Chapter 2 Stoichiometry describes the quantitative relationships among elements


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1 Chapter 2 Stoichiometry describes the quantitative relationships among elements 1. in compounds (composition stoichiometry) 2. in chemical changes (reaction stoichiometry) Dalton s Atomic Theory summarized experimental observations and interpretations in the nature of atoms: 1. an element is composed of extremely small, indivisible particles called atoms 2. Atoms cannot be created, destroyed, or transformed into atoms of another element 3. compounds are formed when atoms of different elements combine with each other in small, wholenumber ratios. 4. the relative numbers and kinds of atoms are constant in a given compound. 1
2 Chemical formula the chemical composition of a substance. Some examples of chemical formulas for elements: Monoatomic elements sodium (Na), copper (Cu), barium (Ba) Diatomic elements N 2, O 2, F 2, Cl 2, Br 2, I 2, H 2 More complex molecules S 8, P 4 For compounds, the formula indicates 1. elements present 2. the ratio in which the atoms of the elements occur Some examples: HCl 1 H atom, 1 Cl atom H 2 O 2 H atoms, 1 O atom NH 31 N atoms, 3 H atoms C 3 H 8 3 C atoms, 8 H atoms 1 chemical formula 1 chemical formula 1 chemical formula 1 chemical formula 2
3 molecular compounds: (Table 2.2) H 2 O 2 hydrogen peroxide NH 3 ammonia CH 3 CH 2 OH methanol CO 2 carbon dioxide Methane CH 4 benzene C 6 H 6 Ion an atom or group of atoms that carries an electrical charge Anion negatively charged ions (Cl , O 2 ) Cation positively charged ions (Na +, Ba 2+ ) Ionic compounds extended array of ions in which the total positive and negative charges are equal (Fig 2.7) Formula unit the simplest whole number ratio of ions in the compound (NaCl) Is PO 4 3 a molecule? No. polyatomic anion. Compound names and formulas (table 23) NaCl sodium chloride NH 4 NO 3 ammonium nitrate Sulfuric acid H 2 SO 4 Na 2 SO 4 sodium sulfate Molecule? ion 3
4 How much does a single atom weigh? Different elements weigh different amounts related to what makes them unique. What units do we use to define the weight of an atom? amu units of atomic weight. (atomic mass unit) 1 amu 1/12 the mass of 1 carbon12 atom. H 1 amu N 14 amu Mg 24.3 amu Ag amu When scientists do experiments with substances (elements, molecules, ionic compounds), they usually  Use the chemical formula to plan (ratio of atoms, NaCl)  Measure the amount needed in grams How can we relate grams to the chemical formula? Mole x atoms, molecules or formula units Avogadro s number **one mole of atoms of an element has a mass in grams numerically equal to the atomic weight of the element** 1 H atom 1 amu 1 mol H 1.01 g 1 C atom 12 amu 1 mol C g 4
5 Why do we use such a weird number? Because it gives us an easy relationship between amus and grams. How many atoms are contained in 1.67 moles of Mg? How many grams does 1.67 moles of Mg weigh? = 1.00 x atoms = 40.6 grams The ratio of atoms in the simplest formula for a compound is the same as the ratio of moles of atoms of the elements in a sample of the compound 1 H 2 O molecule 2 H atoms 1 O atom 1 mol H 2 O 2 mol H 1 mol O How many molecules are there in one mole of propane? x molecules In one mole of chlorine (Cl 2 ), what is the number of a) Cl 2 molecules? x molecules b) Cl atoms? 2(6.022 x molecules) 5
6 In one mole of Calcium nitrate, what is the number of a) formula units? x b) Ca 2+ cations? x c) Nitrate anions? 2(6.022 x ) How much does H 2 O weigh? Formula weights the sum of the atomic weights of the atoms in the formula. Propane C 3 H 8 3(12.01) + 8(1.00) = Chlorine Cl 2 2(35.45) = Calcium nitrate Ca(NO 3 ) [ (16.00)] = formula weight all compounds, molecules and elements atomic weight atoms molecular weight molecules Molar mass the mass of one mole of a substance. (g/mol) Numerically equivalent to formula weight (amu/formula unit) 6
7 Calculate the number of C 3 H 8 molecules in 74.6 g propane. 7
8 Percent composition the % of each element in the compound (by mass) ex. calculate the percent composition of C and H in C 3 H 8. Molar mass of C 3 H 8 = C 3 (12.01 g/mol) = (g/mol) H 8 (1.008 g/mol) = (g/mol) (g/mol) % carbon: = (g/mol) x 100 = 81.71% (g/mol) % hydrogen = (g/mol) x 100 = 18.29% (g/mol) empirical formula the smallest, whole number ratio of elements present NaCl H 2 O H 2 O 2  HO Molecular formula actual number of atoms present in a molecule 8
9 If we have the % composition of a substance, we can determine the empirical formula Ex % K g K 34.76% Mn g Mn 40.50% O g O % 100 g substance g K x 1 mol K = mol K g g Mn x 1 mol Mn = mol Mn g g O x 1 mol O = mol O g divide through by smallest # moles to get the ratio: mol K/ = mol Mn/ = mol O/ = 4 empirical formula is KMnO 4 but we still don t know the molecular formula unless we also have the molar mass of the substance ex. in previous example, if molar mass = 158 g/mol empirical formula = molecular formula 9
10 if a compound has the empirical formula HO and its molar mass = 34.0 g/mol HO = = 17.0 g/mol 34/17 = 2 2(HO) = H 2 O 2 molecular formula Chapter 2: Composition stoichiometry the relative ratios of different elements within one particular compound or molecule Chapter 3: Reaction stoichiometry the relative ratios between different substances as they react with each other 10
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