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CHEMICAL
MEASUREMENTS
Section | 2.2
| 2.3 | 2.4 | Summary
| Review Exercise | Bottom|
2.1 INTERNATIONAL SYSTEM OF UNITS, SI
Measurements
A measurement is a procedure in which an unknown quantity is compared to a known quantity. Usually, the known quantity is obtained from a device calibrated to display the correct value. Each measurement consists of two parts, a number and a unit. Additionally, a scientific measurement expresses its reliability because all measurements are to some degree uncertain. To show the reliability of a measurement, scientists express the measurement to the correct number of significant figures (Sec. 2.4).
Modern Metric System--International System of Units
Scientists throughout the world use the same standardized units of measurements. The standardized system of units is called the International System of Units, SI for short.
Base SI Units
The SI system has seven base units from which all the others are derived. The seven base units of the International System are:
1. Meter (m) - unit of length
2. Kilogram (kg) - unit of mass
3. Second (s) - unit of time
4. Kelvin (K) - unit of temperature
5. Mole (mol) - unit of amount of substance
6. Ampere (A) - unit of electric current
7. Candela (cd) - unit of luminous intensity
All other units in the SI system are termed derived units.
SI Prefixes
Prefixes are placed in front of the base SI unit to change the size of the unit. For example, the meter is a unit of length approximately 39.4 inches (in). While it is convenient to measure the dimensions of a room in meters, it is unwieldy to express distances between cities in meters. To increase the meter 1000 times, the prefix kilo, meaning 1000 times, is added to the meter, producing the unit kilometer, km.
The meter is an awkward unit for measuring small distances. Thus, a prefix is added to specify a unit smaller than a meter. The two most commonly used prefixes are centi, 1/100, and milli, 1/1000. Added to the unit meter, two small units of length are obtained, the centimeter, cm, and millimeter, mm. One meter is equivalent to 100 cm and 1000 mm.
Common SI Prefixes
The most common prefixes encountered in chemistry are:
1. mega (M) = 1,000,000 x = 106 x
2. kilo (k) = 1000 x = 103 x
3. deci (d) = 0.1 x = 10-1 x
4. centi (c) = 0.01 x = 10-2 x
5. milli (m) = 0.001 x = 10-3 x
6. micro () = 0.000001 x = 10-6 x
7. nano (n) = 0.000000001 x = 10-9 x
2.2 LENGTH, MASS, VOLUME, DENSITY, AND SPECIFIC GRAVITY
Length
The base SI unit of length is the meter. A meter, m, equals 39.3701 in, which makes it slightly longer (by 9%) than a yard (36 in). While the meter is a convenient day-to-day measurement, it is generally too large to use for length measurements in chemistry because chemists deal with minute objects like atoms and molecules. Therefore, various prefixes are added to the meter to give more useful units of length. A centimeter, cm, is 0.01 meter, a millimeter, mm, is 0.001 meter, and a micrometer, m, is 10-6 m.
A commonly used prefix for the meter when considering atomic dimensions is nano or 1 x 10-9 . What fraction of an inch is 1.000 nm?
Solution 2.1
A nanometer is 1 x 10-9 m and a meter is 39.37 in. Hence, convert nm to in by using the conversion factor 39.37 in/m. A conversion factor is an equality expressed as a fraction.
1.000 nm x 10-9 m/nm x 39.37 in/m = 3.937 x 10-8
in
Convert 5.30 miles to micrometers, m. One mile is 5280 ft.
Solution 2.2
To perform this conversion, first convert miles to ft, using 5280 ft/mile. Then, convert ft to in, knowing the conversion factor 12 in/ft. Convert to m using the conversion factor 39.37 in/m, and finally obtain micrometers by using 1 x 106 m/m.
5.30 miles x 5280 ft/mile x 12 in/ft 1 m/39.37 in x 1 x 106 m/m = 8.53 x 109 m
A human red blood cell, an erythrocyte, has an average diameter of 8 m. What is the average diameter of an erythrocyte in inches?
Solution 2.3
First, convert m to m, knowing the conversion 1 106 m/m. Then, convert m to in, using 39.37 in/m.
8 m x 1 m/1 x 106 m x 39.37 in/m = 3 x 10-4 in = 0.0003 in
Mass
Mass Versus Weight
The kilogram, kg, is the SI unit of mass. Mass is the quantity of matter contained in an object. The terms "mass" and "weight" are often confused. Weight is the gravitational force of attraction that acts on a mass. The mass of a body is constant no matter where it is located. However, weight is a variable quantity that depends on where the object is located. Mass and weight measurements require different instruments. Mass is measured with a balance; weight is measured with a scale.
