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Chapter 1: Matter, Measurement, and Problem Solving 1. Chemistry the central science the study of matter and its changes 2. The Scientific Method OBSERVATIONEXPLANATION (empirical facts (data), then laws) (hypothesis and theory) OBSERVE Gather data, laws Hypothesis TEST Experiment THEORY 3. Matter Matter occupies space and has mass (mass is the amount of matter, weight is the force of gravitational attraction on the mass)

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Properties of Matter 2. Physical Properties vs. Chemical Properties A physical change does not change the chemical makeup. A chemical change (reaction) changes the substance. 3. Intensive vs. Extensive Properties Extensive depends on sample size, intensive does not. color mp, bp conductivity volume mass ~Which is better for determining an unknown substance? ~Would changing the state be a physical or chemical change? 1. States of matter; solid, liquid, gas

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solution; a homogeneous mixture of substances Compounds are substances formed from two or more elements combined in a fixed proportion by mass. Elements are substances that cannot be decomposed by chemical means into simpler substances. Atom: smallest component of an element Molecule: particle that is combination of two or more atoms smallest component of a compound Classification of Matter

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Energy All reactions require or generate energy. Law of Conservation of Energy –Energy cannot be created or destroyed but can be changed from one form to another. Potential energy in chemicals is called chemical energy. Heat and Temperature Energy PotentialKinetic Energy due to movement, dependent on mass and velocity K.E. = ½ mv 2 Stored energy, can be converted to kinetic energy Heat, or thermal energy, is a form of energy (internal motions of atoms and molecules). Temperature is a measure of the intensity of heat (average kinetic energy).

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Units of Measurement 1.Systèm International Units (Table 1.1) Base Units length m meter (39.37 inches) mass kg kilogram (2.205 lb) time s second temp K kelvin Derived Units e.g. Volume = length x length x length (e.g, m 3 ) 1 mL = cm 3 and 1 L = 1000 mL = 1000 cm 3 2.The Metric System (Table 1.2) KNOW ALL of the Decimal Multipliers and SI Prefixes from tera to atto!!! e.g. k = kilo = 10 3 1 km = 10 3 m n = nano = 10 -9 1 nm = 10 -9 m 3.Metric - English Conversions (Table 1.3) KNOW at least one conversion for each: length 1 inch = 2.54 cm or 1 m = 39.37 inches mass 1 kg = 2.205 lb or 1 lb = 454 g volume 1 L = 1.057 qt or 1 gal = 3.786 L

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Units of Measurement 4. Temperature Scales (Figure 1.12) -- Know how to convert! –Fahrenheit °F T F = (9/5) T C + 32 –Celsius °C T C = (5/9)(T F - 32) –Kelvin K T K = T C + 273.15

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Reliability of Measurements ERROR Uncontrolled conditions Poorly defined measurements Limited reproducibility Uncertainties are indicated through using significant figures. true value - measured value = Using the first thermometer, the temperature is 24.3 ºC (3 significant digits). Using the more precise (second) thermometer, the temperature is 24.32 ºC (4 significant digits) e.g temp in this room…

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Example Problem In analyzing a sample of polluted water a chemist measured out 25.00 mL of water with a pipet. At another point 25 mL was measured in a graduated cylinder. What is the difference between the two measurements?

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Example Problem In analyzing a sample of polluted water a chemist measured out 25.00 mL of water with a pipet. At another point 25 mL was measured in a graduated cylinder. What is the difference between the two measurements? Answer: 25 mL from a graduated cylinder means between 24 and 26 mL. From the pipet 25.00 mL indicates a range between 24.99 to 25.01. Therefore the pipet measures volume with greater precision.

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Calculations and Significant Figures 1.Accuracy and Precision Accuracy - how close to the true value? (systematic errors) Precision - how reproducible is the measurement? (random errors) 2.Significant Figures # of significant figures shows degree of uncertainty in measurement e.g. a certain distance, in inches, could be 11.1 or 11.08, or 11.083 depending on how carefully it was measured (3, 4, or 5 sig figs) 3.Exact Numbers Values that are exactly counted or defined can be assumed to have an infinite number of sig figs, e.g. - 25 people - 1 foot = 12 inches - 1 inch = 2.54 cm

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Rules for Significant Figures Non zero integers always count as sig figs Zeros Leading Zeros - those preceding all non zero integers and do not count. e.g. 00.0035 (2 s.f) Captive Zeros - those between non zero integers, count as sig figs, 1.008 = 4 s.fs Trailing Zeros - are at the end of a number and are significant when a decimal point is present, e.g. 0.120 --trailing zeros before decimal point: ambiguous (e.g. 300 could be 1, 2, or 3 sig figs). Use scientific notation! Exact Numbers – calculations not obtained through use of measuring device are all significant, 4/3 r 3

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Sig Fig Example ~Overall, this number has how many sig fig?

