INTRODUCTION and CHAPTER ONE General Chemistry 200 INTRODUCTION and CHAPTER ONE Read the Introduction and Chapter 1. Chemistry is NOT a spectator sport. Work out complete solutions for all the bold numbered problems.

Welcome to the World of Chemistry

Beginning Chemistry 200 Read the book before class! Study the examples and do the practice exercises. Be prepared for each class session, lecture and laboratory. Don’t get behind.

Outline for Chapter 1 Definitions Homogenous and Heterogeneous Matter ? 3 states of matter Chemical –vs.- Physical Change Calculations Density Temperature Kelvin  Celsius  Fahrenheit Fahrenheit  Celsius  Kelvin Significant Figures (Sig Figs)

Matter

Mixtures (a)Heterogeneous (b)Suspension (c)Homogenous Cookie Blood Salt water

The Language of Chemistry CHEMICAL ELEMENTS - pure substances that cannot be decomposed by ordinary means to other substances. Aluminum Bromine Sodium

The Language of Chemistry The elements, their names, and their symbols are given on the PERIODIC TABLE How many elements are there?

The Periodic Table Dmitri Mendeleev (1834 - 1907)

Sodium Find sodium, Na, on the chart.

Copper atoms on silica surface. An atom is the smallest particle of an element that has the chemical properties of the element. Copper atoms on silica surface. Find copper on the chart.

The Atom An atom consists of a nucleus (of protons and neutrons) and electrons in space about the nucleus. Electron cloud Nucleus

CHEMICAL COMPOUNDS are composed of atoms and so can be decomposed to those atoms. The red compound is composed of • Ni- Nickel • C- Carbon • O- Oxygen • N- Nitrogen Fixed composition

MOLECULE The smallest unit of a compound that retains the chemical characteristics of the compound. MOLECULAR FORMULA Composition of molecules C8H10N4O2 - Caffeine H2O

The Nature of Matter Mercury Gold Hg Au Chemists are interested in the nature of matter and how this is related to its atoms and molecules.

Graphite layer structure of carbon atoms reflects physical properties.

We can write SYMBOLS to describe these worlds. Chemistry & Matter We can explore the MACROSCOPIC world — what we can see — Understand the PARTICULATE world — we cannot see — We can write SYMBOLS to describe these worlds.

A Chemist’s View Macroscopic Particulate Symbolic- 2 H2(g) + O2 (g) --> 2 H2O(g)

STATES OF MATTER SOLIDS — have rigid shape, fixed volume. External shape can reflect the atomic and molecular arrangement. Reasonably well understood. LIQUIDS — have no fixed shape and may not fill a container completely. Not well understood. GASES — expand to fill their container. Good theoretical understanding.

THE THREE STATES OF MATTER Bromine (gas) Aluminum (solid) Water or H2O (liquid)

KINETIC NATURE OF MATTER Matter consists of atoms and molecules in motion.

Physical Properties What are some physical properties? Color Melting and boiling point Odor Conductivity Density

Physical Changes Some physical changes would be boiling of a liquid melting of a solid dissolving a solid in a liquid to give a homogeneous mixture

DENSITY - an important and useful physical property = mass ( g ) volume cm 3 Gold 19.3 g/cm3 13.6 g/cm3 Mercury

Which is more dense?

Relative Densities of the Elements

Sig Figs Sig Fig PPT

Density = mass ( g ) volume cm 3 Problem: A piece of copper has a mass of 57.54 g. It is 9.36 cm long, 7.23 cm wide, and 0.95 mm thick. Calculate density (g/cm3).

95 mm • 1cm 10 mm 0.095 cm3 . = (9.36 cm)(7.23 cm)(0.095 cm) = 6.4 cm3 SOLUTION 1. Get dimensions in common units. 2. Calculate volume in cubic centimeters. 3. Calculate the density. 95 mm • 1cm 10 mm 0.095 cm3 . = (9.36 cm)(7.23 cm)(0.095 cm) = 6.4 cm3 6.4 cm3 57.54g = 9.0g/cm3

PROBLEM: Mercury (Hg) has a density of 13. 6 g/cm3 PROBLEM: Mercury (Hg) has a density of 13.6 g/cm3. What is the mass of 95 mL of Hg? In grams? In pounds? Solve the problem using DIMENSIONAL ANALYSIS.