SI Units of Mass
The standard kilogram, a block of platinum-iridium alloy from which all mass measurements are compared, is located in France. A kilogram is equivalent to 2.2046 pounds, lb. The kilogram is too massive for most routine chemical laboratory measurements. Therefore, smaller multiples of the kilogram are used. Grams (g), milligrams (mg), and micrograms (g) are the most commonly used units of mass in chemistry labs. One pound, 1 lb, is equivalent to 454 g.
The weight of the average human brain is about 3.0 lbs. What is the average mass of a human brain in kilograms, kg?
Solution 2.4
To convert lbs to kg use the conversion factor 2.2 lb/kg.
3.0 lb x 1 kg/2.2 lb = 1.4 kg
The mass of an empty beaker is 60.809 g. The combined mass of a sample placed into the beaker and the beaker is 61.101 g. What is the mass of the sample in mg?
Solution 2.5
To find the mass of the sample in g, the mass of the beaker is subtracted from the mass of the beaker and sample.
mass of sample = mass of beaker + sample - mass of beaker
= 61.101 g - 60.809 g = 0.292 g sample
Then, use the conversion factor 1000 mg/g to convert to mg.
0.292 g sample x 1000 mg sample/g sample = 292 mg sample
Volume
Volume is the amount of space occupied by matter. All matter takes up some space and therefore has volume. The amount of space is measured in cubic units or a length unit to the third power. For example, the volume, V, of a rectangular object is calculated by multiplying its length, l, times its width, w, times its height, h.
V = lwh (2.1)
The base unit of length is the meter; thus, one SI unit of volume is m x m x m or m3 (cubic meter). One cubic meter is equivalent to 264 gallons (gal), and for most chemical measurements the cubic meter is too large.
Cubic Decimeters and Liters
The cubic decimeter, dm3, is often used in chemistry. Deci is the prefix that means 1/10, thus a decimeter is 0.1 m (3.937 in). The dm3 is exactly the same volume as the non-SI unit, the liter, L. Both the cubic decimeter and the liter are one-thousandth of a cubic meter (0.001 m3). Cubic decimeters are much less frequently encountered due to the popularity of the liter. Just remember that the liter and dm3 are different units for the same volume 0.001 m3.
A liter is approximately the same volume as a quart. One liter is 1.057 quarts (qt), which means it is slightly larger than a quart. One liter also contains 1000 mL. One mL of volume is exactly equivalent to one cubic centimeter, (cm3 or cc).
How many L, mL, and cm3 are contained in one quart? One L contains 1.057 qts.
Solution 2.6
To obtain the number of L/qt take the reciprocal of the relationship 1.057 qt/L.
(1 L/1.057 qt) = 0.9461 L/qt
To calculate the number of mL/qt use the conversion factor, 1000 mL/L.
0.9461 L/qt x 1000 mL/L = 946.1 mL/qt
Because 1 mL is equivalent to 1 cm3, the relationship between
cm3 and qt is the same as that for mL and qt, 946.1 cm3/qt.
How many liters of gasoline must be purchased to fill a 15.0-gal gasoline tank?
Solution 2.7
First, convert gal to qt, knowing that there are 4 qt per gal. Then, convert gal to L, using the conversion factor 1.057 qt/L.
15.0 gal x 4 qt/l gal x 1 L/1.057 qt = 56.8 L
How many microliters, L, are in 1.00 m3?
Solution 2.8
First, convert m3 to dm3 using the conversion factor 0.001 m3/dm3.
One dm3 is the same as 1 L, 1 L/1 dm3. Complete the problem by using the conversion factor 1 x 10-6 L/L.
A person is found to have 5.3 qt of blood. (a) What is the volume of blood in L? (b) If the plasma constitutes 55% of the volume of blood and 92% of the volume of plasma is water, what volume of water (in L) is in this person?
Solution 2.9
(a) Use the conversion factor, 1 L/1.057 qt, to convert to L.
5.3 qt x 1 L/1.057 qt = 5.0 L
(b) Fifty-five percent plasma means that for each 100 L blood, 55 L are plasma, 55 L plasma/100 L blood. Similarly, 92% water means 92 L water/100 L plasma. Thus, calculate the volume of water as follows.
5.0 L blood x 55 L plasma/100 L blood x 92 L water/100 L plasma = 2.5
L water
An automobile engine has a displacement of 3.80 L. What is its displacement in cubic inches?
Solution 2.10
To convert L to in3, the following conversion factors must be used: 1 L/1 dm3, 10 dm/m, and 39.37 in/m. To find the relationships between dm3 and m3, and in3 and m3, cube the conversion factors 10 dm/m and 39.37 in/m as follows.
(10 dm)3/(1 m)3 = 103 dm3/1 m3
(39.37 in)3/(1 m)3 = 6.102 x 104 in3/1 m3
Then, calculate the volume in in3 as follows.
Density
The density, d, of an object describes how much mass is contained in a unit volume.
d = mass/volume (2.2)
Substances with higher densities have more mass packed into equivalent volumes than those with lower densities. For example, 1.0 cm3 of gold has a mass of 19 g; in contrast, 1.0 cm3 of water only has a mass of 1.0 g. If equal masses of gold and water are compared, the gold occupies 1/19th the volume of water.
