Chapter 1 Matter and Energy Matter is anything that has mass and occupies space, while mass is a measure of “how much” matter a sample contains. Energy is the capacity to do work. There are many types: kinetic, potential, chemical, electrical, etc. Energy is important since all chemical processes involve a chemical change of some sort. If energy in put into a system to do work, then the process is endothermic, whereas if energy is given off, the process in exothermic. Law of Conservation of Matter: there is no observable change in the quantity of matter during a chemical reaction or a physical change (quantity of matter remains constant). Law of Conservation of Energy: energy cannot be created or destroyed in a chemical reaction or a physical change, instead it is converted to another form. Law of conservation of matter and energy: the total amount of matter and energy in the universe is fixed. States of matter Solids, liquids, gases (more in depth later) – solids have definite shape, not easily compressible; liquids take shape of container, only slightly compressible, flows; gases expand infinitely, completely fill container, compressible. Chemical and Physical properties Chemical properties are observed as substances undergo changes in composition (an element’s chemical properties enable it to react with other substances in specific reactions). Physical properties are observed when there is no chemical reaction or change in composition (melting point, density, boiling point, hardness, etc.). Properties can be further classified as intensive or extensive. Intensive properties do no depend on the amount of substance present (melting point, density) while extensive properties depend on the amount of substance present (volume, mass). Chemical and Physical Changes In a chemical change, one or more substance reacts to form a new substance while absorbing or releasing energy. In a physical change there is no change in chemical composition, however there is a change in state (water freezing) and energy is released or absorbed. Mixtures, Substances, Compounds, Elements Matter can be described as mixtures or pure substances. Substances can be compounds or elements. Elements cannot be broken into other substances by chemical means whereas compounds can be broken to into smaller substances (elements and compounds) by chemical means. Mixtures result when substances combine but retain their characteristic properties (solutions, solid-solid). They can be homogeneous or heterogeneous. Homogeneous mixtures have same composition throughout and the components are indistinguishable, while heterogeneous mixtures do not have same composition throughout and components are distinguishable. Law of definite proportions (Constant Composition) Different samples of any pure compound contain the same elements in the same proportions by mass.

Measurements in Chemistry Unit meter kilogram second ampere Kelvin mole

Quantity length mass time current temperature amt. substance

Sy m b o l m kg s A K mol

Metric System of Measurement Name mega kilo deka deci centi milli micro nano pico femto

Sy m b o l M k da d c m m n p f

Multiplier 106 103 10 10-1 10-2 10-3 10-6 10-9 10-12 10-15

Significant Figures -Nonzero digits are always significant (288 g, 23.5 mL); both have three significant figures -Leading zeroes are never significant; 0.000357 m has three significant figures -Zeroes between nonzero digits are always significant (2.0305 g); five significant figures -Zeroes at the end of a number that contains a decimal point are always significant (38.0 cm has three significant figures); (440.0 mL has four significant figures can be written as 4.400 x 102 mL) -Trailing zeroes at the end of a number may or may not be significant; remove doubt by using scientific notation; (1300 ? 1.3 x 103 has two significant figures; 1.30 x 103 has three etc….) -In addition and subtraction, the last digit retained in the sum or the difference is determined by the position of the first doubtful digit; 23.4 g + 12.26 g = 35.66 g is reported as 36.7 g. -In multiplication and division, the answer contains no more significant figures than the least number of significant figures used in the operation; 3.2 cm x 2.15 cm = 6.88 cm2 , should be reported as 6.9 cm 2.

Dimensional Analysis (Unit Factor Method) The unit must always accompany the numeric value of a measurement!!!! Unit factors can be constructed from any two terms that describe the same or equivalent “amounts” of the item in question.

Example: 1 ft = 12 in. Dividing both sides by 1 ft gives 1ft/1ft = 12 in./1 ft or 1= 12 in./1 ft Other common unit factors are: 1yd/3 ft, 1yd/36 in., 1 mi/5280 ft, 4 qt/1 gal, 2000 lb/1 ton Problem 1: Express 2.3 mi in inches. ? inches = 2.3 miles Choose the appropriate unit factors to convert the given units miles

feet

inches

? in. = 2.3 mi x (5280ft/mi) x (12 in./ft) = 1.46 x 105 in.

