CHAPTER 0

THE ANALYTICAL PROCESS

1

0-1. Qualitative analysis finds out what is in a sample. Quantitative analysis measures

how much is in a sample.

0-2. Steps in a chemical analysis:

(1) Formulate the question: Convert a general question into a specific one that

can be answered by a chemical measurement.

(2) Select the appropriate analytical procedure.

(3) Obtain a representative sample.

(4) Sample preparation: Convert the representative sample into a sample suitable

for analysis. If necessary, concentrate the analyte and remove or mask

interfering species.

(5) Analysis: Measure the unknown concentration in replicate analyses.

(6) Produce a clear report of results, including estimates of uncertainty.

(7) Draw conclusions: Based on the analytical results, decide what actions to

take.

0-3. Masking converts an interfering species to a noninterfering species.

0-4. A calibration curve shows the response of an analytical method as a function of

the known concentration of analyte in standard solutions. Once the calibration

curve is known, then the concentration of an unknown can be deduced from a

measured response.

0-5. (a) A homogeneous material has the same composition everywhere. In a

heterogeneous material, the composition is not the same everywhere.

(b) In a segregated heterogeneous material, the composition varies on a large

scale. There could be large patches with one composition and large patches

with another composition. The differences are segregated into different

regions. In a random heterogeneous material, the differences occur on a fine

scale. If we collect a “reasonable-size” portion, we will capture each of the

different compositions that are present.

(c) To sample a segregated heterogeneous material, we take representative

amounts from each of the obviously different regions. In panel b in Box 0-1,

66% of the area has composition A, 14% is B, and 20% is C. To construct a

2 Chapter 0

representative bulk sample, we could take 66 randomly selected samples

from region A, 14 from region B, and 20 from region C. To sample a

random heterogeneous material, we divide the material into imaginary

segments and collect random segments with the help of a table of random

numbers.

0-6. We are apparently observing interference by Mn2+ in the I- analysis by method

A. The result of the I- analysis is affected by the presence of Mn2+. The greater

the concentration of Mn2+ in the mineral water, the greater is the apparent

concentration of I- found by method A. Method B is not subject to the same

interference, so the concentration of I- is low and independent of addition of

Mn2+. There must be some Mn2+ in the original mineral water, which causes

method A to give a higher result than method B even when no Mn2+ is

deliberately added.

CHAPTER 1

MEASUREMENTS

3

A note from Dan: Don’t worry if your numerical answers are slightly different

from those in the Solutions Manual. You or I may have rounded intermediate

results. In general, retain many extra digits for intermediate answers and save

your roundoff until the end. We'll study this process in Chapter 3.

1-1. (a) meter (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol)

(b) hertz (Hz), newton (N), pascal (Pa), joule (J), watt (W)

1-2. Abbreviations above kilo are capitalized: M (mega, 106), G (giga, 109), T (tera,

1012), P (peta, 1015), E (exa, 1018), Z (zetta, 1021) and Y (yotta, 1024).

1-3. (a) mW = milliwatt = 10-3 watt

(b) pm = picometer = 10-12 meter

(c) k = kiloohm = 103 ohm

(d) F = microfarad = 10-6 farad

(e) TJ = terajoule = 1012 joule

(f) ns = nanosecond = 10-9 second

(g) fg = femtogram = 10-15 gram

(h) dPa = decipascal = 10-1 pascal

1-4. (a) 100 fJ or 0.1 pJ (d) 0.1 nm or 100 pm

(b) 43.172 8 nF (e) 21 TW

(c) 299.79 THz or 0.299 79 PHz (f) 0.483 amol or 483 zmol

1-5. (a) 5.4 Pg = 5.4 × 1015 g. 5.4 × 1015 g × 1 kg

1 000 g = 5.4 × 1012 kg of C

(b) The formula mass of CO2 is 12.010 7 + 2(15.999 4) = 44.009 5

5.4 × 1012 kg C × 44.009 5 kg CO2

12.010 7 kg C = 2.0 × 1013 kg CO2

(c) 2.0 × 1013 kg CO2 × 1 ton

1 000 kg = 2.0 × 1010 tons of CO2

people 10 5

tons10 2.09

10

= 4 tons per person

4 Chapter 1

1-6. Table 1-4 tells us that 1 horsepower = 745.700 W = 745.700 J/s.

100.0 horsepower = (100.0 horsepower745.700 J / s

)horsepower

= 7.457 × 104 J/s.

4 J7.457 10

sJ

4.184cal

× 3 600 s

h = 6.416 × 107

cal

h.

1-7. (a)

6 cal2.2 10

day

J4.184

cal

1 day 24 h

1 h 3600 s

(120 pound

kg) 0.453 6

pound

= 2.0 J/(s.kg)

= 2.0 W/kg

Similarly, 3.4 × 103 kcal

day 3.0 J/(s.kg) = 3.0 W/kg.

(b) The office worker’s power output is

6 cal2.2 10

day

J4.184

cal

1 day 24 h

h 3600 s

= 1.1 × 102 J

s = 1.1 × 102 W

The person’s power output is greater than that of the 100 W light bulb.