Densities of substances are measured by finding the mass of known volume
of the substance. Mass is most commonly measured in g, and volume is measured
in mL, cm3 (cc), dm3, and L. For more compact forms
of matter, liquids and solids, densities are expressed as g/cm3
or g/mL. For less dense forms of matter, mainly gases, densities are expressed
in g/L and g/dm3.
Explain why it is incorrect to say that, in general, gold is "heavier" than iron. Explain.
Solution 2.11
Stated properly, one says "gold is more dense than iron." The density of gold is 19.3 g/cm3, which is higher than the density of iron, 7.86 g/cm3. This means that the mass of an equal volume of gold is greater than that of iron. Gold has more matter packed into the same volume than iron.
An empty 10-mL graduated cylinder has a mass of 53.24 g. When 10.0 mL of an unknown liquid is added to the cylinder, the total mass is 63.12 g. Calculate the density of the unknown liquid.
Solution 2.12
The mass of the unknown liquid is found by subtracting the mass of the graduated cylinder from the mass of the graduated cylinder and liquid.
Mass of liquid = 63.12 g - 53.24 g = 9.88 g
Then, divide the mass of the liquid by its volume, 10.0 mL.
d = mass/volume = 9.88 g/10.0 mL = 0.988 g/mL
If 25.0 cm3 of decane, a colorless organic liquid, is poured into a graduated cylinder, what is the mass of decane in the cylinder? The density of decane is 0.730 g/cm3.
Solution 2.13
Density is a conversion factor for mass and volume of a substance. Therefore, use the density, 0.730 g decane/mL, to convert mL to g.
25.0 cm3 decane x 0.730 g/cm3 = 18.3 g decane
What mass of water in kg fills an aquarium that has the dimensions 1.2 m x 0.74 m x 0.50 m. The density of water, H2O, is 1.0 g/cm3.
Solution 2.14
First, calculate the volume of H2O in m3.
V = lwh = 1.2 m x 0.74 m x 0.50 m = 0.44 m3 H2O
Then, convert m3 to cm3 as follows.
0.44 m3 x 1 x 106 cm3/m3 = 4.4 x 105 cm3 H2O
The conversion factor 1 x 106 cm3/m3 is obtained by cubing the conversion factor for cm to m, 100 cm/m. Finally, use the density, 1.0 g/cm3, and relationship between g and kg, 1000 g/kg, to calculate the mass of water.
4.4 x 105 cm3 H2O x 1.0 g H2O/cm3
x 1 kg/1000 g = 4.4 x 102 kg H2O
Specific Gravity
Specific gravity is the ratio of the density of a substance to the density of water. Stated another way, specific gravity is the ratio of the mass of a substance to the mass of an equal volume of water.
Specific gravity = dsubstance/dwater = (gsubstance/cm3)/(gwater/cm3) = gsubstance/gwater
Because the density of water is 1.0 g/cm3 at 25oC, the numerical values for the specific gravity and density of a substance are the same. However, specific gravity is a unitless quantity because the units cancel when the ratio is calculated. For example, the density of mercury is 13.6 g/cm3 and the specific gravity of mercury is 13.6. If equal volumes of mercury and water are compared, the sample of mercury has 13.6 times the mass of water.
Measuring Specific Gravity
Specific gravity determinations, instead of density determinations, are made in medical and technical labs because they are measured easily with a hydrometer. A hydrometer is usually a glass tube designed to measure specific gravity as it floats in a liquid. The side of the hydrometer is calibrated so the specific gravity of the liquid can be read at the surface of the liquid. In medical laboratories, a special hydrometer, called a urinometer, is used to measure the specific gravity of urine. The specific gravity of urine helps physicians to diagnose abnormal conditions such as kidney disease or diabetes.
The specific gravity of whole blood ranges from 1.052 to 1.064. Explain what this means.
Solution 2.15
Specific gravity is the ratio of the density of a fluid to the density
of water, which is 1.0 g/cm3. Therefore, if the density of blood
plasma ranges from 1.052 to 1.064. The density of blood plasma ranges from
1.052 g/cm3 to 1.064 g/cm3. This shows that blood
plasma is more dense than pure water. Blood plasma has a higher density
than water because it contains formed elements (erythrocyte, leucocytes,
and thrombocytes) and many dissolved substances in the blood plasma.
2.3 TEMPERATURE, ENERGY, HEAT, AND SPECIFIC HEAT
Temperature Versus Heat
Temperature is a measure of the degree of hotness of matter. If an object at a higher temperature contacts one at a lower temperature, the temperature of the hotter object decreases and the temperature of the colder object increases until the temperatures of both become equal. It is a common practice to say that heat flows from the hotter object to the colder object. Heat is a form of energy that is usually detected only when objects of different temperatures are in contact with each other.
