A mole, which we symbolize as “mol,” is the amount of a substance that consists of as many entities (molecules, ions, particles, etc.) as there are atoms in exactly 12 grams of the element 12C (carbon-12).
Although that definition is just a bit convoluted, the number in question is 6.022137 * 1023 or 602,213,700,000,000,000,000,000
We call this Avogadro’s number.1 In other words, 12 grams of 12C consists of
602,213,700,000,000,000,000,000 atoms. The classic illustration of this number’s unimaginable enormity is to try to appreciate an Avogadro’s number of dollars. If a person started with 6.022137 * 1023 dollars,
and spent a billion dollars every second for 75 years, more than 99.99% of the starting balance would still be available.
Like the prefixes on basic units of measurement, the term “mole” simplifies quantities by deleting zeros. After all, it is much easier to talk about frequent-flyer “miles” than it is to talk about frequent-flyer “inches.” To erase even more zeros, we can affix those same metric prefixes to “mole” or “mol.” For example,
0.001 mol = 1 millimole = 1 mmol 0.000001 mol = 1 micromole = 1 μmol 0.000000001 mol = 1 nanomole = 1 nmol 0.000000000001 mol = 1 picomole = 1 pmol
If the masses of two atoms differ, then a mole of one substance weighs more than a mole of the other. Consequently, a mole of 12C atoms weighs 12.0 grams, and a mole of Fe (iron) atoms weighs
55.8 grams.
The formula weight of a substance is the sum of all the atomic weights in the formula. For example, the formula weight of water (H2O) is 18.0 g:
2 * 1.0 g + 1 * 16.0 g H atoms O atom
Therefore, one mole of water molecules weighs 18.0 grams. Likewise, a mole of glucose (C6H12O6) weighs
180 grams:
6* 12.0 g + 12 * 1.0 g + 6 * 16.0 g C atoms H atoms O atoms
The molar mass of a substance is numerically equal to the formula weight and is defined as the mass of one mole of the substance. Table 4-2 H lists several selected elements and their average atomic weights.
1The value of 6.023 is also used widely as the significand. Because the difference between 6.023 and 6.022 is only 0.02%, this book uses both values.
element average atomic Weight (g/mol)
Hydrogen (H) 1.01 Calcium (Ca) 40.08 Carbon (C) 12.01 Chlorine (Cl) 35.45 Cobalt (Co) 58.93 Magnesium (Mg) 24.31 nitrogen (n) 14.01 oxygen (o) 16.00 Phosphorus (P) 30.97 Potassium (K) 39.10 Sodium (na) 22.99 Sulfur (S) 32.07
H TabLe 4-2 Selected Elements and Their Atomic Weights
f
f
CheCkpoint 4-2
Compute the formula weight of each of the following chemical compounds. (a) Co2 (b) H2S (c) MgSo4 (d) na2o
(a) (1 * 12.01 g/mol) + (2 * 16.00 g/mol) = 44.01 g/mol (b) 34.09 g/mol (c) 120.38 g/mol (d) 61.98 g/mol
exaMpLe 1
To convert “3.0 in” to “cm,” we employ the equivalence of 1 in = 2.54 cm From this equivalence, we can create two unit factors:
1 in
2.54 cm and 2.54 cm
1 in
next, we multiply our original value, “3.0 in,” by the unit factor that gives us our target unit, “cm”:
3.0 in * 2.54 cm1 in = 7.6 cm
The units of “in” cancel, leaving the units of “cm”, and we conclude that 3.0 inches is the same as 7.6 centimeters.
To convert in the reverse direction, say, from “9.1 cm” to “in,” we follow the same procedure by multiplying our original value by the unit factor that gives us our target unit:
9.1 cm * 2.54 cm1 in = 3.6 in The units of “cm” cancel, leaving the units of “in.”
diMenSionaL anaLySiS
Mentioned earlier, dimensional analysis is a unit-conversion technique based on the fact that any quan- tity can be multiplied by “1” without its value being changed. We use this technique to convert one unit into another by multiplying by what we call “unit factors.” Let us consider four examples.
exaMpLe 2
It is sometimes necessary to string several unit factors together. For example, let us calculate the number of ounces in one ton. We start with “1 ton,” identify our target as “ounces”, and then note any unit factors that link “tons” to “ounces.”
conversion: 1 tonS ? oz
relevant Unit factors:
2000 lb ton ton 2000 lb 16 oz lb lb 16 oz
Solution: 1 ton * 2000 lbton * 16 ozlb = 32,000 oz
We solve the problem by multiplying the original value (“1 ton” in this case) by unit fac- tors in such a way that only the target units emerge at the end of the calculation. The units of “ton” cancel, as do the units of “lb.” We conclude that there are 32,000 ounces in one ton.
