Presentation on theme: "1 Quantities of Reactants and Products Chapter 4 Antoine Lavoisier 1743-1794. “Father of modern chemistry.” Recognized true elements. Used quantitative."— Presentation transcript:
1 Quantities of Reactants and Products Chapter 4 Antoine Lavoisier “Father of modern chemistry.” Recognized true elements. Used quantitative measurements in chemical reactions.
2 Lavoisier: mass is conserved in a chemical reaction. Chemical equations: descriptions of chemical reactions. Two parts to an equation: reactants and products: Chemical Equations 2H 2 + O 2 2H 2 O ReactantsProduct
3 H2H2 O2O2 H2OH2O
4 Stoichiometric coefficients: numbers in front of the chemical formulas give numbers of molecules or atoms reacting (and numbers being produced).
5 CH 4 + O 2 CO 2 + H 2 O Count atoms: Reactants: Products: 1 C 4 H2 H 2 O 3 O is not balanced. (Why?) Law of Conservation of Mass: All reactions must be balanced
6 Balance reactions only by changing coefficients, not by altering chemical formula
7 Combustion is the burning of a substance in oxygen: C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l)
8 Which is correct? (Blue=A; Red=B) a) A 2 + B A 2 B b) A 2 + 4B 2 AB 2 c) 2A + B 4 2 AB 2 d) A + B 2 AB 2
9 Percentage Composition from Formulas Percent composition is the atomic weight for each element divided by the formula weight of the compound multiplied by 100: Atomic and Molecular Weights
10 Percentage Composition from Formulas FW= (2x1) + (1x32)+ (4x16)= 98 amu FW of O in H 2 SO 4 = 4 x 16 = 64 amu What is % O in H 2 SO 4 (by mass)? %O = 64 x 100 = 65.3% 98
11 The Mole* The “amu” is an “atomic mass unit.” O has a mass of 16 amu – but we can’t weigh out anything in amu If we want to keep the number “16” for the mass of oxygen in some real units (like grams) then we are dealing with a whole bunch of atoms (in 16 g of oxygen). That bunch of atoms is called a mole. Experimentally, 1 mole = 6.02 x things (atoms) This number is called Avogadro’s number. *MSJ Ch 3 pp
12 This amount of P or Ca each contains Avogadro’s number (6.02 x ) of atoms of P and Ca. The mole is defined so that one mole of a substance has a mass equal to its AW or MW in grams Basically, you are replacing amu with grams, e.g. The mass of a P atom is 31 amu. The molar mass of P is 31 grams. The mass of a Ca atom is 40 amu The molar mass of Ca is 40 grams
13 The Mole Experimentally, 1 mole of 12 C has a mass of exactly 12 g. (recall from Ch. 2) Molar Mass Molar mass: mass in grams of 1 mole of substance Units: g/mol or g.mol -1. Mass of 1 mole of 12 C = 12 g exactly
15 This photograph shows one mole of : solid NaCl (58.5 g), liquid H 2 O (18 g), and gaseous N 2 (28 g).
16 Molar Mass Molar mass: sum of the molar masses of the atoms: The Mole What is Molar Mass of H 2 SO 4 ?
17 Interconverting Masses, Moles, and Numbers of Particles The Mole
18 Example 1: 5.00 g of P (a) contains mol of P (b) contains atoms of P Example 2: 5.00 x atoms of C (a) equals mol of C (b) has mass equal to grams.
19 Example 3: 3.5 mol CO 2 : (a) has what mass? (b) contains how many molecules of CO 2 ? (c) contains how many atoms of O?
20 Start with mass % of elements (i.e. empirical data) and calculate a formula. Empirical Formulas from Analyses
21 Empirical Formulas from Analyses* N: 25.9 g x mol = 1.85 mol 14 g Example: compound of N and O Given analysis: N: 25.9%; O: 74.1% Assume 100g; N: 25.9 g; O: 74.1 g Change to mol: Preliminary emp. Formula: N 1.85 O 4.63 Clean it up: divide both by 1.85: Get N 1 O 2.5 ; get rid of fractions, multiply both by 2: Get N 2 O 5 which is the empirical (simplest) formula O: 74.1 g x mol = 4.63 mol 16 g What are some possible molecular formulas? *MSJ Ch 3 pp
22 Molecular Formula from Empirical Formula Once we know the empirical formula, we need the MW to find the molecular formula. Subscripts in the molecular formula are always whole- number multiples of subscripts in the empirical formula. Empirical Formulas from Analyses Example: suppose compound of C and H has empirical formula of C 3 H 8 and a MW = 176 g/mol. What is the molecular formula?
23 Balanced chemical equation gives number of molecules (or moles) that react to form products. Quantitative Information from Balanced Equations Interpretation: balanced equation gives us the ratio of number of moles of reactant to product (or v.v.).These ratios are called stoichiometric ratios. Example: 2 H 2 + O 2 2 H 2 O Molecules: Moles: Ratio of O 2 :H 2 O = 1:2 (either molecules or moles)
24 The ratio of grams of reactant cannot be directly related to the grams of product. Quantitative Information from Balanced Equations
25Stoichiometry Problem: aluminum sulfide + water aluminum hydroxide + hydrogen sulfide (a) Write balanced reaction: (b) How many g aluminum hydroxide obtained from 10.5 g of aluminum sulfide?
26 Stoichiometry Problem: 2 NaN 3 (s) 2Na(s) + 3 N 2 (g) (a) how many mol N 2 produced from 2.50 mol NaN 3? (b) how many g NaN 3 needed to form 6.00 g N 2 (c) how many g NaN 3 needed to produce 10.0 ft 3 of N 2 ? (1.00 ft 3 = 28.3 L; density of N 2 = 1.25 g/L)
27 If the reactants are not present in stoichiometric amounts, at end of reaction some reactants are still present (in excess). Limiting Reactant: one reactant that is consumed. Limiting Reactants O 2 INXS O2O2 H2H2
28 RECIPE: 3 cups flour + 4 eggs + 2 cups sugar cake How many cakes can be made from these amounts? What is the “limiting reactant” (LR) What’s left over and how much of it is left over? On Hand:cups floureggs cups sugar
29 RECIPE: 3 cups flour + 4 eggs + 2 cups sugar cake How many cakes can be made from these amounts? What is the “limiting reactant” (LR) What’s left over and how much of it is left over? On Hand:cups floureggs cups sugar
30 Limiting Reagent (a)Assume a reactant (any one) is LR. Calculate stoichiometric amount of product (any product) formed. (b) Pick another reactant and make it the LR. Calculate the stoichiometric amount of same product formed. (c) Whichever reactant gives the smaller amount of product is the Limiting Reagent
31 Limiting Reagent The LR is always used up in a chemical reaction. Everything else is in excess (INXS).
32 Limiting Reagent (a) which is LR? (b) how many grams NO formed? (c) how much of excess reactant remains? Problem: 4 NH O 2 4 NO + 6 H 2 O 2.25 g NH 3 mixed with 3.75 g O 2 and allowed to react
33 Theoretical Yields Limiting Reactants The amount of product predicted from stoichiometry taking into account limiting reagents is called the theoretical yield. The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield:
34 Theoretical Yield Problem: C 6 H 6 + Br 2 C 6 H 5 Br + HBr (a) theoretical yield of C 6 H 5 Br when 30.0 g of C 6 H 6 reacts with 65.0 g of Br 2 ? (b) if actual yield of C 6 H 5 Br is 56.7 g, calculate %yield