A 0.4494 g sample of pewter, containing tin, lead, copper, and zinc, was dissolved in acid. Tin was precipitated as Sno, . 4 H, O and removed by filtration. The resulting filtrate and washings were diluted to a total volume of 200.0 mL. A 15.00 mL aliquot of this solution was buffered, and titration of the lead, copper, and zinc in solution required 35.07 mL of 0.001463 M EDTA. Thiosulfate was used to mask the copper in a second 20.00 mL aliquot. Titration of the lead and zinc in this aliquot required 34.08 mL of the 0.001463 M EDTA solution. Finally, cyanide was used to mask the copper and the zinc in a third 25.00 mL aliquot. Titration of the lead in this aliquot required 26.37 mL of the 0.001463 M EDTA solution. Determine the percent composition by mass of each metal in the pewter sample. Cu : % Zn : % Pb : % Sn : %

 

Percent of Copper in sample = 2.60%

Percent of zinc sample = 2.78%

Percent of lead sample = 14.27%

Percentage of tin in sample = 80.35%

Step-by-step explanation

Given:

 

Molarity of EDTA solution = 0.001463 M

Molarity of EDTA solution = 0.001463 mol/L

 

Volume of EDTA solution required by lead = 26.37 mL (1L / 1000mL)

Volume of EDTA solution required by lead = 0.02637 L

 

Thus,

number of moles of Pb²⁺ present in 25 mL solution = number of moles of EDTA required = (Molarity) x (Volume)

number of moles of Pb²⁺ present in 25 mL solution = ( 0.001463 mol/L)(0.02637 L)

number of moles of Pb²⁺ present in 25 mL solution = 0.0000386 mol

 

Volume required by zinc and lead in 20 mL solution = 34.08 mL ( 1L/1000mL)

Volume required by zinc and lead in 20 mL solution = 0.03408 L

 

 

Similarly,

Total number of lead and zinc ions in 20 mL solution = (0.001463 mol/L) (0.03408 L)

Total number of lead and zinc ions in 20 mL solution = 0.0000499 mol

 

 

So,

number of moles of lead in 20 mL solution = ( 20 mL)( 0.0000386 mol / 25 mL)

number of moles of lead in 20 mL solution = 0.0000309 mol

 

Moles of zinc in 20 mL solution = 0.0000499 mol - 0.0000309 mol

Moles of zinc in 20 mL solution = 0.0000190 mol

 

 

Volume of EDTA required by 15 mL lead, copper and zinc = 35.07 mL (1L/1000mL)

Volume of EDTA required by 15 mL lead, copper and zinc = 0.03507 L

 

Moles of copper, lead and zinc in 15 mL solution = (0.001463 mol/L) (0.03507 L)

Moles of copper, lead and zinc in 15 mL solution = 0.0000513 mol

 

 

Thus,

moles of lead in 15 mL solution = (0.0000386 mol) (15 mL / 25 mL)

moles of lead in 15 mL solution = 0.0000232 mol

 

 

Moles of zinc in 15 mL solution = (0.0000190 mol)(15 mL / 20 mL)

Moles of zinc in 15 mL solution = 0.0000143 mol

 

Hence,

moles of copper present in 15 mL solution = total moles - number of moles of lead and zinc

moles of copper present in 15 mL solution = 0.0000513 mol - 0.0000232 mol - 0.0000143 mol

moles of copper present in 15 mL solution = 0.0000138 mol

 

Moles of copper in 200 mL solution = (200 mL)(0.0000138 mol / 15mL)

Moles of copper in 200 mL solution = 0.000184 mol

 

Mass of copper in 200 mL solution = (moles)(molar mass)

Mass of copper in 200 mL solution = (0.000184 mol)(63.55 g/mol)

Mass of copper in 200 mL solution = 0.0117 g

 

Percent of Copper in sample = (mass of copper / total mass) x 100%

Percent of Copper in sample = (0.0117 g / 0.4494 g) x 100%

Percent of Copper in sample = 2.60%

 

 

Mass of Zinc in 200 mL solution = (200 mL)(0.0000143 mol / 15 mL)

Mass of Zinc in 200 mL solution = 0.000191 mol

 

Mass of zinc 200 mL solution = (moles)(molar mass)

Mass of zinc 200 mL solution = (0.000191 mol)(65.38 g/mol)

Mass of zinc 200 mL solution = 0.0125 g

 

Percent of zinc sample = (mass of zinc/total mass) x 100%

Percent of zinc sample = (0.0125g / 0.4494 g) x 100%

Percent of zinc sample = 2.78%

 

 

Mass of lead 200 mL solution = (200 mL)(0.0000232 mol/ 15 mL)

Mass of lead in 200 mL solution = 0.000309 mol

 

Mass of lead 200 mL solution = (moles)(molar mass)

Mass of lead 200 mL solution = (0.000309 mol)(207.2 g/mol)

Mass of lead 200 mL solution = 0.0641 g

 

Percent of lead sample = (mass of lead /total mass) x 100%

Percent of lead sample = (0.0641g/ 0.4494 g) x 100%

Percent of lead sample = 14.27%

 

Thus,

Percentage of tin in sample = 100% - ( Percent of copper in sample + Percent of zinc sample + Percent of lead sample)

Percentage of tin in sample = 100% - (2.60% - 2.78%- 14.27%)

Percentage of tin in sample = 80.35%