Unit III: Quantitative Composition of Compounds A. Atoms and Isotopes B. Atomic Composition of Chemical Compounds C. Formula and Molecular Mass D. Calculations using Moles of Atoms E. Calculations using Moles of Compounds F. Mass Percent Composition of Compounds G. Empirical and Molecular Formulas
3-A The Structure of an Atom e e n p p n n p p n e e
3-A What is an atom? An atom is a fundamental piece of matter, composed of subatomic particles. An single atom is the smallest amount of an element that maintains properties of that element. Atoms are not created, changed, or destroyed during a chemical reaction
3-A The Structure of an Atom Proton (+1) e pn n p p n e p Neutron (0) 1.673 x 10-24 g e n 1.674 x 10-24 g Electron (-1) e 9.109 x 10-28 g
3-A Atomic Mass Protons and neutrons are regarded as having one atomic mass unit (1 AMU) Electrons have a mass of zero AMU The total number of protons and neutrons in an atom is called the atomic mass
3-A Isotopes While all atoms of the same element behave similarly in chemical reactions, not all atoms are identical Atoms always have the same number of protons, but may have different masses.
3-A Isotopes of hydrogen N P N e 12 3 1 H Atomic mass = 12 3 Hydrogen Deuterium Tritium
3-A Isotopes Mass number Atomic number A Z E Element symbol 30 Zn The isotope zinc-65 zinc 65.38 65 30 Zn
3-A Practice with isotopes Name the following isotopes 13 6 C carbon-13 15 7 N nitrogen-15 How many protons are in each of the following isotopes? 25 12 Mg 30 15 P
3-A Fill in the gaps in the following table assuming each column represents a neutral atom: Isotopic Symbol Number of Protons 34 Number of Neutrons 28 42 Number of Electrons Mass Number 95 23 241
3-A Kinds of isotopes Stable and radio isotopes Oxygen for example has isotopes with between 4 and 16 neutrons. Only those atoms with 8, 9, or 10 neutrons are stable.
Atomic mass vs. Atomic weight Atomic Mass refers to the mass of a particular isotope of a particular element Atomic weight refers to the weighted average mass of all the isotopes of an element. (you might also see this as relative atomic mass)
3-B Atomic Composition of Chemical Compounds To calculate formula weights we need to know the number of atoms of each element in a compound. Example: Hexanoic acid CH 3 (CH 2 ) 3 CH 2 COOH 6 carbon atoms 2 oxygen atoms 12 hydrogen atoms
3-C UNIT 3-C: Formula and Molecular weights Molecular weight refers to covalent compounds Formula weight refers to the weight of an ionic compound s formula unit.
3-C UNIT 3-C: Lecture Problem 1 (pg 95) A. Find the formula weight of barium nitrate. B. Find the molecular weight of diphosphorus pentaoxide
3-D Unit 3-D: Calculations Using Moles of Atoms When we talk about practical amounts of elements or compound we are talking about huge numbers. We need a convenient term that describes a large group of things (similar to a dozen or gross). We use the term: mole
3-D Moles A mole of anything always contains the same number, 6.022 x 10 23 This special number is called Avogadro s number. It just happens that there are 6.022 x 10 23 atoms of carbon-12 in exactly 12 grams of carbon-12.
3-D Moles and masses The atomic mass of C-12 is 12 amu If the atomic mass of He-4 (for example) is 4 amu.. (1/3 the mass of C-12).and there are 12 g of C-12 in one mole of C- 12.. Then there will be 4 g of He-4 in one mole of He-4
3-D Lecture Problem 3-2 A. What is the atomic weight of zinc (in amu)? B. How many grams are in a mol of zinc? C. How many atoms are present in a mol of zinc? Page 97
3-D Lecture Problem 3-3 A. How many grams are in 2.20 mol of iron? B. Calculate the number of mol in 2.45 x 10-5 g of calcium. Pg. 97
Unit 3 Part E: Calculations using Moles of Compounds
3-E Calculating molecular weights and molar masses If we add up the atomic weights of all the atoms in a compound we get the molecular weight. (units of amu) If we want to talk about the mass of 1 mol of a compound we replace the units of amu with the units g/mol.