Temperature Units--Kelvin, Celsius, and Fahrenheit
The SI unit of temperature was named after Lord Kelvin, the title given to William Thomson (1824-1907), a brilliant British physicist. The Kelvin temperature scale has units called kelvins, K, equal in magnitude to Celsius degrees, oC, but are displaced by 273 degrees (actually 273.15 degrees). Therefore, the zero point on the Kelvin scale, 0 K, called absolute zero, is equivalent to -273.15oC. Because absolute zero is the lowest possible temperature, the Kelvin scale does not have negative values. To change Celsius degrees to kelvins it is only necessary to add 273.15 to the Celsius temperature.
K = oC + 273.15 (2.3)
The normal freezing and boiling points of water are 0.00oC and 100.00oC, respectively. Calculate the freezing and boiling points of water in K.
Solution 2.16
Freezing PointH2O = 0oC + 273.15 = 273.15 K
Boiling PointH2O = 100oC + 273.15 = 373.15 K
On the Kelvin scale the freezing and boiling points of water are 273.15 K and 373.15 K, respectively.
Temperature Conversion Formulas
While oC and K are universally used in science, in the United States the Fahrenheit scale predominates among nonscientists. To convert temperatures from Fahrenheit to Celsius, use the following formula.
oC = 5/9(oF - 32) (2.4)
To convert a Celsius temperature to Fahrenheit, use the rearranged form of this equation.
oF = 9/5oC + 32 (2.5)
Normal human body temperature is 98.6oF. What is this temperature in oC and K?
Solution 2.17
Use Eq. 2.4 to convert oF to oC.
oC = 5/9(oF - 32) = 5/9(98.6oF - 32) = 37.0oC
Use Eq. 2.3 to convert oC to K.
K = oC + 273.15 = 37.0oC + 273.15 = 310.2 K
Convert 300.0 K to oF.
Solution 2.18
First, convert kelvins to oC, and then convert to oF. To find the temperature in oC, rearrange the equation and substitute the Kelvin temperature in the equation.
K = oC + 273.15
oC = K - 273.15
oC = 300.0 - 273.15 = 26.8oC
The answer to this equation, 26.8oC, must be rounded to one decimal place because the temperature was measured to one decimal place. Then, substitute this value into Eq. 2.5 and solve for oF,
oF = 9/5oC + 32
oF = 9/5(26.8oC) + 32 = 80.2oF
One temperature is the same on both the Celsius and Fahrenheit scales. Calculate this temperature.
Solution 2.19
For oF, substitute oC in the following temperature conversion equation because they are equal. Then solve for oC.
oC = 5/9 (oF - 32) = 5/9 (oC - 32)
= -40oC = -40oF
Measurement of Energy and Heat
Recall from Chap. 1 that energy is the capacity to do work and heat is a form of energy that always flows spontaneously from warmer to colder objects. The two units used to measure energy and heat are joules and calories.
The Joule
The SI unit for energy is the joule, J. One joule is 1 kg m2/s2. A 1 kg mass that travels at 1 m/s has a kinetic energy of 1 kg(m/s)2 or 1 J. A joule is a rather small unit of energy for chemical reactions. Therefore, the kilojoule, kJ, is most often encountered.
The Calorie
The nonSI unit for energy is the calorie, cal. Today, one calorie is
exactly equivalent to 4.184 J. The old definition of the calorie was the
amount of heat needed to raise one gram of water by one degree Celsius.
Because the calorie is a small unit of energy, the kilocalorie is most
frequently encountered in chemistry. One kilocalorie is exactly equivalent
to 4.184 kJ.
Convert 34.5 kcal to joules and kilojoules.
Solution 2.20
First, convert kcal to J.
34.5 kcal x 1000 cal/1 kcal x 4.184 J/1 cal = 1.44 x 105 J
Then, convert J to kJ.
1.44 x 105 J x 1 kJ/1000 J = 1.44 x 102 kJ
Specific Heat
When substances that are not undergoing state changes are heated, they increase in temperature. Specific heat, c, is the amount of heat required to increase the temperature of one g of substance by 1oC (1 K). Each substance has its own characteristic specific heat. Substances with higher specific heats require more heat to increase the temperature of equal mass samples by a fixed temperature than do substances with lower specific heats. Specific heat is calculated as follows
c = q/g x (delta T) (2.6)
in which c is specific heat in J/(goC) or cal/(goC),
q is the heat in J or cal, g is mass of the substance in
grams, deltaT is the change in temperature (T2
- T1), T2 is the final temperature,
and T1 is the initial temperature.
A 215-g sample of an unknown solid requires 1.448 kJ of heat to raise its temperature from 25.3oC to 44.9oC. Calculate the specific heat of the solid in J/(goC).