1-8. 3 Btu5.00 10

h

J1055.06

Btu

1 h 3600 s

= 1.47 × 103 J

s= 1.47 × 103 W

1-9. (a) m 1 inch

1 000km

0.025 4 m

1 foot 12 inch

1 mile

5 280 foot

= 0.621 37 milekm

(b) 100 km 0.621 37 miles 3.785 4 L

4.6 L km gallon

= 51

milesgallon

(c) The diesel engine produces 223 g CO2/km, which we will convert into g/mile:

2g CO 1 km223

km 0.621 37 mile

= 359

g CO2

mile

In 15 000 miles, CO2 = (15 000 miles)(359 g/mile) = 5.38 106 g or 5.38

103 kg = 5.38 metric tons. The gasoline engine produces 266 g CO2/km,

Measurements 5

which we convert into 428 g/mile or 6.42 metric tons in 15 000 miles.

1-10. Newton = force = mass × acceleration = kg2

m

s

Joule = energy = force × distance = kg2

m

s

. m = kg2

2m

s

Pascal = pressure = force / area = kg 2

m

s

/m2 = 2

kg

m s

1-11. mg

0.03m

2day

m1000

km

2

(535 km

2 1 g)

1000 mg

×

1 kg

1000 g

1 ton

1000 kg

day365

year

= 6 ton

year

1-12. (a) molarity = moles of solute / liter of solution

(b) molality = moles of solute / kilogram of solvent

(c) density = grams of substance / milliliter of substance

(d) weight percent = 100 × (mass of substance/mass of solution or mixture)

(e) volume percent = 100 × (volume of substance/volume of solution or mixture)

(f) parts per million = 106 × (grams of substance/grams of sample)

(g) parts per billion = 109 × (grams of substance/grams of sample)

(h) formal concentration = moles of formula/liter of solution

1-13. Acetic acid (CH3CO2H) is a weak electrolyte that is partially dissociated. When

we dissolve 0.01 mol in a liter, the concentrations of CH3CO2H plus CH3CO2-

add to 0.01 M. The concentration of CH3CO2H alone is less than 0.01 M.

1-14. 32.0 g / [(22.990 + 35.453) g/mol] = 0.548 mol NaCl

0.548 mol / 0.500 L = 1.10 M

1-15. 3mol CH OH1.71

L solution(0.100 L solution

) = 0.171 mol CH3OH

6 Chapter 1

3(0.171 mol CH OH3

32.04 g)

mol CH OH

= 5.48 g

1-16. (a) 19 mPa = 19 × 10-3 Pa. 19 × 10-3 Pa × 5

1bar

10 Pa = 1.9 × 10-7 bar

(b) T (K) = 273.15 + ˚C = 273.15 – 70 = 203 K

nV =

PRT =

71.9 10 bar

L bar0.08314

mol K203 K

= 1.1 × 10-8 M = 11 nM

1-17. 1 ppm = 1 g solute

106 g solution. Since 1 L of dilute solution ≈ 103 g,

1 ppm = 10-3 g solute/L ( = 10-3 g solute / 103 g solution).

Since 10-3 g = 103 g, 1 ppm = 103 g/L or 1 g/mL.

Since 10-3 g = 1 mg, 1 ppm = 1 mg/L.

1-18. 0.2 ppb means 0.2 × 10-9 g of C20H42 per g of rainwater

= 0.2 × 10-6 g C20H42

1 000 g rainwater ≈ 0.2 × 10-6 g C20H42

L rainwater .

60.2 10 g / L

282.55 g / mol = 7 × 10-10 M

1-19. 4g HClO0.705

g solution(37.6 g solution

) = 26.5 g HClO4

37.6 g solution – 26.5 g HClO4 = 11.1 g H2O

1-20. (a) g solution

1.67mL

mL1000

L

= 1.67 × 103 g solution

(b) 4g HClO0.705

g solution3(1.67 10 g solution

) = 1.18 × 103 g HClO4

(c) (1.18 × 103 g ) / (100.46 g /mol) = 11.7 mol

1-21. molality = mol KI

kg solvent

20.0 wt% KI = 200 g KI

1 000 g solution = 200 g KI 800 g H2O

To find the grams of KI in 1 kg of H2O, we set up a proportion:

Measurements 7

200 g KI 800 g H2O =

x g KI1 000 g H2O x = 250 g KI

But 250 g KI = 1.51 mol KI, so the molality is 1.51 m.

1-22. (a) 150 × 10–15 mol/cell

2.5 × 104 vesicles/cell = 6.0

amolvessicle

(b) 18(6.0 10 mol 23 molecules) 6.022 10

mol

= 3.6 × 106 molecules

(c) Volume = 4

3 (200 × 10-9 m)3 = 3.35 × 10-20 m3;

203.35 10 m

3

310 m 3/ L

= 3.35 × 10-17 L

(d) 18

1710 10 mol

3.35 10 L

= 0.30 M

1-23. -380 10 g

180.2 g / mol = 4.4 × 10-4 mol;

-4

-34.4 10 mol

100 10 L

= 4.4 × 10-3 M;

Similarly, 120 mg/100 L = 6.7 × 10-3 M.

