# UNIT (1) MEASUREMENTS IN CHEMISTRY

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1 UNIT (1) MEASUREMENTS IN CHEMISTRY Measurements are part of our daily lives. We measure our weights, driving distances, and gallons of gasoline. As a health professional you might measure blood pressure, temperature, pulse rate, drug dosage, or percentage of body fat. A measurement contains a number and a unit. A unit specifies the physical property and the size of a measurement, while the number indicates how many units are present. A number without a unit is usually meaningless. 1.1 Units of Measurement In the United States most measurements are made with the English system of units which usually contain fractions (a collection of functionally unrelated units.) The metric system is a decimal-based system of units of measurement which is used most often worldwide. Around 1960, the international scientific organization adopted a modification of the metric system called International System or SI (from Systèm International). Quantity English Unit Metric Unit SI Unit Mass pound (lb) gram (g) kilogram (kg) Length foot (ft) meter (m) meter (m) Volume quart (qt) liter (L) cubic meter (m 3 ) Temperature degree Fahrenheit ( F) degree Celsius ( C) Kelvin (K) Energy calorie (cal) calorie (cal) Joule (J) 1.2 Scientific Notation Scientific notation is a common method used to conveniently represent very small or very large numbers. There are two parts to any number expressed in scientific notation, a coefficient, and a power of 10. The number 683 is written in scientific notation as 6.83 x The coefficient is 6.83 and 10 2 shows the power of 10 (the superscript 2 is called an exponent). A number less than one would contain a negative exponent. For example: the number is written as 7.5 x 10-3 (note the negative exponent). The coefficient must always be a number greater than or equal to 1 but less than 10 or 1 coefficient <

2 Worked Example 1-1 Express the following numbers in scientific notation: a) b) Apply the following: Place the decimal point after the first nonzero digit in the number. Indicate the number of places the decimal was moved using the power of 10. If the decimal is moved to the left, the power of 10 is positive. If moved to the right, it is negative. a) x 10 2 (coefficient = , exponent = +2) b) x 10-3 (coefficient = 7.956, exponent = -3) Practice 1-1 Express each of the following values in scientific notation: a) There are 33,000,000,000,000,000,000 molecules of water in one milligram of water. b) A single molecule of sucrose weighs g. a) 3.3 x (coefficient = 3.3, exponent = 19) b) 5.7 x (coefficient = 5.7, exponent = -22) Practice 1-2 Convert each the following scientific notation to decimal notation. a) 8.54 x 10 3 b) 6.7 x 10-5 c) 1.29 x 10 4 d) x 10-2 a) 8540 b) c) d)

3 Scientific Notation and Calculators Numbers in scientific notation can be entered into most calculators using the EE or EXP key. As an example try 9.7 x Enter the coefficient (9.7) into calculator. 2. Push the EE (or EXP) key. Do NOT use the x (times) button. 3. Enter the exponent number (3). Number to Enter Method Display Reads 9.7 x EE or EXP or 9.7E03 or 9700 Now try 8.1 x 10-5 : 1. Enter the coefficient (8.1) into calculator. 2. Push the EE (or EXP) key. Do NOT use the x (times) button. 3. Enter the exponent number (5). Use the plus/minus (+/-) key to change its sign. Number to Enter Method Display Reads 8.1 x EE or EXP 5 +/ or 8.1E Metric Prefixes The metric system is a decimal-based system of units of measurement used by most scientists worldwide. In the metric system, a prefix can be attached to a unit to increase or decrease its size by factors (powers) of 10. Prefix Value tera- (T) = 1,000,000,000,000 giga- (G) 10 9 = 1,000,000,000 mega- (M) 10 6 = 1,000,000 kilo- (k) 10 3 = 1,000 deci- (d) 10-1 = 0.1 centi- (c) 10-2 = 0.01 milli- (m) 10-3 = micro- (μ) 10-6 = nano- (n) 10-9 = pico- (p) =

4 Practice 1-3 Give the metric prefix that corresponds to each of the following a) 1,000,000,000 b) 10-6 c) 1000 d) 0.01 e) 10-9 f) a) giga b) micro c) kilo d) centi e) nano f) tera 1.4 Significant Figures in Measurements A student is asked to determine the mass of a small object using two different balances available in the lab. The lower priced model reports masses to within ±0.01 g (one-one hundredth), while the higher priced one reports to within ± g (one-ten thousandth). The student measures the mass three times on each balance and completes the following table. First balance Second balance Three measurements 2.16, 2.14, 2.15 g , , g Average 2.15 g g Reproducibility ±0.01 g ± g Which digit is the uncertain digit in the average? Which digits are certain digits in the average? How many significant digits are in the average? The last digit; 5 The last digit; 8 2, 1 2, 1, 5, 3 Three significant digits Five significant digits Significant figures (sig figs) are the digits that are known with certainty plus one digit that is uncertain. All nonzero digits in measurements are always significant. Are zeroes significant? YES: zeros between nonzero digits (20509). YES: zeros at the end of a number when a decimal point is written (3600.). NO: zeros at the end of a number when no decimal point is written (3600). NO: zeros at the beginning of a number (0.0047). 1-4