Example 2: Unit Conversions: The Angstrom (Å) is a unit of length equal to 1 x 10-10 m, typically used to express atomic radii. Convert 1.10 Å to cm and nm. ? cm = 1.10 Å x (1 x 10-10 m/1 Å) x (1 cm/1 x 10-2) m = 1.10 x 10-8 cm ? nm = 1.10 Å x (1 x 10-10 m/1 Å) x (1 nm/1 x 10-9) m = 1.10 x 10-1 nm

Example 3: Mass Conversion: A sample of gold (Au) has a mass of 0.234 mg. What is the mass in g? kg? Use the relationship 1 g = 1000 mg and 1000 g = 1 kg. ? g = 0.234 mg x(1 g/1000 mg) = 2.34 x 10-4 g ? kg = 2.34 x 10-4 g x (1 kg/1000 g) = 2.34 x 10-7 kg Example 4: Area is two dimensional thus units must be in squared terms Express 2.61 x 104 cm2 in ft2. ? ft 2 = 2.61 × 10 4 cm 2 (

1in 2 1ft 2 ) ( ) 2.54cm 12in

= 28.0938061 9 ft 2 ≈ 28.1 ft 2 Example 5: Percentage Percentage is the parts per hundred of a sample

A 335 g sample of ore yields 29.5 g of iron. What is the percent of iron in the ore? ? % iron = =

grams of iron x 100% grams of ore 29.5 g Fe x 100% 335 g ore

= 8.81% Density and Specific Gravity Density = mass/volume or D = m/V Densities may be used to distinguish between two substances or in identifying a particular substance. Expressed in g/cm3 or g/mL for liquids and solids and g/L for gases. Example 6: Density, mass, volume: A 47.3-mL sample of ethanol has a mass of 37.32 g. What is its density? Use D = m/V = 37.32g/47.3 mL = 0.789 g/mL Example 7: Density, mass, volume: If 102 g of ethanol is needed for a chemical reaction, what volume of liquid would you use? D = m/V, so V = m/D = 102 g/0.789 g/mL = 129 mL Example 8: Density, Specific Gravity The specific gravity (Sp. Gr.) of a substance is the ratio of its density to the density of water, at a particular temperature. Sp. Gr. = Dsubstance/Dwater What is the specific gravity of table salt at 20 oC if its density is 2.16 g/mL? Sp. Gr. = Dsalt/Dwater = 2.16 g/mL/1.00 g/mL = 2.16 Example 9: Specific Gravity, volume, percentage by mass Battery Acid is 40% sulfuric acid, H2SO4 and 60% water by mass. Its specific gravity is 1.31. Calculate the mass of pure H2SO4 in 100.00 mL of battery acid. Using the Sp. Gr., we get the density: D = 1.31 g/mL Since the solution is 40% H2SO4 by mass we can write the following unit factor: 40.0 g H2SO4/100 g soln. Solving, we get: ? H2SO4 = 100.00 mL soln x(1.31 g/1 mL soln) x (40.0 g H2SO4/100 g soln) = 52.4 g H2SO4 Heat and Temperature Temperature measures the “hotness” or “coldness” of a body while heat is a form of energy that flows from a hotter object to a cooler object. Temperature scales (Kelvin (K), Celsius (oC), Fahrenheit (oF)):

Converting from Kelvin to oCelsius or vice versa is straightforward or ? oC = K – 273.15o ? K = oC + 273.15o Converting from Celsius to Fahrenheit is not as trivial 180 oF/100 oC or 1.8 oF/1.0 oC and 100 oC/180 oF or 1.0 oC/1.8 oC ? oF = (x oC x (1.8 oF/1.0 oC)) + 32 oF and ? oC = 1.0 oC/1.8 oC (x oF - 32 oF) Specific Heat The amount of heat required to raise the temperature of one gram of the substance one degree Celsius or Kelvin with any phase change Specific heat = amt. of heat in joules (J)/(mass of substance in g)(temp. change in oC); the units are J.g-1.o C-1 Heat capacity of a body is the amount of heat required to raise its temperature1oC. The heat capacity of a body is its mass in grams times its specific heat; the units are J /oC or J.oC-1 Example 10: Specific heat How much heat, in joules is required to raise the temperature of 205g of water from 21.2 oC to 91.4oC? Specific heat = amt. of heat in joules (J)/(mass of substance in g)(temp. change in oC) Rearrange the equation to get, amt. of heat = (mass of substance in g)(temp. change in oC)(Specific heat) amt. of heat = (205 g)(70.2 oC)(4.18 J/g.oC) = 6.02 x 104 J Problems: 7, 11, 13, 15, 18, 19, 21, 22, 23, 24, 27, 29, 32, 38, 45, 49, 57, 61, 62