Express each final result to an appropriate number of significant figures.
1. You have a sample containing chromate, CrO42-, to be determined by precipitation with AgNO3. You are told the sample will contain between 10% and 30% by weight of Chromium. You want both the sample weighed and the silver chromate obtained to weigh no less than 0.2 grams for accurate mass measurement.
A. Write the balanced chemical reaction between silver nitrate and chromate.
Ans to 1.A. 2Ag+ + 2NO3- + CrO42- = Ag2CrO4 (s) + 2NO3-
B. What sample size should you weigh out? [Hint: choose a value, then do (C) below and see if you like the result.]
Ans to 1.B. Try 0.2 g of unk and do the calculation in C (too small, 0.3 g needed -- see C)
C. What mass range of silver chromate will you obtain?
Ans to 1.C.
0.2 g unk
0.1 g Cr mole Cr 1 mole Ag2CrO4 33.73 g Ag2CrO4 g unk 51.996 g Cr 1 mole Cr mole Ag2CrO4 = 0.128 g Ag2CrO4
Which is smaller than 0.2 g, so not enough unk was taken. Calculate how much unknown it would take to generate the minimum 0.2 g silver chromate required.
0.2 g Ag2CrO4
mole Ag2CrO4 1 mole Cr 51.996 g Cr g unk 331.73 g 1 mole Ag2CrO4 mole Cr 0.1 g Cr = 0.313 g unk minimum
So if 0.313 g of 10% Cr gives 0.2 g of silver chromate, then a 30% sample will give 0.6 grams.
D. How many mL of 0.08995 M AgNO3 should you use to ensure precipitation of all the chromate in your sample?
Ans to 1.D.
0.313 g unk
0.3 g Cr max mole Cr 1 mole Ag2CrO4 2 mole Ag+ g unk 51.996 g Cr 1 mole Cr 1 mole Ag2CrO4
liter
1000 mL 0.08995 mole Ag+ liter = 40.15 mL
add 10% excess for a total of 44 mL.
2. A mixture containing only Al2O3 (FW 101.96) and Fe2O3 (FW 159.69) weighs 2.019 g. When heated under a stream of H2, the Al2O3 is unchanged, but the Fe2O3 is converted to metallic Fe plus H2O (g). If the residue weighs 1.774 g, what is the weight percent of Fe2O3 in the original mixture? (From Harris, 4th ed.)
Ans to 2. Let x = moles Fe2O3Let y = moles Al2O3
The sample contains only Fe2O3 and Al2O3, so
x moles Fe2O3 x 159.69 g/mole + y moles Al2O3 x 101.96 g/mole = 2.019 g
Then only the Fe2O3 is converted to Fe
x moles Fe2O3 x 2 moles Fe/1 mole Fe2O3 x 55.847 gFe/mole + y moles Al2O3 x 101.96 g/mole = 1.774 g
This gives the system of equations:
159.69 x + 101.96 y = 2.019
111.694 x + 101.96 y = 1.774
-----------------------------------
47.996 x + 0 = 0.245; x = 0.245/47.996 = 5.105 x 10-3 moles Fe2O3
5.105 x 10-3 moles Fe2O3
159.59 g 100% mole 2.019 g sample = 40.4%
A short-cut is possible if we note that the 0.245 g lost is all oxygen. Compute moles of oxygen, then moles of Fe2O3 and % as above. But it is best to learn methods that can work on a wide variety of problems.
3. You are running three replicates of an unknown. The directions call for dissolving each unknown in 150 mL of 0.02 M HNO3, then carrying out a precipitation. Following the precipitation you are to wash each sample three times with 25 mL portions of 0.02 M HNO3.
A. How much wash solution should you prepare?
Ans to 3.A. Each unknown needs 150 ml to dissolve, plus 3 x 25 = 75 ml wash. 225 ml x 3 unks = 675 ml total needed. Make 800 mL in a 1 liter erlenmeyer flask.
B. Does this solution have to be made EXACTLY 0.02000 M or is substantially less accuracy OK? Explain, stating what accuracy is called for here (+/-20%, +/-10%, +/-5%, +/-1%, +/-0.1%?)
Ans to 3.B. This solution is not a calibration standard, nor does it have to be a precise pH, so 10-20% accuracy is good enough.
C. Describe how you will make your wash solution from the reagent nitric acid in the hood. Can you make it in one step, or will you have to prepare a lower concentration of nitric acid first? Explain.
Ans to 3.C.
800 mL
0.02 mmole HNO3 mL conc HNO3 mL 16 mmole conc HNO3 = 1.00 mL conc. HNO3
We can probably get 10% accuracy measuring 1 ml with a small graduated cylinder. For more precison onother solutions we would need an intermdiate dilution. If we need to make more wash solution a dilution of the 16M would be safer to handle too. This is why the procedure in lab tells you to use 6M HNO3.