1-24. (a) Mass of 1.000 L = 1.046g

mL × 1 000

mL

L × 1.000 L = 1 046 g

Grams of C2H6O2 per liter = 6.067mol

L × 62.07

g

mol = 376.6

gL

(b) 1.000 L contains 376.6 g of C2H6O2 and 1 046 – 376.6 = 669 g of H2O

= 0.669 kg

Molality = 6.067 mol C2H6O2

0.669 kg H2O = 9.07 mol C2H6O2

kg H2O = 9.07 m

1-25. Shredded wheat: 1.000 g contains 0.099 g protein + 0.799 g carbohydrate

0.099 g × 4.0 Cal

g + 0.799 g × 4.0

Cal

g = 3.6 Cal

Doughnut: 1.000 g contains 0.046 g protein + 0.514 g carbohydrate + 0.186 g fat

0.046 g × 4.0 Cal

g + 0.514 g × 4.0

Cal

g + 0.186 g × 9.0

Cal

g = 3.9 Cal

In a similar manner, we find 2.8 Calg for hamburger and 0.48

Calg for apple.

There are 16 ounces in 1 pound, which Table 1-4 says is equal to 453.592 37 g

8 Chapter 1

28.35 g

ounce.

To convert Cal/g to Cal/ounce, multiply by 28.35:

Shredded Wheat Doughnut Hamburger Apple

Cal/g 3.6 3.9 2.8 0.48

Cal/ounce 102 111 79 14

1-26. Mass of water = (225 m)2 (10.0 m )

1 000 kg

m 3 = 1.59 × 109 kg

1.6 ppm = 1.6 × 10-3 g F–

kg H2O

Mass of F– required =

3

2

g F1.6 10

kg H O

9

2(1.59 10 kg H O

) = 2.5 × 106 g F–.

(If we retain three digits for the next calculation, this last number is 2.54 ×

106.)

The atomic mass of F is 18.998 and the formula mass of H2SiF6 is 144.09. One

mole of H2SiF6 contains 6 moles of F.

mass of F–

mass of H2SiF6 =

6 18.998144.09 =

2.54 × 106 g Fx g H2SiF6

x = 3.2 × 106 g H2SiF6

1-27. (a) PV = nRT

(1.000 bar)(5.24 × 10-6 L) = n L bar

0.08314 (298.15 )mol K

K

n = 2.11 × 10-7 mol 2.11 × 10-7 M

(b) Ar: 0.934% means 0.009 34 L of Ar per L of air

(1.000 bar)(0.009 34 L) = n

0.083 14 L . bar

mol . K (298.15 K)

n = 3.77 × 10-4 mol 3.77 × 10-4 M

Kr: 1.14 ppm 1.14 L Kr per L of air 4.60 × 10-8 M

Xe: 87 ppb 87 nL Xe per L of air 3.5 × 10-9 M

1-28. 2.00 L ×mol

0.050 0L

×g

61.83mol

= 6.18 g in a 2 L volumetric flask

1-29. Weigh out 2 × 0.050 0 mol = 0.100 mol = 6.18 g B(OH)3 and dissolve in 2.00

kg H2O.

Measurements 9

1-30. Mcon . Vcon = Mdil . Vdil

mol

0.80L

(1.00 L

) = mol

0.25L

Vdil Vdil = 3.2 L

1-31. We need 1.00 L × 0.10 mol

L = 0.10 mol NaOH = 4.0 g NaOH

4.0 g NaOH

g NaOH0.50

g solution

= 8.0 g solution

1-32. (a) Vcon = Vdil MdilMcon

= 1 000 mL

1.00 M

18.0 M = 55.6 mL

(b) One liter of 98.0% H2SO4 contains (18.0 mol )(98.079 g/ mol ) = 1.77 × 103

g of H2SO4. Since the solution contains 98.0 wt% H2SO4, and the mass of

H2SO4 per mL is 1.77 g, the mass of solution per milliliter (the density) is

2 41.77 g H SO

2 4

/ mL

0.980 g H SO / g solution = 1.80 g solution/mL

1-33. 2.00 L of 0.169 M NaOH = 0.338 mol NaOH = 13.5 g NaOH

density = g solution

mL solution

= 13.5 g NaOH

g NaOH(16.7 mL solution) 0.534

g solution

= 1.52 g

mL

1-34. FM of Ba(NO3)2 = 261.34 4.35 g of solid with 23.2 wt% Ba(NO3)2 contains

(0.232)(4.35 g) = 1.01 g Ba(NO3)2

mol Ba2+ = 3 2(1.01 g Ba(NO )

3 2

)

(261.34 g Ba(NO ) / mol) = 3.86 × 10-3 mol

mol H2SO4 = mol Ba2+ = 3.86 × 10-3 mol

volume of H2SO4 = (3.86 × 10-3 mol)

(3.00 mol/L) = 1.29 mL

1-35. 25.0 mL of 0.023 6 M Th4+ contains

(0.025 0 L)(0.023 6 M) = 5.90 × 10-4 mol Th4+

mol HF required for stoichiometric reaction = 4 × mol Th4+ = 2.36 × 10-3 mol

50% excess = 1.50(2.36 × 10-3 mol) = 3.54 × 10-3 mol HF

10 Chapter 1

Required mass of pure HF = (3.54 × 10-3 mol)(20.01 g/mol) = 0.070 8 g

Mass of 0.491 wt% HF solution = (0.070 8 g HF )

(0.004 91 g HF / g solution) = 14.4 g

1-36. Concentrations of reagents used in an analysis are determined either by weighing

out supposedly pure primary standards or by reaction with such standards. If the

standards are not pure, none of the concentrations will be correct.