PROBLEM: Mercury (Hg) has a density of 13. 6 g/cm3 PROBLEM: Mercury (Hg) has a density of 13.6 g/cm3. What is the mass of 95 mL of Hg? First, note that 1 cm3 = 1 mL Then, use dimensional analysis to calculate mass. 95 cm 3 • 13.6 g = 1.3 x 10 g See next slide What is the mass in pounds?

The milliliter and the cubic centimeter are equivalent The milliliter and the cubic centimeter are equivalent. Notice the units of 10’s. back

PROBLEM: Mercury (Hg) has a density of 13. 6 g/cm3 PROBLEM: Mercury (Hg) has a density of 13.6 g/cm3. What is the mass of 95 mL of Hg? What is the mass in pounds? (1 lb = 454 g) 1.3 x 10 3 g • 1 lb 454 g = 2.8 lb

Density PROBLEM: An object weighing 15.67 g is placed in water starting at 6.8 mL, then displaces water to 20.2 mL. What is the density of the object? 20.2 6.8 20.2mL - 6.8mL = 13.4mL

PROBLEM: An object weighing 15. 67 g is placed in water starting at 6 PROBLEM: An object weighing 15.67 g is placed in water starting at 6.8 mL, then displaces water to 20.2 mL. What is the density of the object? 1 15.67g = 1.17g/cm3 (20.2 - 6.8)ml

Chemical Properties and Chemical Change Burning hydrogen (H2) in oxygen (O2) gives H2O.

Chemical Properties Similar to Physical Properties only with reference to a Chemical reactions Heat and or light produced Color Oder

Chemical Properties and Chemical Change Burning hydrogen (H2) in oxygen (O2) gives H2O. Chemical change or chemical reaction involves the transformation of one or more atoms or molecules into one or more different molecules.

Chemical Change 2 Al + 3 Br2 Al2Br6

Electrolyzing water

UNITS OF MEASUREMENT We make QUALITATIVE observations of reactions — changes in color and physical state. We also make QUANTITATIVE MEASUREMENTS, which involve numbers. Use SI units — based on the metric system

UNITS OF MEASUREMENT Use SI units — based on the metric system length (meter, m) mass (kilogram, kg, and gram, g) time (second)

Units of Length 1 kilometer (km) = ? meters (m) 1 meter (m) = ? centimeters (cm) 1 centimeter (cm) = ? millimeter (mm) 1 nanometer (nm) = 1.0 x 10-9 meter O—H distance = 9.4 x 10-11 m 9.4 x 10-9 cm 0.094 nm 94 pm

Measurement Learn the prefixes in Table 1.4 Other Relationships 1 cm3 = 1 mL = 0.001 L 1.00 lb = 454 g 1.00 in = 2.54 cm 1.06 qt = 1.00 L Significant figures Page 47 Precision and accuracy Page 43 Examples

Temperature Scales Fahrenheit Celsius Kelvin Anders Celsius 1701-1744 Lord Kelvin (William Thomson) 1824-1907

Temperature Scales Notice that 1 Kelvin degree = 1 degree Celsius

Calculations Using Temperature Generally require temp’s in Kelvin T (K) = t (°C) + 273 Body temp = 37 oC + 273 = 310. K Liquid nitrogen = -196 oC + 273 = 77 K

In Class Problems A rectangular box has dimensions of 20.0 cm  15.0 cm  8.00 mm. Calculate the volume of the box in liters. A standard sheet of paper has dimensions of 8.5 inch by 11 inch. A sheet of paper weighs on the average 0.150 g and has a density of 0.710 g/cm3. Calculate the thickness of the paper in cm. A gallon (3.78 L) of latex paint can cover 385 ft2 of the surface of a wall. What is the average thickness of one coat of paint (in micrometers)?