Solution 2.21
Use Eq. 2.6 to calculate the specific heat of this solid.
c = q/(g deltaT) = q/(g (T2
- T1))
= 1.448 kJ/(215 g x (44.9oC - 25.3oC)) = 1.448
kJ/(215 g x 19.6oC) = 3.44 x 10-4 kJ/(goC)
The answer has the units kJ/(goC); hence, convert kJ to J as follows.
3.44 x 10-4 kJ/(goC) x 1000 J/kJ = 0.344 J/(goC)
How many joules of heat are required to increase the temperature of 1.0 kg of silver, Ag, from 22.1oC to 71.9oC? The specific heat of Ag is 0.23 J/(goC).
Solution 2.22
Use Eq. 2.6 to calculate the number of joules.
q = g x deltaT x c
= 1.0 kg Ag x 1000 g Ag/kg Ag x (71.9oC - 22.1oC) x 0.23 J/(goC) = 11,454 J = 1.1 x 104 J
What is the final temperature of sand when 3.55 kJ is added to 375 g sand that is originally at 29.4oC? The specific heat of sand is 0.80 J/(goC).
Solution 2.23
Use Eq. 2.6 to calculate the final temperature. All values are known except for the final temperature, T2. Therefore, substitute the known values and solve for T2 as follows.
q = g x (T2 - T1) x csand
3.55 kJ x 1000 J/kJ = 375 g x (T2 - 29.4oC) x 0.80 J/(goC)
12oC = T2 - 29.4oC
T2 = 41oC
2.4 UNCERTAINTY IN MEASUREMENTS
Accuracy
Both precision and accuracy are considered in all chemical measurements. The accuracy of a measurement is how close the measured value is to a standard or "true" value. A more accurate measurement is one closer to the standard value than a less accurate measurement. Accuracy is measured in terms of the deviation of the measurement(s) (called the error) from the "true" value.
Precision
Precision is how closely repeated measures are grouped. In other words, how reproducible the measurements are. The smaller the range of values obtained when measuring the same quantity, the greater the precision. Most frequently, good precision is an indication of high accuracy, but not always.
Errors in Measurement
Measurement errors account for the range of different values that are obtained when making the same measurement repeatedly. Two types of errors are generally found in chemical measurements: systematic and random errors.
Systematic Errors
Systematic errors result from: poor procedures and methods; malfunctioning and uncalibrated instruments; human error; impure samples; and some unrecognized factors that influence the results. For example, suppose that you measure the weight of an object on a scale but fail to adjust the scale to the zero point before weighing the object. The measurement will be either high or low, depending on the initial incorrect setting of the scale. Systematic errors are reduced by finding their causes and eliminating them.
Random Errors
Random errors occur in all chemical measurements. Even if every precaution is taken to avoid systematic errors, small deviations or random errors arise that are unavoidable and not identifiable. Random errors, by definition, are impossible to illustrate. If random errors could be identified, they would be corrected, and thus would not be random errors.
Errors and Measurement Uncertainty
Collectively, systematic and random errors introduce uncertainty--or lack of confidence--in all measured values. Hence, all reported measurements should indicate the degree of certainty and uncertainty of the measurement. In chemistry, this is most frequently accomplished through the use of significant figures, which are described in the following section.
What are Significant Figures?
Significant figures, also called significant digits, are measured digits in a number that are known with certainty plus one uncertain digit. Stated differently, significant figures are all known digits plus the first doubtful or estimated digit.
It is important to note that significant figures only apply to measured values, and do not apply to exact numbers. Significant figures only apply to measurements that are to some degree uncertain.
Significant Figures and Uncertainty
Usually, the last significant figure is thought to be uncertain by 1. For example, by stating the volume of a liquid is 35.5 mL, this means that the measured volume is at most 35.6 mL (+0.1) and at least 35.4 mL (-0.1). If the same liquid is totally transferred to a more precise volumetric instrument--let's say one that has a scale with 0.1-mL marks etched accurately on the side--it is possible to obtain an additional significant figure.
The Number of Significant Figures
Whenever a measurement is encountered, always remember that besides the numerical value and units, the number also indicates the precision with which the measurement was made or stated differently indicates the number of significant figures in the measurement. Consider the following three measurements and the number of significant figures they represent.
1.25 m indicates three significant figures (1, 2, 5) (range 1.26 m - 1.24 m)
434.56 K indicates five significant figures (4, 3, 4, 5, 6) (range 434.57 K - 434.55 K)
8.913477 cm indicates seven significant figures (8, 9, 1, 3, 4, 7, 7)
(range 8.913478 cm - 8.913476 cm)
Zeros and Significant Figures
All nonzero numbers in measurements are always significant, but zeros pose a special problem because a zero in a number that only acts as a placeholder is not significant. Placeholders are not measured quantities; therefore, they are not significant figures.
The following rules summarize and illustrate all possible cases in which zeros are found in measurements:
Rule 1. Zeros located in the middle of a number.