1-37. The equivalence point occurs when the exact stoichiometric quantities of reagents

have been mixed. The end point, which comes near the equivalence point, is

marked by a sudden change in a physical property brought about by the

disappearance of a reactant or appearance of a product.

1-38. In a blank titration, the quantity of titrant required to reach the end point in the

absence of analyte is measured. By subtracting this quantity from the amount of

titrant needed in the presence of analyte, we reduce the systematic error.

1-39. In a direct titration, titrant reacts directly with analyte. In a back titration, a

known excess of reagent that reacts with analyte is used. The excess is then

measured with a second titrant.

1-40. Primary standards are purer than reagent-grade chemicals. The assay of a

primary standard must be very close to the nominal value (such as 99.95–

100.05%), whereas the assay on a reagent chemical might be only 99%. Primary

standards must have very long shelf lives.

1-41. Since a relatively large amount of acid might be required to dissolve a small

amount of sample, we cannot tolerate even modest amounts of impurities in the

acid for trace analysis. Otherwise, the quantity of impurity could be greater than

quantity of analyte in the sample.

1-42. 40.0 mL of 0.040 0 M Hg2(NO3)2 = 1.60 mmol of Hg 22+ , which will require 3.20

mmol of KI. This is contained in volume = 3.20 mmol

0.100 mmol/mL = 32.0 mL.

1-43. 108.0 mL of 0.165 0 M oxalic acid = 17.82 mmol, which requires

2 mol MnO4

-

5 mol H2C2O4 (17.82 mol H2C2O4) = 7.128 mmol of MnO4

- .

Measurements 11

7.128 mmol / (0.165 0 mmol/mL) = 43.20 mL of KMnO4.

Another way to see this is to note that the reagents are both 0.165 0 M. Therefore,

volume of MnO4- =

25(volume of oxalic acid) .

For second question, volume of oxalic acid = 52(volume of MnO4

-) = 270.0 mL.

1-44. 1.69 mg of NH3 = 0.099 2 mmol of NH3. This will react with 32 (0.099 2) =

0.149 mmol of OBr-. The molarity of OBr- is 0.149 mmol/1.00 mL = 0.149 M.

1-45. mol sulfamic acid = 0.333 7 g

97.094 g/mol = 3.4369 mmol

molarity of NaOH = 3.4369 mmol

34.26 mL = 0.100 3 M

1-46. HCl added to powder = (10.00 mL)(1.396 M) = 13.96 mmol

NaOH required = (39.96 mL)(0.100 4 M) = 4.012 mmol HCl consumed by carbonate = 13.96 – 4.012 = 9.948 mmol

mol CaCO3 = 12 mol HCl consumed = 4.974 mmol = 0.4978 g CaCO3

wt% CaCO3 = 0.4978 g CaCO3

0.541 3 g limestone × 100 = 92.0 wt%

CHAPTER 2

TOOLS OF THE TRADE

12

2-1. The primary rule is to familiarize yourself with the hazards of what you are about

to do and not to do something you consider to be dangerous.

2-3. Dichromate (Cr2O 72-) is soluble in water and contains carcinogenic Cr(VI).

Reducing Cr(VI) to Cr(III) decreases the toxicity of the metal. Converting

aqueous Cr(III) to solid Cr(OH)3 decreases the solubility of the metal and

therefore decreases its ability to be spread by water. Evaporation produces the

minimum volume of waste.

2-4. The upper “0” means that the reagent has no fire hazard. The right hand “0”

indicates that the reagent is stable. The “3” tells us that the reagent is corrosive

or toxic and we should avoid skin contact or inhalation.

2-5. The lab notebook must: (1) state what was done; (2) state what was observed; and

(3) be understandable to a stranger.

2-6. See Section 2.3.

2-7. The buoyancy correction is 1 when the substance being weighed has the same

density as the weights used to calibrate the balance.

2-8. m = (14.82 g)

1 –

0.001 2 g/mL8.0 g/mL

1 –

0.001 2 g/mL0.626 g/mL

= 14.85 g

2-9. The smallest correction will be for PbO2, whose density is closest to 8.0 g/mL.

The largest correction will be for the least dense substance, lithium.

2-10. m = 4.236 6 g

1 –

0.001 2 g/mL8.0 g/mL

1 –

0.001 2 g/mL1.636 g/mL

= 4.239 1 g

Without correcting for buoyancy, we would think the mass of primary standard is

less than the actual mass and we would think the molarity of base reacting with

the standard is also less than the actual molarity. The percentage error would be

true mass – measured mass

true mass × 100 = 4.239 1 – 4.236 6

4.239 1 × 100 = 0.06%.