exaMpLe 3
Let us now try a slightly longer conversion, one that is more germane to laboratory work. How many micromoles are there in 6.0 milligrams of glucose (molar mass = 180 g/mol)?
conversion: 6.0 mgS ? 𝛍mol
relevant Unit factors:
1000 mg g g 1000 mg 180 g mol mol 180 g 1 * 106μmol mol mol 1 * 106μmol Solution: 6.0 mg* 1000 mg *g 180 g *mol 1* 10 6 μmol mol = 33 μmol
As in the earlier example, we solve this problem by multiplying the original value (6.0 mg) by unit factors in such a way that only the target units emerge at the end of the calcu- lation. The units of “mg,” “g,” and “mol” cancel. For glucose, then, we conclude that 6.0 mg is the equivalent of 33 μmol.
exaMpLe 4
This example of dimensional analysis involves a concentration. Suppose we must convert a test result of 0.739 mmol/L to “nmol/mL.” Although the strategy is the same as in the preceding two examples, we now have two units to convert.
conversion: 0.739 mmol/LS ? nmol/mL
relevant Unit factors:
1000 mmolmol 1000 mmolmol 1* 10mol9 nmol mol
1 * 109 nmol 1000 mL L L 1000 mL Solution: 0.739 mmol L * mol 1000 mmol * 1 * 109 nmol mol * L 1000 mL = 739 nmol/mL
CheCkpoint 4-3
1. Using dimensional analysis, convert “890 nmol/L” to “μmol/dL.” 890 nmol L * mol 1* 109 nmol * 1 * 106 μmol mol * L 10 dL = 0.089 μmol/dL 2. Using dimensional analysis, convert “0.864 g of Co2” to “mmol of Co2.”
0.864 g CO2 * 44.01 g COmol CO2
2 *
1000 mmol CO2
TeMperaTUre ScaLeS
There are three temperature scales, each with its own history and utility. On the Fahrenheit scale, the freezing point of water (at sea level) is 32 degrees (°F) and the boiling point is 212°F, setting these two temperatures 180F° apart. On the Celsius scale, however, water freezes at 0°C and boils at 100°C, the interval being 100C°.
Thus, 180 Fahrenheit degrees covers the same range as 100 Celsius degrees, making each Celsius degree 1.8 times a Fahrenheit degree:
180 Fo
100 Co = 1.8 Fo/Co
The raTio MeThod
The ratio method is another way to convert between units. For example, if we want to convert “13 yards ” to “feet,” we can set up an equation of ratios and then cross-multiply:
3 ft yd = x 13 yd (3 ft)(13 yd) = (1 yd)x (3 ft)(13 yd) yd = x 39 ft = x
For longer conversions, a sequence of ratio calculations can substitute for dimensional analysis. Consider example 3 above, the conversion of “6.0 mg” of glucose to “micromoles.” The order in which we convert the units is
mgS g S mol S μmol
This is the same order in which we converted them using dimensional analysis above. Conversion 1 (mg S g) 1000 mg
g =
6.0 mg x x = 0.0060 g Conversion 2 (g S mol) 180 gmol = 0.0060 gx
x = 3.3 * 10-5 mol Conversion 3 (mol S mmol ) 1 * 106 μmol
mol =
x 3.3 * 10-5 mol x = 33 μmol
CheCkpoint 4-4
Using the ratio method, carry out the following conversions.
(a) 6.2 mS yd (b) 535 μLS mL (c) 0.5844 g naClS μmol naCl (a) 1.094 yd
m =
x
6.2 m x = 6.8 yd
There are two formulas for converting between Fahrenheit and Celsius temperatures: oF = ¢oC * 9 5≤ + 32 o oC = 5 9 (oF - 32o)
The third temperature scale bears the name of its developer, Lord Kelvin. On the Kelvin scale, the zero-degree point corresponds to the theoretical absence of all thermal energy, “absolute zero.” A tem- perature on the Kelvin scale is represented by “K,” with no degree symbol.
Each degree, or increment, on the Kelvin scale is a “kelvin” (all lowercase). Each increment on the Kelvin scale is the same as an increment on the Celsius scale. Therefore, the two scales have a simple relationship, and the formula for interconverting them is
K = °C + 273.15
Practice Problems
1. (LO 1, 2, 4, 5) Carry out each of the following unit conversions.
(a) 3.1 mLS μL (b) 420 ngS mg (c) 0.002 mLS μL (d) 0.78 μgS ng (e) 8.5 dLS L (f) 1445 mgS g (g) 0.0364 LS mL (h) 13 pgS ng
(i) 620 nmolS μmol (j) 9.7 * 10-5 molS mmol (k) 4.0 dLS mL (l) 73 μgS mg
(m) 400 μLS mL (n) 2.5 * 10-4 μgS pg (o) 6.09 * 104 μmolS mol (p) 0.62 LS dL (q) 705 μmolS pmol (r) 2 ngS pg
2. (LO 6) Calculate the molar mass of each of the following substances.