In all cases, zeros in the middle of a number are significant. In each of the following, the zero is a significant figure:
10.004 g indicates five significant figures (1, 0, 0, 0, 4)
47,000.15 mL indicates seven significant figures (4, 7, 0, 0, 0, 1, 5)
Zeros in the middle of measured quantities are measured digits and are not placeholders; accordingly, they are significant in all cases.
Rule 2. Zeros located in front of a number.
Zeros in front of numbers are usually to the right of the decimal point. These zeros act as placeholders (they are not measured), so they are not significant figures. Consider each of the following examples of such cases.
.0005 L indicates one significant figure (5)
.0000477 m indicates three significant figures (4, 7, 7)
Sometimes a zero is placed in front of the decimal point to show that no other digit is present. Similarly this zero is not significant. The following examples serve to illustrate this point.
0.00831 mm indicates three significant figures (8, 3, 1)
0.1 mL indicates one significant figure (1)
2.C Rule 3. Zeros located after a number to the right of the decimal point.
For this specific case, the zero is either a measured quantity (certain) or a good estimate (the first uncertain digit); consequently, zeros after a number and to the right of the decimal point are all significant. Two examples of this case are as follows
.650 dm indicates three significant figures (6, 5, 0)
.108300 ms indicates six significant figures (1, 0, 8, 3, 0, 0)
For each of the above measurements, the zeros were measured by some instrument. All the zeros are certain except the last zero, which is uncertain but still significant.
Rule 4. Zeros located after a number to the left of the decimal point.
Zeros found after a number and to the left of the decimal point are significant if they are measured, and are not significant if they are placeholders.
An object with a measured mass of 500 g has a questionable number of significant figures because more information is required to determine the correct number. The measurement, 500 g, contains three significant figures only if the second zero (units place) is the first uncertain figure. It contains two significant figures if the first zero (tens place) is the first uncertain figure. Lastly, the 5 could be the uncertain digit; if this is the case, the measurement only has one significant figure.
To avoid the confusion generated by the ambiguous nature of zeros to the left of decimal points, such measurements are often expressed in scientific notation in which the decimal factor represents the correct number of significant figures. Thus, 500 grams is expressed as 5 x 102 g (one significant figure) or 5.0 x 102 g (two significant figures) or 5.00 x 102 g (three significant figures), depending on the proper number of significant figures.
Write the number of significant figures indicated by each of the following measurements: (a) 55,9977 g, (b) 0.2937 cm, (c) 1000.0200 L, d. 0.0000500 kg, e. 2.001010 x 1012 mg, f. 0.0000000070 m, g. 7.000 x 106 mm
Solution 2.24
(a) 6 significant figures, (b) 4 significant figures, (c) 8 significant
figures, d. 3 significant figures, e. 7 significant figures, f. 2 significant
figures, g. 4 significant figures
A college chemistry professor counts the number of students in his laboratory class. The count indicates the presence of 21 students. How many significant figures does this represent?
Solution 2.25
Significant figures do not apply in this case because there is no uncertainty
in counting students.
Addition and Subtraction of Significant Figures
When measured quantities are added and subtracted, the answer can have
no more digits to the right of the decimal point than does the measured
quantity with the least number of decimal places.
Add the masses 2.0965 g and 1.41 g and express the answer to the correct number of significant figures.
Solution 2.26
The answer can only have two decimal places. First calculate the sum of the two numbers:
2.0965 g
+ 1.41 g
3.5065 g = 3.51 g
Then round off the answer to the correct number of decimal places. Because
the second mass was only measured to two decimal places, the answer can
only have two decimal places. The answer to this problem is 3.51 g.
Rules for Rounding Off Measurements
When rounding off, look at the first nonsignificant figure--one place to the right of the least significant figure. The least significant figure is the last figure in the number retained after rounding off. Then apply the following three rounding rules.
Rule 1. If the value of the first nonsignificant figure is greater than 5, add 1 to the least significant figure and drop all nonsignificant digits.
Rule 2. If the value of the first nonsignificant figure is less than 5, retain the least significant figure and drop all nonsignificant digits.
Rule 3. a. If the first nonsignificant figure is 5 and is followed by nonzero digits increase the value of the least significant figure by one and drop all nonsignificant digits.
b. If the 5 is followed by zeros or nothing, add 1 to the least significant figure if it is an odd number and drop all nonsignificant digits.
c. If the least significant figure is an even number, retain the least significant figure and drop all nonsignificant figures.
In Problem 2.26, the first nonsignificant figure is 6, so 1 is added to 0 which gives 1 as the second decimal place. The final answer is expressed as 3.51 g.
What is the sum of 10.0043 mL + 5.5 mL + 9.250 mL?