Tools of the Trade 13

2-11. (a) One mol of He (= 4.003 g) occupies a volume of

V = nRT

P =

(1 molL bar

) 0.08314

mol K(293.15 K

)

1 bar = 24.37 L

Density = 4.003 g / 24.37 L = 0.164 g/L = 0.000 164 g/mL

(b) m =

0.000164 g/mL(0.823 g) 1

8.0 g/mL

0.000164 g/mL1

0.97 g/mL

= 0.823 g

2-12. (a) (0.42) (2 330 Pa) = 979 Pa

(b) Air density =

(0.003 485)(94 000) – (0.001 318)(979)

293.15 = 1.11 g/L = 0.001 1 g/mL

(c) mass = 1.000 0 g

0.0011 g/mL1

8.0 g/mL

0.0011 g/mL1

1.00 g/mL

= 1.001 0 g

2-13. mb = ma ra2

rb2 = (100.0000 g)

(6 370 000 m)2

(6 370 030 m)2 = 99.999 1 g

2-14. TD means “to deliver” and TC means “to contain.”

2-15. Dissolve (0.250 0 L)(0.150 0 mol/L) = 0.037 50 mol of K2SO4 (= 6.535 g, FM

174.26 g/mol) in less than 250 mL of water in a 250-mL volumetric flask. Add

more water and mix. Dilute to the 250.0 mL mark and invert the flask many

times for complete mixing.

2-16. The plastic flask is needed for trace analysis of analytes at ppb levels that might

be lost by adsorption on the glass surface.

2-17. (a) With a suction device, suck liquid up past the 5.00 mL mark. Discard one or

two pipet volumes of liquid to rinse the pipet. Take up a third volume past

the calibration mark and quickly replace the bulb with your index finger.

(Alternatively, use an automatic suction device that remains attached to the

14 Chapter 2

pipet.) Wipe excess liquid off the outside of the pipet with a clean tissue.

Touch the tip of the pipet to the side of a beaker and drain liquid until the

bottom of the meniscus reaches the center of the mark. Transfer the pipet to a

receiving vessel and drain it by gravity while holding the tip against the wall.

After draining stops, hold the pipet to the wall for a few more seconds to

complete draining. Do not blow out the last drop. The pipet should be nearly

vertical at the end of delivery.

(b) Transfer pipet.

2-18. (a) Adjust the knob for 50.0 L. Place a fresh tip tightly on the barrel. Depress

the plunger to the first stop, corresponding to 50.0 L. Hold the pipet

vertically, dip it 3–5 mm into reagent solution, and slowly release the plunger

to suck up liquid. Leave the tip in the liquid for a few more seconds.

Withdraw the pipet vertically. Take up and discard three squirts of reagent to

clean and wet the tip and fill it with vapor. To dispense liquid, touch the tip

to the wall of the receiver and gently depress the plunger to the first stop.

After a few seconds, depress the plunger further to squirt out the last liquid.

(b) The procedure in (a) is called forward mode. For a foaming liquid, use

reverse mode. Depress the plunger beyond the 50.0 L stop and take in

more than 50.0 L. To deliver 50.0 L, depress the plunger to the first stop

and not beyond.

2-19. The trap prevents liquid filtrate from being sucked into the vacuum system. The

watchglass keeps dust out of the sample.

2-20. Phosphorus pentoxide

2-21. 20.214 4 g – 10.263 4 g = 9.951 0 g. Column 3 of Table 2-7 tells us that the true

volume is (9.951 0 g)(1.002 9 mL/g) = 9.979 9 mL.

2-22. Expansion = 0.999 102 60.997 047 9 = 1.002 060 8 ≈ 0.2%. Densities were taken from Table

2-7. The 0.500 0 M solution at 25˚ would be (0.500 0 M)/(1.002) = 0.499 0 M.

2-23. Using column 2 of Table 2-7, mass in vacuum =

(50.037 mL )(0.998 207 1 g/ mL ) = 49.947 g.

Using column 3, mass in air = 50.037 mL

1.0029 mL /g = 49.892 g.

Tools of the Trade 15

2-24. When the solution is cooled to 20˚C, the concentration will be higher than the

concentration at 24˚C by a factor of density at 20°Cdensity at 24°C. Therefore, the concentration

needed at 24˚ will be lower than the concentration at 20°C.

Desired concentration at 24˚C = (1.000 M) 0.997 299 5 g/mL

0.998 207 1 g/mL

= 0.999 1 M

(using the quotient of densities from Table 2-7). The true mass of KNO3

needed is 0.5000 L mol0.9991

L

g101.103

mol

= 50.506 g.

m' = (50.506 g)

1 –

0.001 2 g/mL2.109 g/mL

1 –

0.001 2 g/mL8.0 g/mL

= 50.484 g

2-25. (a) Fraction within specifications = e-t(ln 2)/tm. If tm = 2 yr and t = 2 yr, then

fraction within specifications = e-2(ln 2)/2 = e-ln 2 = ½.