Sample problems Calculate the volume of 525 g of mercury, d=13.534g/cm3. The melting point of tin is 505.5 K. Calculate the Celsius temperature. Find the symbol for gold and the element name for K. An iron sheet is 3.50 cm square and has a mass of 15.396 g. The density of iron is 7.87 g/cm3. Calculate the thickness of the iron sheet in mm. The End!

Dimensional Analysis “English-English” (one conversion) 1) 6 in = ? ft 2) 3.5 gal = ? qt 3.5 gal 4 qt 1 gal = 14 qt

Dimensional Analysis “Metric-Metric” (one conversion) 1) 5.0 cm = ? mm 2) 4.0 dg = ? kg 4.0 dg 1 kg 104 dg = 4.0 x 10-4 kg

Dimensional Analysis “Metric-English” (one conversion) 1) 200.0 cm = ? in 200.0 cm 1.00 in 2.54 cm = 78.74 in 2) 34 qt = ? L 34 qt 1.00 L 1.06 qt = 32 L

Dimensional Analysis “English-English” (two or more conversions) 1) 6 in = ? mile 6 in 1 ft 12 in 1 mile 5280 ft = 9 x 10-5 mile 2) 3.5 gal = ? oz 32 oz 1 qt 3.5 gal 4 qt 1 gal = 450 oz

Dimensional Analysis “Metric-Metric” (two or more conversions) 1) 5.0 cm = ? km 5.0 cm 1 m 100cm 1 km 1000 m = 5.0 x 10-5 km 2) 4 kg = ? pg 103 g 1 kg 1012 pg 1 g 4 kg = 4 x 1015 pg

Dimensional Analysis “Metric-English” (two or more conversions) 1) 200 m = ? in 200 m 100 cm 1 m 1.00 in 2.54 cm = 8000 in 2) 34 qt = ? mL 103 mL 1 L 34 qt 1.00 L 1.06 qt = 3.2 x 104 mL

Dimensional Analysis “Derived Unit Conversions” Area 1) 8.0 ft2 = ? cm2 (2.54)2 cm2 1.00 in2 8.0 ft2 144 in2 1 ft2 = 7400 cm2 2) 2.3 cm2 = ? nm2 2.3 cm2 1014 nm2 1 cm2 = 2.3 x 1014 nm2

Dimensional Analysis “Derived Unit Conversions” Volume 1) 445 dm3 = ? mL 445 dm3 103 cm3 1 dm3 1 mL 1 cm3 = 4.45 x 105mL 2) 5 cm3 = ? mm3 5 cm3 103 mm3 1 cm3 = 5 x 103 mm3

Dimensional Analysis “Other Conversions Problems” 1) 100. km/hr = ? mile/hr 1 mile 5280 ft 105 cm 1 km 1.00 in 2.54 cm 1 ft 12 in 100. km hr = 62.1 mile/hr 2) 25 m/gal = ? nm/qt 109 nm 1 m 1 gal 4 qt 25 m gal = 6.2 x 109 nm/qt

Dimensional Analysis “Other Conversions Problems” (continue) 3) Calculate the density of a material if 45 mL of it has a mass of 128 g. 128 g 45 mL = 2.8 g/mL 4) Calculate the volume in mL of 2.5 g of a material that has a density of 3.65 g/mL. 2.5 g mL 3.65 g = 0.68 mL

Dimensional Analysis “Other Conversions Problems” (continue) 5) Calculate the mass of 20 L of a material that has a density of 8.54 g/mL. 20 L 103 ml 1 L 8.54 g ml = 2 x 105 g 6) How many grams of gold are in 48 g of an alloy that is 22.1% gold? 48 g alloy 22.1 g gold 100.0 g alloy = 11 g gold

Practice Problems 1) 6.45 m = ? cm 2) 12.4 kg = ? mg 3) 184 oz = ? g 4) 24 oz/hr = ? L/day 5) Determine the volume (in L) of 2 kg of sodium chloride. (density = 2.17 g/mL) 6) How many grams of brass contain 50.0 g of zinc? (This brass contains 15% Zn) 1) 645 cm 2) 1.24 x 107 mg 3) 5220 g 4) 17 L/day 5) 0.9 L 6) 330 g Return