(a) KCl (b) nH3 (c) K3Po4 (d) na2S (e) MgCl2
Summary
without its value being changed. We use this technique to convert one unit into another by multiplying by “unit factors.”
8. The ratio method is useful in the conversion of units. An
equation of ratios is established, and then cross-multipli- cation gives the value of the unknown quantity. For longer conversions, a sequence of ratio calculations can substitute for dimensional analysis.
9. There are three temperature scales: Fahrenheit, Celsius,
and Kelvin. Their values are interconvertible by these equations: oF = ¢oC* 9 5≤ + 32 o oC = 5 9 (°F - 32o) K = °C + 273.15
1. The metric system has only one basic unit for length, one
for volume, and one for weight. The United States Custom- ary System of Units has several units for each quantity. 2. The metric system is built on the number “10.” Prefixes
create decimal multiples and submultiples of basic units.
3. The International System of Units (SI) has seven basic units
but permits the use of some non-SI units.
4. U.S. customary units and metric units are interconvertible. 5. The mole (“mol”) is the amount of a substance that
consists of as many entities (molecules, ions, particles, etc.) as there are atoms in exactly 12 grams of the ele- ment 12C (carbon-12). The number is 6.022137 * 1023 or 602,213,700,000,000,000,000,000. It is called Avogadro’s number. We can affix metric prefixes to “mole” or “mol.” 6. The formula weight of a substance is the sum of all the
atomic weights in the formula. The molar mass of a sub- stance is numerically equal to the formula weight and is defined as the mass of one mole of the substance.
7. Dimensional analysis is a problem-solving technique based
3. (LO 4, 5, 7) For each of the following substances, suppose you have either the actual mass or the actual number of moles specified. Complete the table by supplying the missing information.
5. (LO 8) Convert each of the following Fahrenheit tempera- tures to its Celsius equivalent.
(a) 72°F (b) 230°F (c) -7°F (d) 45°F (e) 10°F 6. (LO 8) Convert each of the following Celsius temperatures
to its Kelvin equivalent.
(a) 0°C (b) 100°C (c) -273.15°C 7. (LO 3, 4, 5) Carry out the following conversions. (a) 53 kmS mi (b) 5.6 in S cm (c) 128 lbS kg (d) 2 kgS oz (e) 31 ft S m (f) 3500 ydS km (g) 8.9 LS gal (h) 0.20 cups S mL (i) 6.2 qt S L
Contextual Problems
1. (LO 1, 2, 4) If the typical red blood cell has a volume of 90 fL, then how many such cells could theoretically occupy a volume of 1 mL?
2. (LO 1, 2, 4, 5) One of the automated instruments in your laboratory requires the preparation of a special clean- ing solution, made by mixing 30 mL of concentrate with enough water to bring the final volume to 250 mL. How much concentrate do you use to make 1 L of the cleaning solution?
3. (LO 2, 4, 5) On a balance that reads in “grams,” you must weight out 450 mg of a substance. How many grams does this represent?
4. (LO 2, 4, 5) Using a pipet that reads in “microliters,” you must deliver 0.080 mL of a solution. How many microliters does this represent?
5. (LO 2, 4, 5) Some of your laboratory’s instruments return test results in units that must be converted into other units before being released into the hospital information sys- tem. Convert each of the following results into the units specified. (a) 628 pg/mLS ng/L (b) 0.0198 μmol/LS pmol/mL (c) 1.74 μg/dLS mg/mL (d) 0.49 mmol/LS μmol/dL PEARSON
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Substance Molar Mass (g/mol) Mass (g)actual actual Moles sodium chloride (naCl) 58.44 3.50 (a) sucrose (C12H22o11) 342.3 (b) 0.0004 urea (CH4n2o) 60.06 0.70 (c) potassium hydrogen phosphate (KHPo4) 174.18 (d) 0.062 glucose (C6H12o6) 180.2 5.66 (e)
4. (LO 8) Convert each of the following Celsius temperatures to its Fahrenheit equivalent.
aqueous concentration dissolve equivalent equivalent weight immiscible insoluble miscible molality molarity normality pH
5
Solutions and
Concentrations
Chapter Outline
Key Terms 80 Expressing Concentration 81 Specific Gravity 84 The pH Scale 85Converting Between Units 87
Learning Objectives
At the end of this chapter, the student should be able to do the following: 1. Explain the nature of solutions and distinguish between the solute and
solvent
2. Explain the various expressions of concentration
3. Calculate and interpret concentrations expressed in any of these four sys- tems: percentage, molarity, molality, and normality
4. Use specific gravity in diluting concentrated solutions 5. Describe the relative convenience of the pH scale 6. Calculate pH values and interconvert them with molarity 7. Interconvert expressions of concentration