(b) Fraction within specifications = 0.95 = e-t(ln 2)/2 yr

ln(0.95) = –t(ln 2)/2 t = –2 ln(0.95)/ln 2 = 0.148 yr = 54 days 8 weeks

To solve for t, take the natural logarithm of both sides:

2-26. Al extracted from glass = (0.200 L)(5.2 10-6 M) = 1.04 10-6 mol

mass of Al = (1.04 10-6 mol)(26.98 g/mol) = 28.1 g

This much Al was extracted from 0.50 g of glass, so

wt% Al extracted = 100 28.1 10-6 g

0.50 g = 0.005 62 wt%

Fraction of Al extracted = 0.005 62 wt%

0.80 wt% = 0.007 0 (or 0.70% of the Al)

16 Chapter 2

2-27.

Graph of van Deemter Equation

Flow rate Plate heightConstants (mL/min) (mm)A = 4 8.194

1.65 6 6.092B = 8 5.064

25.8 10 4.466C = 20 3.412

0.0236 30 3.21840 3.23950 3.34660 3.49670 3.67180 3.86190 4.061

100 4.268Formula:C5 = $A$6+$A$8/B5+$A$10*B5

van Deemter Equation

0

1

2

3

4

5

6

7

8

9

0 20 40 60 80 100

Flow rate (mL/min)P

late

hei

gh

t (m

m)

CHAPTER 3 EXPERIMENTAL ERROR

17

3-1. (a) 5 (b) 4 (c) 3

3-2. (a) 1.237 (b) 1.238 (c) 0.135 (d) 2.1 (e) 2.00

3-3. (a) 0.217 (b) 0.216 (c) 0.217

3-4. (b) 1.18 (3 significant figures) (c) 0.71 (2 significant figures)

3-5. (a) 3.71 (b) 10.7 (c) 4.0 × 101 (d) 2.85 × 10-6

(e) 12.625 1 (f) 6.0 × 10-4 (g) 242

3-6. (a) BaF2 = 137.327 + 2(18.998 403 2) = 175.324 because the atomic mass of Ba

has only 3 decimal places.

(b) C6H4O4 = 6(12.010 7) + 4(1.007 94) + 4(15.999 4) = 140.093 6

(The fourth-decimal place in the atomic mass of C has an uncertainty of ± 8

and the fourth-decimal place of O has an uncertainty of ± 3. These uncer-

tainties are large enough to make the fourth-decimal place in molecular mass of C6H4O4 insignificant. Therefore, another good answer is 140.094.)

3-7. (a) 12.3 (b) 75.5 (c) 5.520 × 103 (d) 3.04

(e) 3.04 × 10-10 (f) 11.9 (g) 4.600 (h) 4.9 × 10-7

3-9. All measurements have some uncertainty, so there is no way to know the true

value.

3-10. Systematic error is always above or always below the “true value” if you make

replicate measurements. In principle, you can find the source of this error and

eliminate it in a better experiment so the measured mean equals the true mean.

Random error is equally likely to be positive or negative and cannot be

eliminated. Random error can be reduced in a better experiment.

3-11. The apparent mass of product is systematically low because the initial mass of

the (crucible plus moisture) is higher than the true mass of the crucible. The error

is systematic. There is also always some random error superimposed on the

systematic error.

3-12. (a) 25.031 mL is a systematic error. The pipet always delivers more than it is

18 Chapter 3

rated for. The number ± 0.009 is the random error in the volume delivered.

The volume fluctuates around 25.031 by ±0.009 mL.

(b) The numbers 1.98 and 2.03 mL are systematic errors. The buret delivers too

little between 0 and 2 mL and too much between 2 and 4 mL. The observed

variations ±0.01 and ±0.02 are random errors.

(c) The difference between 1.9839 and 1.9900 g is random error. The mass will

probably be different the next time I try the same procedure.

(d) Differences in peak area are random error based on inconsistent injection

volume, inconsistent detector response, and probably other small variations

in the condition of the instrument from run to run.

3-13. (a) Carmen (b) Cynthia (c) Chastity (d) Cheryl

3-14. 3.124 (±0.005), 3.124 (±0.2%). It would also be reasonable to keep an additional digit: 3.1236 (±0.0052), 3.1236 (±0.17%)

3-15. (a) 6.2 (±0.2)

– 4.1 (±0.1) 2.1 ± e e2 = 0.22 + 0.12 e = 0.224 Answer: 2.1 ± 0.2 (or 2.1 ± 11%)

(b) 9.43 (±0.05) 9.43 (±0.53%)

× 0.016 (±0.001) × 0.016 (±6.25%) %e2 = 0.532 + 6.252

0.150 88 (± %e) %e = 6.272

Relative uncertainty = 6.27%; Absolute uncertainty = 0.150 88 × 0.062 7

= 0.009 46; Answer: 0.151 ± 0.009 (or 0.151 ± 6%)

(c) The first term in brackets is the same as part (a), so we can rewrite the problem as 2.1 (±0.224) ÷ 9.43 (±0.05) = 2.1 (±10.7%) ÷ 9.43 (±0.53%)

%e = 10.72 + 0.532 = 10.7%

Absolute uncertainty = 0.107 × 0.223 = 0.023 9 Answer: 0.223 ± 0.024 (±11%)

