몰질량 = 472.09 g/mol 이므로,

0.0005880 mol × (472.09 g/mol) = 0.2776 g Hg2Cl2

 

 

 

시료 중 Hg2Cl2의 함량(%)을 계산하면,

(0.2776 g / 0.7152 g) × 100 = 38.81%

 

 

 

답: 38.81%

 

 

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[Hg2Cl2의 질량 계산 공식]

> 침전물, Hg5(IO6)2의 질량 = x g

 

Hg2Cl2의 질량(g)

= x / (1448.75) × 5 / 2 × (472.09)

= 0.3408 / (1448.75) × 5 / 2 × (472.09)

= 0.2776 g

 

 

 

 

[예제]

> 시료의 질량 = 1.0451 g

> 침전물, Hg5(IO6)2의 질량 = 0.5718 g

 

 

Hg2Cl2의 질량(g)

= 0.5718 / (1448.75) × 5 / 2 × (472.09)

= 0.4658 g

 

 

Hg2Cl2의 함량(%)

= (0.4658 / 1.0451) × 100

= 44.57%

 

 

 

 

[키워드] 1.0451 g sample excess H5IO6 0.5718 g Hg2Cl2 percent