5 Worked Example 1-2 How many significant figures does each number have? a) b) 600. c) 93,000 d) 2.08 x 10-5 e) 600 f) g) h) x 10 2 i) x 10 6 sf sf sf , x x x Significant Figures in Exact Numbers Exact numbers have an unlimited number of significant figures. Exact numbers are obtained by counting items or by definition. Counting: 24 students mean students. 8 pennies means pennies. Definition: 1 m = 100 cm means m = cm 1.5 Calculations Involving Significant Figures Rules for Rounding off Numbers If the first digit to be deleted is 4 or less, leave the last reported digit unchanged. If the first digit to be deleted is 5 or greater, increase the last reported digit by one. In some cases you need to add significant zeros. The number 2, reported in four significant figures, is Practice 1-4 Round off each of the following to three significant figures. a) b) c) d) 5 e) f) 8000 g) 2.4 x 10-5 h) 670 a) (9.17) b) (9.18) c) (9.18) d) 5 (5.00) e) ( ) f) 8000 (8.00 x 10 3 ) g) 2.4 x 10-5 (2.40 x 10-5 ) h) 670 (670.) 1-5

6 Rules for Rounding off in Calculations A. Multiplication and Division The answer carries the same number of significant figures as the factor with the fewest significant figures. Practice 1-5 Perform each of the following calculations to the correct number of significant figures. a) x b) (2.50 x 10-3 ) x ( x 10 5 ) c) d) (6.56 x ) (7.8 x 10 9 ) a) x = b) (2.50 x 10-3 ) x ( x 10 5 ) = 4.63 x 10 2 c) = 20.7 d) (6.56 x ) (7.8 x 10 9 ) = 8.4 B. Addition and Subtraction The answer should have the same number of decimal places as the quantity with the fewest decimal places. Practice 1-6 Perform each of the following calculations to the correct number of significant figures: a) b) c) d) e) (6.8 x 10-2 ) + (2.04 x 10-2 ) f) (5.77 x 10-4 ) - (3.6 x 10-4 ) a) = 75.7 b) = c) = 3.0 d) = e) (6.8 x 10-2 ) + (2.04 x 10-2 ) = 8.8 x 10-2 f) (5.77 x 10-4 ) - (3.6 x 10-4 ) = 2.2 x

7 1.6 Writing Conversion Factors Many problems in chemistry require converting a quantity from one unit to another. To perform this conversion, you must use a conversion factor or series of conversion factors that relate two units. This method is called dimensional analysis. Any equality can be written in the form of a fraction called a conversion factor. A conversion factor is easily distinguished from all other numbers because it is always a fraction that contains different units in the numerator and denominator. Converting kilograms to pounds can be performed using the equality 1 kg = 2.20 lb. The two different conversion factors that may be written for the equality are shown below. Note the different units in the numerator and denominator, a requirement for all conversion factors. Conversion Factors: umerator Denominator 1 g 2.20 or g Some common units and their equivalents are listed in Table 1.1. You should be able to use the information, but you will not be responsible for memorizing the table. The Table will be given to you during quizzes and exams. Table 1.1 Some Common Units and Their Equivalents Length 1 m = 100 cm 1 m = 1000 mm 1 cm = 10 mm 1 km = 1000 m 1nm = 10-9 m 1Å = m 1 in = 2.54 cm 1 ft = cm 1 mi = 1.61 km 1 yd = 0.91 m 1 ft = 12 in. Mass 1 kg = 1000 g 1 g = 1000 mg 1 lb = 454 g 1kg = 2.20 lb 1 oz = g 1 lb = 16 oz Volume 1L = 1000 ml 1 ml = 1 cm 3 1qt = L 1 gal = 3.78 L Energy 1 cal = 4.18 J Temperature F = 1.8 C + 32 C = ( F -32)/1.8 K = C

