a-terpinolene Ethyl butanoate 3-Carene Ethyl acetate Ethyl 2-butenoate a-terpinene a-thujene Dimethyl sulfide Limonene b-phellandrene Myrcene p-cymen-8-ol b-caryophyllene cis-3-hexene-1-ol hexadecyl acetate 5-Butyldihydro-3H-2- furanone trans-2-hexenal Ethyl tetradeconaoate a-humulene Sabinene 2-Carene Camphene Ethyl octanoate 4-Isopropenyl-1- methylbenzene 1-Hexanol g-terpinene Hexanal Ethyl hexadecanoate a-copaene Hexadecanal Ethanol Ethyl propionate Dihydro-5-hexyl-3H-2- furanone Carveol Geranial Ethyl decanoate Furfural Butyl acetate Methyl butanoate 2,3, Pentanedione 1,1, diethoxyethane Pentadecanal Butyl formate 1-Butanol 5-Methylfurfural Ethyl dodecanoate 2-Acetylfuran 2 Methyl-1-butanol 4-Methylacetophenoen Acetaldehyde Cyclohexane
Would you eat this? a-terpinolene Ethyl butanoate 3-Carene Ethyl acetate Ethyl 2-butenoate a-terpinene a-thujene Dimethyl sulfide Limonene b-phellandrene Myrcene p-cymen-8-ol b-caryophyllene cis-3-hexene-1-ol hexadecyl acetate 5-Butyldihydro-3H-2- furanone trans-2-hexenal Ethyl tetradeconaoate a-humulene Sabinene 2-Carene Camphene Ethyl octanoate 4-Isopropenyl-1- methylbenzene 1-Hexanol g-terpinene Hexanal Ethyl hexadecanoate a-copaene Hexadecanal Ethanol Ethyl propionate Dihydro-5-hexyl-3H-2- furanone Carveol Geranial Ethyl decanoate Furfural Butyl acetate Methyl butanoate 2,3, Pentanedione 1,1, diethoxyethane Pentadecanal Butyl formate 1-Butanol 5-Methylfurfural Ethyl dodecanoate 2-Acetylfuran 2 Methyl-1-butanol 4-Methylacetophenoen Acetaldehyde Cyclohexane
The Components of Matter 2-5 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Definitions for Components of Matter Element - the simplest type of substance with unique physical and chemical properties. An element consists of only one type of atom. It cannot be broken down into any simpler substances by physical or chemical means. Molecule - a structure that consists of two or more atoms that are chemically bound together and thus behaves as an independent unit. Figure 2.1 2-6
Definitions for Components of Matter Compound - a substance composed of two or more elements which are chemically combined. Figure 2.1 cont d Mixture - a group of two or more elements and/or compounds that are physically intermingled. 2-7
2-8
2-9 Law of Conservation of Matter
Law of Mass Conservation: The total mass of substances does not change during a chemical reaction. reactant 1 + reactant 2 product total mass = total mass calcium oxide + carbon dioxide calcium carbonate (Words) CaO + CO 2 CaCO 3 (Symbols) 56.08 g + 44.01 g 100.09 g (Quantities) 2-10
2-11 Which is the better Vitamin C?
Law of Definite (or Constant) Composition: No matter the source, a particular compound is composed of the same elements in the same parts (fractions) by mass. Calcium carbonate Figure 2.2 Analysis by Mass (grams/20.0 g) 8.0 g calcium 2.4 g carbon 9.6 g oxygen 20.0 g Mass Fraction (parts/1.00 part) 0.40 calcium 0.12 carbon 0.48 oxygen 1.00 part by mass Percent by Mass (parts/100 parts) 40% calcium 12% carbon 48% oxygen 100% by mass 2-12
Sample Problem 2.2 Calculating the Mass of an Element in a Compound PROBLEM: PLAN: Pitchblende is the most commercially important compound of uranium. Analysis shows that 84.2 g of pitchblende contains 71.4 g of uranium, with oxygen as the only other element. How many grams of uranium can be obtained from 102 kg of pitchblende? The mass ratio of uranium/pitchblende is the same no matter the source. We can use the ratio to find the answer. 2-13
Sample Problem 2.2 Calculating the Mass of an Element in a Compound continued SOLUTION: Mass (kg) of uranium = mass (kg) uranium in pitchblende mass (kg) pitchblende x mass (kg) pitchblende 71.4 kg uranium = 102 kg pitchblende x = 86.5 kg uranium 84.2 kg pitchblende 86.5 kg uranium x 1000 g kg = 8.65 x 10 4 g uranium 2-14
2-15 Law of Multiple Proportions
Law of Multiple Proportions: If elements A and B react to form two compounds, the different masses of B that combine with a fixed mass of A can be expressed as a ratio of small whole numbers. Example: Carbon Oxides A & B Carbon Oxide A : 57.1% oxygen and 42.9% carbon Carbon Oxide B : 72.7% oxygen and 27.3% carbon Assume that you have 100 g of each compound. In 100 g of each compound: g O = 57.1 g for oxide A & 72.7 g for oxide B g C = 42.9 g for oxide A & 27.3 g for oxide B Carbon oxide A Carbon oxide B g O g C = 57.1 42.9 = 1.33 g O 72.7 = g C 27.3 2.66 g O/g C in B 1.33 g O/g C in A = 2.66 2 = 1 2-16