Solution 2.27
10.0043 mL (four decimal places)
5.5 mL (one decimal places)
+ 9.250 mL (three decimal places)
24.7543 mL (round to one decimal place)
The answer, 24.7543 mL, must be rounded to one decimal place because
the second measured quantity, 5.5 mL, only contains one decimal place.
Because the first nonsignificant figure is 5 followed by nonzero digits
and the least significant figure is 7, an odd number, add one to the least
significant figure and drop all nonsignificant digits. This gives a final
answer of 24.8 mL.
Multiplication and Division of Significant Figures
When measurements are multiplied and divided, the answer can have no more significant figures than the measurement with the least number of significant figures. If two numbers are multiplied, one with six and the other with three significant figures, the answer can only have three significant figures.
Find the area of a surface that is 5.82131 cm by 4.11 cm.
Solution 2.28
To find the area multiply the length times the width.
5.82131 cm (six significant figures)
x 4.11 cm (three significant figures)
23.92558410 cm2
The first nonsignificant digit in the answer is 2, which is less than
5; thus, Rule 2 is applied and the answer is rounded off to 23.9 cm2
(three significant figures).
Perform the following arithmetic operations and express the answer to the correct number of significant figures.
(7.290 m x 2.0400 m)/ 0.95 m =
Solution 2.29
The denominator contains a measurement with only two significant figures; hence, this limits the answer to two significant figures. Perform the math operations and round off the resulting answer to two significant figures.
The first nonsignificant figure is 6; hence, the answer is rounded off
by adding 1 to 5 and dropping the nonsignificant figures, which leaves
16 m as the answer.
Perform the following arithmetic operations and express the answer to the correct number of significant figures.
(11.2050 mm - 10.322 mm) x 6.030000 mm =
Solution 2.30
Both rules regarding significant figures apply in this example. After subtracting, the answer can only have three decimal places in the answer because 10.322 mm only contains three decimal places.
11.2050 mm - 10.322 mm = 0.8830 mm = 0.883 mm
Apply the multiplication rule when multiplying 0.883 mm (three significant figures) by 6.030000 mm (seven significant figures). Three significant figures are the maximum number allowed in the answer.
0.883 mm x 6.030000 mm = 5.32449 mm2 = 5.32 mm2
The International System of Units, SI, is the system of measurement used by scientists. All SI units of measurement are either base or derived units. The seven base units SI are the meter, kilogram, kelvin, second, mole, ampere, and candela. All other units are derived from these seven base units; thus, they are called derived SI units. To increase or decrease the magnitude of SI units, it is only necessary to place the appropriate prefix in front of the unit. Frequently encountered SI prefixes include kilo (1000 x), centi (1/100 x), milli (1/1000 x), and micro (10--6 x).
Mass is the quantity of matter in an object and is usually measured in kilograms, grams, and milligrams. In SI, volume is measured in cubic meters, cubic decimeters, or cubic centimeters. They are derived from the base units of length--meters, decimeters, and centimeters. Liters and milliliters, two non-SI volume units, are commonly used in chemistry. Density is the ratio of mass to volume and is most frequently measured in g/cm3 or g/L (g/dm3). Specific gravity is the ratio of the density of a substance to the density of water.
Temperature is a measure of the degree of hotness of matter. Kelvin (K) is the SI unit of temperature. The zero point on the Kelvin scale is absolute zero, the lowest possible temperature. Kelvin degrees are the same magnitude as Celsius degrees, the nonSI unit of temperature. The SI unit of energy is the joule, J, and the nonSI unit is the calorie, cal. One calorie is equivalent to 4.184 J. The quantity of heat flow depends on the difference in temperature of two objects that contact each other.
All measurements are to some degree uncertain because of measurement errors. Systematic and random errors are the two general types of errors found in measurements. Systematic errors result from improper techniques, uncalibrated equipment or human error: they may be corrected. Random errors are those that cannot be identified but are present--they cannot be corrected.
Chemists use significant figures to express the degree of certainty for measurements. Significant figures are the measured digits plus one estimated digit.
1. Write the SI unit that corresponds to the following measurements. (a) density, (b) quantity of substance, (c) length, (d) electric current, (e) energy, (f) temperature.
2. What prefix is used for each of the following: (a) 0.01 x, (b) 1,000,000 x, (c) 10--6 x, (d) 0.1 x, (e) 0.001 x, (f) 10--9 x?