(d) The term in brackets is

6.2 (±0.2) × 10-3 e = 0.22 + 0.12 e = 0.224

+ 4.1 (±0.1) × 10-3 10.3 (±0.224) × 10-3 = 10.3 × 10-3 (±2.17%)

9.43 (±0.53%) × 0.010 3 (±2.17%) = 0.097 13 ± 2.23% = 0.097 13 ± 0.002 17

Answer: 0.0971 ± 0.0022 (± 2.2%)

Experimental Error 19

3-16. (a) Uncertainty = 0.032 + 0.022 + 0.062 = 0.07

Answer: 10.18 (±0.07) (±0.7%)

(b) 91.3 (±1.0) × 40.3 (±0.2)/21.1 (±0.2)

= 91.3 (±1.10%) × 40.3 (±0.50%)/21.1 (±0.95%)

% uncertainty = 1.102 + 0.502 + 0.952 = 1.54%

Answer: 174 (±3) (±2%)

(c) [4.97 (±0.05) – 1.86 (±0.01)]/21.1 (±0.2)

= [3.11 (±0.0510)]/21.1 (±0.2) = [3.11 (±1.64%)]/21.1 (±0.95%)

= 0.147 (±1.90%) = 0.147 (±0.003) (±2%)

(d) 2.016 4 (±0.000 8)

1.233 (±0.002)

+ 4.61 (±0.01)

7.8594 (0.000 8)2 + (0.002)2 + (0.01)2 = 0.0102

Answer: 7.86 (±0.01)(±0.1%)

(e) 2 016.4 (±0.8)

+ 123.3 (±0.2)

+ 46.1 (±0.1)

2 185.8 (0.8)2 + (0.2)2 + (0.1)2 = 0.8

Answer: 2 185.8 (±0.8) (±0.04%)

(f) For y = xa, %ey = a%ex

x = 3.14 ± 0.05 %ex = (0.05 / 3.14) × 100 = 1.592%

%ey = 13 (1.592%) = 0.531%

Answer: 1.4643 ± 0.0078 (±0.53%)

(g) For y = log x, ey = 0.434 29 exx

x = 3.14 ± 0.05 ey = 0.434 29

0.05

3.14 = 0.006 915

Answer: 0.4969 ± 0.0069 (± 1.39%)

3-17. (a) y = x1/2 %ey = 12

100 ×

0.001 13.141 5 = 0.017 5%

(1.75 × 10-4) 3.141 5 = 3.1 × 10-4 Answer: 1.772 43 ± 0.000 31

(b) y = log x ey = 0.434 29

0.001 1

3.141 5 = 1.52 × 10-4

Answer: 0.497 14 ± 0.000 15

20 Chapter 3

(c) y = antilog x = 10x ey = y × 2.302 6 ex

= (103.141 5)(2.302 6)(0.001 1) = 3.51 Answer: 1.3852 ± 0.0035 × 103

(d) y = ln x ey = 0.001 13.141 5 = 3.5 × 10-4 Answer: 1.144 70 ± 0.000 35

(e) Numerator of log term: y = x1/2 ey = 12

0.006

0.104 × 100 = 2.88%

0.3225 2.88%

0.0511 0.0009

= 0.3225 2.88%

0.0511 1.76%

= 6.311 ± 3.375% = 6.311 ± 0.213

For y = log x, ey = 0.434 29 exx = 0.434 29

0.213

6.311 = 0.015

Answer: 0.800 ± 0.015

3-18. (a) Standard uncertainty in atomic mass is equal to the uncertainty listed in the

periodic table divided by 3 because atomic mass has a rectangular

distribution of values.

Na = 22.989 769 28 ± 0.000 000 02/ 3 g/mol

Cl = 35.453 ± 0.002/ 3 g/mol ______________________________________________________________________________________

58.442 770 [(2 × 10-8)2]/3 + [(2 × 10-3)2]/3 = 1.2 × 10-3

58.443 ± 0.0012 g/mol

(b) molarity = molL =

[2.634 (±0.002)g] / [58.443 (±0.0012)g/mol]0.100 00 (±0.000 08) L

= 2.634 (±0.076%) / [58.443 (±0.002 1%)

0.100 00 (±0.08%)

relative error = (0.076%)2 + (0.002 1%)2 + (0.08%)2 = 0.11%

molarity = 0.450 7 (±0.000 5) M

3-19. m = m'

1 – dadw

1 – dad

m = [1.034 6 (±0.000 2) g]

1 –

0.001 2(±0.000 1) g/mL8.0 (±0.5) g/mL

1 – 0.001 2(±0.000 1) g/mL

0.997 299 5 g/mL

Experimental Error 21

m = [1.034 6 (±0.019 3%)]

1 –

0.001 2 (±8.33%)8.0 (±6.25%)

1 – 0.001 2 (±8.33%)0.997 299 5 (±0%)

m = [1.034 6 (±0.019 3%)][1 – 0.000 150 (±10.4%)]

[1 – 0.001 203 (±8.33%)]

m = [1.034 6 (±0.019 3%)] [1 – 0.000 150 (±0.000 015 6)]