8 Worked Example 1-3 Write conversion factors for each of the following equalities or statements: a) 1 g = 1000 mg b) 1 foot = 12 inches c) 1 quart = liter d) The accepted toxic dose of mercury is 0.30 mg per day. Equality Conversion factor Conversion factor 1 g = 1000 mg 1 g 1000 mg 1000 mg 1 g 1 foot = 12 inches 1 quart = liter The accepted toxic dose of mercury is 0.30 mg per day. 1 ft. 12 in. 1 t mg 1 da 12 in. 1 ft t. 1 da 0.30 mg 1.7 Problem Solving in Chemistry - Dimensional Analysis Dimensional analysis is a general method for solving numerical problems in chemistry. In this method we follow the rule that when multiplying or dividing numbers, we must also multiply or divide units. Solving problems by dimensional analysis is a three-step process. 1. Write down the given measurement; number with units. 2. Multiply the measurement by one or more conversion factors. The unit in each denominator must cancel (match) the preceding unit in each numerator. 3. Perform the calculation and report the answer to the proper significant figures based on numbers given in the question (data), not conversion factors used. 1-8

9 Worked Example 1-4 Convert km to meters. To convert kilometers to meters, we could use the following equality: 1 km = 1000 m (See Table 1.1) The corresponding conversion factors would be: We select the conversion factor to cancel kilometers, leaving units of meters. The number of significant figures in your answer reflect km. The exact conversion factor does not limit the number of significant figures in your answer. Worked Example 1-5 Convert 4.5 weeks to minutes. Worked Example 1-6 Convert 2.7 g/ml to lb/l. (45360 rounded to 2 sig figs.) We need two conversion factors: one to convert g to lb and the other to convert ml to L. We know that 1 lb= 454 g and 1 L = 1000 ml (See Table 1.1) Remember that the number of significant figures in your answer reflect 2.7. The conversion factors do not limit the number of significant figures in your answer. 1-9

10 Practice 1-7 Perform each of the following conversions: a) Convert 14.7 lb to ounces. b) Convert 19.8 lb to kilograms. c) Convert 23 m/sec to mi/hr. Conversion Convert 14.7 lb to ounces Convert 19.8 lb to kilograms Convert 23 m/sec to mi/hr. Calculation 14.7 lb x 16 oz 1 lb = 235 oz 1 kg 19.8 lb x 2.20 lb = 9.00 kg 23 m 1 km 1.0 mi 3600 sec x x x 1.0 sec 1000 m 1.61 km 1 hr = 51 mi hr 1.8 Density and Specific Gravity Density is the ratio of the mass of a substance to the volume occupied by that substance. Density is expressed in different units depending on the phase (form) of the substance. Solids are usually expressed in grams per cubic centimeter (g/cm 3 ), while liquids are commonly grams per milliliter (g/ml). The density of gases is usually expressed as grams per liter (g/l). 1-10

11 Worked Example 1-7 If 10.4 ml of a liquid has a mass of g, what is its density? Density can be used as a conversion factor that relates mass and volume, note the different units in the numerator and denominator. Densities can be used to calculate mass if volume is given or calculate volume given mass. For example, we can write two conversion factors for a given density of 1.05 g/ml: 1.05 g 1.00 ml or 1.00 ml 1.05 g Worked Example 1-8 The density of a saline solution is 1.05 g/ml. Calculate the mass of a ml sample. Practice 1-8 The density of rubbing alcohol is g/ml. What volume of rubbing alcohol would you use if you needed 32.0 g? We use the density as a conversion factor: V = 32.0 g x 1.00 ml g = 40.7 ml 1-11

12 Specific Gravity is the ratio of the density of liquid to the density of water at 4 C, which is 1.00 g/ml. Since specific gravity is a ratio of two densities, the units cancel. An instrument called a hydrometer is used to measure the specific gravity of liquids. Worked Example 1-9 What is the specific gravity of jet fuel if the density is g/ml? Practice 1-9 A 50.0 ml sample of blood has a mass of 53.2 g. a) Calculate the density of the blood. b) Calculate the specific gravity of the blood. d = m V a) d = 53.2 g 50.0 ml = 1.06 g/ml specific gravity = density of blood density of water = 1.06 g/ml 1.00 g/ml =

13 1.9 Temperature Scales Temperature, reported in Fahrenheit ( F) or Celsius ( C), is used to indicate how hot or cold an object is. The SI unit for reporting temperature is Kelvin (K). See the comparison of the three scales: Freezing point Boiling point Normal body of water of water temperature Fahrenheit 32 F 212 F 98.6 F Celsius 0 C 100 C 37 C Kelvin 273 K 373 K 310 K The following formulas show the conversions: Fahrenheit to Celsius: C = ( F - 32) 1.8 Celsius to Fahrenheit: F = 1.8 C + 32 Celsius to Kelvin: K = C Practice 1-10 Complete the following table. Fahrenheit Celsius Kelvin 88 F -55 C 469K Fahrenheit Celsius Kelvin 88 F 31 C 304 K -67 F -55 C 218 K 385 F 196 C 469 K 1-13