3. What is the meaning of each of the following prefixes. (a) milli, (b) deca, (c) nano, (d) deci, (e) kilo, (f) centi
4. Convert 172 ft to mm.
5. The distance from the earth to the sun is 9.3 x 107 miles. Express this distance in kilometers.
6. Convert 9.800 x 1012 g to kg and mg.
7. Convert each of the following Celsius temperatures to Kelvin. (a) -141oC, (b) -235.6oC, (c) 204oC
8. Convert 100oF to oC and K.
9. A block of wood is 2.1 cm long, 0.54 cm wide, and 934 mm high. What is the volume of the block in liters?
10. Convert 188.2 L to m3.
11. Convert 8.1 ft3 to cm3.
12. An object has a mass of 7.24 g and a volume of 6.55 cm3. What is the density of the object?
13. Vanadium, V, has a density of 6.11 g/mL. Calculate the mass in kg of V contained in 0.000671 m3.
14. The density of pure silicon, Si, is 2.33 g/mL. Calculate the volume in mL of a 900.5-mg sample of Si.
15. Air has a density of 1.29 g/L. Express the density of air in mg/mm3.
16. A 1.28-qt bottle is used to store liquid mercury (density = 13.6 g/cc). What is the mass of the mercury in the bottle in kg?
17. A bar of metallic osmium, Os, has the following dimensions: 0.35 m x 2.43 cm x 453.9 mm. If the density of Os is 22.6 g/cm3, what is the mass of the Os bar in kilograms?
18. Ethyl alcohol is also called grain or drinking alcohol. Calculate the specific gravity of ethyl alcohol, if the density of ethyl alcohol is 0.789 g/cm3 and the density of water is 1.00 g/cm3.
19. One hectare is a unit of area equal to exactly 10,000 m2. How many square miles are contained in one hectare? One mile equals 5280 ft.
20. The mass of a cylinder is 124.54 g. Its height is 3.22 cm and has a circular diameter of 1.88 cm. If the formula for the volume of a cylinder is V = r2h, calculate the density of the cylinder.
21. What is the specific heat of a substance that requires 52.7 kJ to raise the temperature of a 1.05-kg sample from 15.2oC to 27.2oC?
22. (a) How many kJ of heat are required to raise the temperature of a 215-g sample of Cu from 0.0oC to 100.0oC? The specific heat of Cu is 0.39 J/(goC). (b) How many kJ of heat are required to raise the temperature of a 215-g sample of H2O from 0.0oC to 100.0oC? The specific heat of H2O is 4.18 J/(goC).
23. What is the final temperature of water after 35 kJ of heat are transferred to a 150-g sample of water that is initially at 25oC?
24. By what means is the accuracy of a measurement determined?
25. How many significant figures are contained in each of the following measured quantities? (a) 24.084 mg, (b) 128.0700 L, (c) 0.098 cm, (d) 0.0012000 dm3, (e) 0.0005 kg, (f) 0.00000000191010 mol, (g) 1.0010 J
26. Round off each of the following to four significant figures: (a) 194.645 g, (b) 10.998 g, (c) 962.1539 g
27. Use scientific notation to express 6,000 cm to one, two, three, and four significant figures.
28. Add each of the following and express the answer to the correct number of significant figures.
(a) 2.345 g + 2.5 g =
(b) 32.0030 mL + 11.87 mL =
29. Multiply each of the following and express the answer to the correct number of significant figures.
(a) 732.2 s x 8.1 s =
(b) 5.5050 cm x 9.9 cm =
30. Perform the following arithmetic operations and express the answer to the correct number of significant figures.
(a) (133 g x 534.00 g)/(9.1 g + 0.4543 g) =
(b) (154.7325 m - 154.7036 m) x 9.89892 m =
(c) ((0.000100 cm3)/(0.0027711 cm2) - 0.0000522
cm + 0.00046 cm)/5.198 cm =
1. (a) g/cm3, (b) mole, (c) m, (d) A, (e) J, (f) K
2. (a) centi, (b) mega, (c) micro, (d) deci, (e) milli, (f) nano
3. (a) 0.001 x, (b) 10 x, (c) 10--9 x, (d) 0.1 x, (e) 1000 x, (f) 0.01 x
4. 5.24 x 104 mm
5. 1.5 x 108 km
6. 9800 kg, 9.800 x 109 mg
7. (a) 132 K, (b) 37.6 K, (c) 477 K
8. 37.8oC, 311.0 K
9. 0.11 L
10. 0.1882 m3
11. 2.3 x 105 cm3
12. 1.11 g/cm3
13. 4.1 kg
14. 0.386 mL
15. 0.00129 mg/mm3
16. 16.5 kg
17. 87 kg
18. 0.789
19. 3.863 x 10--3 mile2
20. 13.9 g/cm3
21. 4.18 J/(goC)
22. (a) 8.4 kJ, (b) 90 kJ
23. 81oC
24. Comparing the value to the true value
25. (a) 5, (b) 7, (c) 2, (d) 5, (e) 1, (f) 6, g. 5
26. (a) 194.6 g, (b) 11.00 g, (c) 962.2 g
27. 6 x 103 cm, 6.0 x 103 cm, 6.00 x 103 cm, 6.000 x 103 cm
28. (a) 4.8 g, (b) 43.87 g
29. (a) 5.9 x 103 s2, (b) 54 cm2
30. (a) 7.4 x 103 g, (b) 0.286 m2, (c) 0.00702
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