[1 – 0.001 203 (±0.000 100)]

m = [1.034 6 (±0.019 3%)] [0.999 850 0 (±0.000 015 6)]

[0.998 797 (±0.000 100)]

m = [1.034 6 (±0.019 3%)] [0.999 850 0 (±0.001 56%)]

[0.998 797 (±0.010 0%)]

m = 1.035 7 (±0.021 8%) = 1.035 7 (±0.000 2) g

3-20. mol Fe2O3 = 0.2774 ± 0.0018 g

159.688 g/mol = 0.2774

159.688 ± 0.0018

159.688

= 1.7371 ± 0.0113 mmol Fe2O3;

mass of Fe = 2[1.7371 (±0.0113) × 10-3 mol][55.845 g/mol] = 0.19402 ± 0.00126 g

mass of Fe per tablet = (0.19402 ± 0.00126 g)/12 = 16.168 ± 0.105 mg

= 16.2 ± 0.1 mg

3-21. mol H+ = 2 × mol Na2CO3

mol Na2CO3 = 0.967 4 (±0.000 9) g

105.988 4 (±0.000 7) g/mol = 0.967 4 (±0.093%) g

105.988 (±0.000 66%) g/mol

= 0.009 127 4 (±0.093%) mol

mol H+ = 2(0.009 127 4 (±0.093%)) = 0.018 255 (±0.093%) mol

(Relative error is not affected by the multiplication by 2 because mol H+

and uncertainty in mol H+ are both multiplied by 2.)

molarity of HCl = 0.018 255 (±0.093%) mol

0.027 35 (±0.000 04) L = 0.018 255 (±0.093%) mol

0.027 35 (±0.146%) L

= 0.66746 (±0.173%) = 0.667 46 (±0.001 155)

= 0.667 ± 0.001 M

3-22. To find the uncertainty in co3, we use the function y = xa in Table 3-1,

where x = co and a = 3. The uncertainty in co3 is

%ey = a %ex = 3 × 0.000 000 335.431 020 36 × 100 = 1.823 × 10-5%

So co3 = (5.431 020 36 × 10-8 cm)3 = 1.601 932 796 0 × 10-22 cm3 with a

22 Chapter 3

relative uncertainty of 1.823 × 10-5%. We retain extra digits for now and round

off at the end of the calculations. (If your calculator cannot hold as many digits

as we need for this arithmetic, you can do the math with a spreadsheet set to

display 10 decimal places.)

The value of Avogadro's number is computed as follows:

NA = mSi

(co3)/8 =

28.085 384 2 g/mol(2.329 031 9 g/cm3 × 1.601 932 7960 × 10-22 cm3)/8

= 6.022 136 936 1 × 1023 mol-1

The relative uncertainty in Avogadro's number is found from the relative uncertainties in mSi, , and co3. (There is no uncertainty in the number 8

atoms/unit cell.)

percent uncertainty in mSi = 100 (0.000 003 5/28.085 384 2) = 1.246 × 10-5%

percent uncertainty in = 100 (0.000 001 8/2.329 031 9) = 7.729 × 10-5% percent uncertainty in co3 = 1.823 × 10-5% (calculated before)

percent uncertainty in NA = %emSi2 + %er2 + (%eco3)2 =

= (1.246 × 10-5)2 + (7.729 × 10-5)2 + (1.823 × 10-5)2 = 8.038 × 10-5%

The absolute uncertainty in NA is (8.038 × 10-5%)(6.022 136 936 1 × 1023)/100

= 0.000 004 841 × 1023. Now we will round off NA to the second digit of its

uncertainty to express it in a manner consistent with the other data in this

problem:

NA = 6.022 136 9 (±0.000 004 8) × 1023 or 6.022 136 9 (48) × 1023

3-23. C: 12.010 7 ± 0.000 8/ 3; H: 1.007 94 ± 0.000 07/ 3

O: 15.999 4 ± 0.000 3/ 3; N: 14.006 7 ± 0.000 2/ 3

+9C: 9(12.010 7 ± 0.000 46) = 108.096 3 ± 0.004 2

+9H: 9(1.007 94 ± 0.000 040) = 9.071 46 ± 0.000 36

+6O: 6(15.999 4 ± 0.000 17) = 95.996 4 ± 0.001 0

+3N: 3(14.006 7 ± 0.000 12) = 42.020 1 ± 0.000 35 C9H9O6N3: 255.184 26 ± ?

Uncertainty = 0.004 22 + 0.000 362 + 0.001 02 + 0.000 352 = 0.004

Answer: 255.184 ± 0.004

Experimental Error 23

3-24. Relative uncertainties:

Large volume: 0.000 5 L/5.013 82 L = 0.010%

Small volume: 0.000 9 mL/3.793 0 mL = 0.024%

Pressure: 0.03 mm/400 mm = 0.008%

Temperature: 0.03 K/300 K = 0.01%

The largest uncertainty is in the volume of the small vessel = 0.024%.

Uncertainty in CO2 0.024% of 400 ppm = 0.000 24 400 ppm = 0.1 ppm.