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Calculations with Sig Figs –multiplication and division Look for factor with fewest # of sig figs –addition and subtraction Look for value with fewest # of decimal places Example: Sig Fig rules are applied in the same order as the mathematical operations! During calculations, keep the number in your calculator; only round off at the end!

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Example Problems (One decimal place)

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Sample Problem Complete the following calculation. Write the answer in proper scientific notation, rounded to the correct number of significant figures, and with proper units.

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Sample Problem Complete the following calculation. Write the answer in proper scientific notation, rounded to the correct number of significant figures, and with proper units.

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Unit Conversions UNIT ANALYSIS given quantity x conversion factor(s) = desired quantity (starting units)(target units) e.g. 48 inches is 4 feet how is this shown in a calculation? (48 inches) x (1 ft/12 inches) = 4.0 ft Now, what is 48 inches in meters? (48 in) x (2.54 cm/in) x (1 m/100 cm) = 1.2 m Another example: convert 25 miles/gallon to km/L: Finally, correct sig fig! 11 km/L

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Density and Specific Gravity density d = mass/volume (usually g/cm 3 or g/mL) –e.g. density of water is 1.00 g/cm 3 or 1.00 g/mL –density of iron is 7.86 g/mL specific gravity d substance /d water (a dimensionless quantity) –e.g. specific gravity of iron is 7.86 –(i.e. iron is 7.86 times more dense than water) equivalence factor -- just like a conversion factor, used in unit analysis, e.g 7.86 g/mL given quantity x equivalence factor(s) = desired quantity (starting units)(target units)

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Sample Problems 1. An ocean dwelling dinosaur has an estimated body volume of 1.38 x 10 6 cm 3 and a mass of 1.24 x 10 6 g. What is its density? Answer = 0.899 g/cm 3 (note sig fig!!) 2. The density of table salt is 2.16 g/mL at 20 o C. What is its specific gravity? specific gravity = (2.16 g/mL)/(1.00 g/mL) = 2.16

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Sample Problems At the Athens Olympics, Justin Gatlin won the gold medal in the mens 100 (treat as 3 sig fig) meter race in a time of 9.85 seconds. The temperature at the time was 23 °C. Calculate his average speed over that distance in units of miles/hour. Butane (the fuel in your typical gas grill), has a specific gravity of 0.579. The new tank you just bought from Wal- Mart contains 5.00 gallons of butane. Calculate the mass of the butane in pounds.

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Sample Problems At the Athens Olympics, Justin Gatlin won the gold medal in the mens 100 (treat as 3 sig fig) meter race in a time of 9.85 seconds. The temperature at the time was 23 °C. Calculate his average speed over that distance in units of miles/hour. Answer: 22.7 miles/hour Butane (the fuel in your typical gas grill), has a specific gravity of 0.579. The new tank you just bought from Wal- Mart contains 5.00 gallons of butane. Calculate the mass of the butane in pounds. Answer: 24.1 lb

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Sample Problem Extremely tiny clusters of atoms or molecules are often called nano-particles due to the small scale of their dimensions. The specific gravity of gold is 19.3 and the mass of one gold atom is 197 amu (where 1 amu = 1.66 x 10 -24 g). Calculate the number of gold atoms in a sphere- shaped nanoparticle of gold that is 2.000 nanometers in diameter. [Hint: recall that the volume of a sphere is (4/3) r 3 where r is the spheres radius and = 3.14159]

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Sample Problem Extremely tiny clusters of atoms or molecules are often called nano-particles due to the small scale of their dimensions. The specific gravity of gold is 19.3 and the mass of one gold atom is 197 amu (where 1 amu = 1.66 x 10 -24 g). Calculate the number of gold atoms in a sphere- shaped nanoparticle of gold that is 2.000 nanometers in diameter. [Hint: recall that the volume of a sphere is (4/3) r 3 where r is the spheres radius and = 3.14159] Answer: 247 atoms per nanoparticle

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