3-E Lecture Problem III-4 (pg. 99) What is the molar mass of ammonium sulfide? (NH 4 ) 2 S Molar mass of (NH 4 ) 2 S is 68.17 g/mol
3-E Unit factors The concept of the mol is very useful in constructing unit factors.. Using the example of glucose (C 6 H 12 O 6 ) With a molar mass of 180.18g/mol Write as many unit factors as possible utilizing the mol concept.
3-E Lecture Problem 3-5 (pg. 101) A. How many mol of hydrogen atoms are in 83.5 g of C 6 H 12 O 6? B. How many water molecules are in 0.946 mol of water? Extra practice: How many total atoms are in 55.3 g of sulfur dioxide?
3-E Using mols to describe composition of compounds Lecture problem III-6 (pg. 102) A sample of ammonium dichromate contains 3.07 mol of nitrogen atoms. What is the mass of the sample of ammonium dichromate? (NH 4 ) 2 Cr 2 O 7 has a molar mass of 252.10 g/mol 386.97 g of (NH 4 ) 2 Cr 2 O 7 rounded to 387 g
3-E Lecture problem III-7 (pg. 103) How many grams of bromine are present in a sample of Ca(BrO 3 ) 2 that is found to contain 10.0 g of calcium? Ca has a molar mass of 40.08 g/mol and Br has a molar mass of 79.90 g/mol 39.87 g of Br rounded to 39.9 g
Unit 3-F: Mass Percent Composition 1. Calculate the formula or molecular weight of the compound. 2. Determine the total weight of each element in the compound. % composition of an element = Total weight of that element x 100 formula or molecular weight
Unit 3-F: Mass Percent Composition Lecture problem III-8 (pg. 105). Calculate the percent by mass of calcium in calcium phosphate. Ca 3 (PO 4 ) 2 molar mass of 310.2 g/mol 38.76% Ca by mass in CaPO 4
Unit 3-G: Empirical and Molecular Formulas The empirical formula of a compound represents the smallest ration of atoms that are present in a compound. The formula for an ionic compound is in essence an empirical formula The molecular formula is the true formula, representing the number of atoms of each element that are present in a single molecule of a compound
Unit 3-G: Empirical and Molecular Formulas Consider dinitrogen tetraoxide Based on the compound name we can write its molecular formula: N 2 O 4 The empirical formula describes the smallest ratio of atoms that are present in a compound. There is 1 nitrogen atom for every 2 oxygen atoms So the smallest ratio, or empirical formula) is NO 2
Unit 3-G: Empirical and Molecular Formulas Given the percent composition of an unknown compound, how can we determine its identity? (sometimes actual masses for each element may be given) The first step is to turn the mass percentage into a ratio of atoms in the compound, called an empirical formula.
Solving Empirical Formula Problems 1.If starting with % composition assume exactly 100 grams of sample. Then percentages become mass quantities. (If starting with mass, skip to step 2) 2.Calculate the number of mol of each element present using our unit analysis method. 3.Find the simplest mol ratio by dividing all of the mol quantities by the smallest number of mol of an element present.
Lecture problems III-9 (page 108-109) A. Calculate the empirical formula of a compound that contains 56.4% phosphorus and 43.6% oxygen B. A compound contains 1.79g of phosphorus and 2.29g of oxygen. What is the empirical formula of the compound? C. It is found that 2.231 grams of selenium react with fluorine to form a selenium fluoride with a mass of 5.452 grams. What is the empirical formula of this compound?
Finding the Molecular Formula Find the empirical formula for the compound. Calculate the empirical molar mass. Divide the molecular molar mass by the empirical molar mass. This is our multiplying factor Multiply the quantity of each element in the empirical formula by this factor. The result is the molecular formula.
Molecular Formula Determination Lecture problem (III-10 pg. 111): A. What is the molecular formula for galactose if the empirical formula is CH 2 O and the molar mass is 180.2 g/mol? B. A molecular compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. A 0.100 mol sample of this compound weighs 6.00 grams. What is the molecular formula of the compound?
Extra practice with mystery metal problems 0.125 mol of chlorine gas react with 0.0625 mol of a metal to form 16.3 grams of a binary ionic compound. What is the metal, and what is the formula for the compound? 5.11 grams of chlorine gas reacts with 0.0480 mol of a metal to form 14.56 grams of a new solid compound with the formula MCl 3. What is the metal (symbol)?