Dalton s Atomic Theory The Postulates 1. All matter consists of atoms. 2. Atoms of one element cannot be converted into atoms of another element. 3. Atoms of an element are identical in mass and other properties and are different from atoms of any other element. 4. Compounds result from the chemical combination of a specific ratio of atoms of different elements. 2-17
Dalton s Atomic Theory Mass conservation explains the mass laws Atoms cannot be created or destroyed or converted into other types of atoms. postulate 1 postulate 2 Since every atom has a fixed mass, postulate 3 during a chemical reaction atoms are combined differently and therefore there is no mass change overall. 2-18
Dalton s Atomic Theory Definite composition explains the mass laws Atoms are combined in compounds in specific ratios and each atom has a specific mass. postulate 3 postulate 4 So each element has a fixed fraction of the total mass in a compound. 2-19
Dalton s Atomic Theory Multiple proportions explains the mass laws Atoms of an element have the same mass postulate 3 and atoms are indivisible. postulate 1 So when different numbers of atoms of elements combine, they must do so in ratios of small, whole numbers. Figure 2.3 2-20
Identify the mass law that each of the following demonstrates: (a) A sample of potassium chloride from Chile contains the same percent by mass potassium as one from Poland. (b) A flashbulb contains magnesium and oxygen before use and magnesium oxide afterward, but its mass does not change. (c) Arsenic and oxygen form one compound that is 65.2 mass% arsenic and another that is 75.8 mass% arsenic. 2-21
(a) Law of Definite Composition The compound potassium chloride, KCl, is composed of the same elements and same fraction by mass, regardless of its source (Chile or Poland). (b) Law of Mass Conservation The mass of the substance inside the flashbulb did not change during the chemical reaction (formation of magnesium oxide from magnesium and oxygen). (c) Law of Multiple Proportions Two elements, O and As, can combine to form two different compounds that have different proportions of As present. 2-22
(a) Does the mass percent of each element in a compound depend on the amount of compound? (b)does the mass of each element in a compound depend on the amount of compound? 2-23
a) No The mass percent of each element in a compound is fixed. The percentage of Na in the compound NaCl is 39.34% (22.99 amu/58.44 amu), whether the sample is 0.5000 g or 50.00 g. b) Yes The mass of each element in a compound depends on the mass of the compound. A 0.5000 g sample of NaCl contains 0.1967 g of Na (39.34% of 0.5000 g), whereas a 50.00 g sample of NaCl contains 19.67 g of Na (39.34% of 50.00 g). 2-24
Fluorite, a mineral of calcium, is a compound of the metal and fluorine. Analysis show that a 2.76- g sample of fluorite contains 1.42 g calcium. (a) Calculate the mass of fluorine in the sample. (b)calculate the mass % of each element in fluorite. 2-25
Fluorite is a mineral containing only calcium and fluorine. a) Mass of fluorine = mass of fluorite mass of calcium = 2.76 g 1.42 g = 1.34 g F 1.42 g Ca b) Mass % Ca = x 100 = 51.4% 2.76 g fluorite 1.34 g F Mass % F = x 100 = 48.6% 2.76 g fluorite 2-26
A compound of copper and sulfur contains 88.39 g of metal and 44.61 g of nonmetal. How many grams of copper and sulfur are in 5264 kg of the compound? 2-27
Mass of compound = 88.39 g copper + 44.61 g sulfur = 133.00 g compound Mass of copper = Mass of sulfur = 3 10 g compound 88.39 g copper (5264 kg compound) 1 kg compound 133.00 g compound = 3.49838 x 106 = 3.498 x 106 g copper 3 10 g compound 44.61 g sulfur (5264 kg compound) 1 kg compound 133.00 g compound = 1.76562 x 106 = 1.766 x 106 g sulfur 2-28
Atomic Symbols, Isotopes, Numbers A Z X The Symbol of the Atom or Isotope X = atomic symbol of the element A = mass number; A = Z + N Z = atomic number (the number of protons in the nucleus) N = number of neutrons in the nucleus Isotopes = atoms of an element with the same number of protons, but a different number of neutrons 2-29 Figure 2.8