14 1.10 Heat and Specific Heat Heat and temperature are both a measure of energy. Heat, however, is not the same as temperature. Heat measures the total energy, whereas temperature measures the average energy. A gallon of hot water at 200 F has much more heat energy than a teaspoon of hot water at same temperature. Heat can be measured in various units. The most commonly used unit is calorie (cal). The calorie is defined as the amount of heat required to raise the temperature of 1 gram of water by 1 C. This is a small unit, and more often we use kilocalories (kcal). 1 kcal = 1000 cal utritionist use the word Ca orie (with a capita C ) to mean the same thing as kilocalorie. 1 Cal = 1000 cal = 1 kcal The unit of energy in SI unit is joule (pronounced jool ), which is about four times as big as the calorie: 1 cal = J Specific Heat Substances change temperature when heated, but not all substances have their temperature raised to the same extent when equal amounts of heat are added. Specific Heat is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. It is measured in units of cal/g C or J/g C. (Recall; 1 cal is required to raise the temperature of 1 gram of water by 1 C, the specific heat of water is therefore: 1.00 cal/g C, or J/g C). Specific heats for some substances in various states are listed in the following table. A substance with a high specific heat is capable of absorbing more heat with a small temperature change than a substance with lower specific heat. 1-14

15 Specific Heats for Some Common Substances Substance Specific Heat (J/g C) Solids Liquids Gases gold copper aluminum ice 2.06 mercury methanol 1.77 ethanol 2.42 water 4.18 argon oxygen nitrogen steam 2.03 We can calculate the amount of heat gained or lost by a substance using its specific heat, its measured mass, and the temperature change. Amount of heat = mass x specific heat x change in temperature* q = m x SH x (T final T initial ) * The temperature change cou d a so e written as Δ (delta T). If any three of the four quantities in the equation are known, the fourth quantity can be calculated. Worked Example 1-10 Determine the amount of heat that is required to raise the temperature of g of water from 29.0 C to 46.0 C. The specific heat of water is 4.18 J/g C. q = m x SH x ΔT q = g x 4.18 J/g C x 17.0 C = 526 J 1-15

16 Practice 1-11 What mass of lead is needed to absorb 348 J of heat if the temp of the sample rises from 35.2 C to 78.0 C? The specific heat of lead is J/g C. q = m x SH x T so m = q SH x T m = 348 J J/g C x 42.8 C = 63.0 g Practice 1-12 It takes 87.6 J of heat to raise the temp of 51.0 g of a metal by 3.9 C. Calculate the specific heat of the metal. q = m x SH x T so SH = q m x T SH = 87.6 J 51.0 g x 3.9 C = 0.44 J/g C 1-16

17 Practice x 10 3 J of energy is transferred to 56.0 g of water at 19 C. Calculate the final temperature of water. SH = 4.18 J/g C x 10 3 J ΔT = = 17.1 C 56.0 g x 4.18 J/g C ΔT = T final - T initial 17.1 C = T f - 19 C T f = 36 C 1-17

18 Homework Problems 1.1 Complete the following table. Decimal notation Scientific notation Number of significant figures 400, ,995, x x x x Perform the following calculations to correct number of significant figures. a. 4.6 x x 193 b c. (7.120 x 10-3 ) (6.000 x 10-5 ) d. (5.92 x 10 3 ) x Perform the following calculations to correct number of significant figures. a b. (3.42 x 10-4 ) + (5.007 x 10-4 ) c (8.3 x 10-2 ) d. (3.8 x 10 6 ) - (8.99 x 10 6 ) 1.4 Perform the following conversions. Show your set ups. a. 683 nanometer (nm) to angstrom (Å) b. 520 mi/h to m/sec c g/cm 3 to lb/ft 3 d C to F 1.5 A physician has ordered 37.5 mg of a particular drug over 15 minutes. If the drug was available as 2.5 mg/ml of solution, how many ml would you need to give every 15 seconds? 1-18

19 1.6 What is the density of a metal sample if a15.12-g sample is added into a graduated cylinder increased the liquid level from ml to ml? 1.7 The density of copper is 8.96 g/cm 3. You have three different solid samples of copper. One is rectangular with dimensions 2.3 cm x 3.1 cm x 8.0 cm. The second is a cube with edges of 3.8 cm. The third is a cylinder with a radius of 1.5 cm and a height of 8.4 cm. Calculate the mass of each sample. 1.8 A g sample of metal at 78.0 C is dropped into cold water. If the metal sample cools to 17.0 C and the specific heat of metal is cal/g C, how much heat is released? 1-19

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