1965 to date   Clad CoinageComposition:  75% copper, 25% nickel Weight:  2.27g Diameter:  17.9mm    Edge:  Reeded

A.P. Chemistry.

Gravimetric Analysis using PPT reaction information…a review and extension

Numismatics is the scientific study of money and its history in all its varied forms.

"All science is either Physics or stamp collecting." - Ernest Rutherford physicist and Noble Laureate.

Pre-1965 Composition:  silver and copper…but how much of each typically???

This Roosevelt dime is different though.

A.P. Chem

Gravimetric Analysis:

PURPOSE:

Procedure:

1.

2.

3.

4.

5.

6.

7.

Calculations:

Data

To determine the silver composition of a pre-1965 dime

Mass dime 2.8357 g

Dissolve the dime in HNO3

Ag + NO3- + H+ --> Ag + + NO + H2O

Add HCl Ag+ + Cl- --> AgCl

Filter and dryMass of Filter and Sample = 4.1860 g (Filter paper = 0.7942 g)

Mass AgCl AgCl = 3.3918 g

Find % comp. of AgCl % Ag = 75.265 %

Find mass of Ag Ag = 2.5528 g

6.) 107.87 / 143.32 = 0.75265

7.) 3.3918 x 0.75265 =

Answer: ( 2.5528 g / 2.8357 g ) x 100% = 90.024%

A type of quantitative analysis in which the amount of one species in a material is determined by isolating and massing it

What volume of 6.0 M HCl would have been needed to completely precipitate out all of the silver ions from the solution if the concentration of the silver ion was 0.473 M and there was 50.0 mL of the solution?

What volume of 6.0 M HCl would have been needed to completely precipitate out all of the silver ions from the solution if the concentration of the silver ion was 0.473 M and there was 50.0 mL of the solution?

50.0 ml Ag+ 0.473 mol = 1000.0 mL

0.0237 mol Ag+

0.0237 mol Ag+ 1.0 mol HCl = 1.0 mol Ag+

0.0237 mol HCl

0.0237 mol HCl 1000.0 mL HCl = 4.0 mL HCl sol’n

Ag+ + Cl- --> AgClAg+ + HCl(aq) --> AgCl + H+

6.0 mol HCl

50.0 ml Ag+ 0.473 mol Ag+ 1 mol HCl 1000.0 mL HCl

1 1000.0 mL Ag+ 1 mol Ag+ 6.0 mol HCl

Ag+ + H++ Cl---> AgCl + H+

=

What mass of Fe(OH)3 is produced when 50.0 mL of 6.0 M LiOH sol’n reacts with an excess of 2.0 M Fe(NO3)3 sol’n?

Fe(NO3)3(aq) + LiOH(aq)→ Fe(OH)3(s) + LiNO3(aq)3 3

50.0 ml LiOH 6.0 mol LiOH 1 mol Fe(OH)3 106.8 g Fe(OH)3

1 1000.0 mL LiOH 3 mol LiOH 1.0 mol Fe(OH)3

=

11 g Fe(OH)3

What volume of 6.0 M HCl would have been needed to completely precipitate out all of the silver ions from the solution if the concentration of the silver ion was 0.473 M and there was 50.0 mL of the solution?

What mass of Fe(OH)3 is produced when 50.0 mL of 6.0 M LiOH sol’n reacts with an excess of 2.0 M Fe(NO3)3 sol’n?

What volume of 6.0 M HCl would have been needed to completely precipitate out all of the silver ions from the solution if the concentration of the silver ion was 0.473 M and there was 50.0 mL of the solution?

What mass of Fe(OH)3 is produced when 50.0 mL of 6.0 M LiOH sol’n reacts with an excess of 2.0 M Fe(NO3)3 sol’n?