Presentation on theme: "Stoichiometry “In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy."— Presentation transcript:
1 Stoichiometry“In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.”Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet
2 The Mole 1 dozen = 12 1 gross = 144 1 ream = 500 1 mole = 6.022 x 1023 There are exactly 12 grams of carbon-12 in one mole of carbon-12.
3 I didn’t discover it. Its just named after me! Avogadro’s Number6.022 x 1023 is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro ( ).I didn’t discover it. Its just named after me!Amadeo Avogadro
4 Calculations with Moles: Converting moles to grams How many grams of lithium are in 3.50 moles oflithium?3.50 mol Li6.94 g Li= g Li45.11 mol Li
5 Calculations with Moles: Converting grams to moles How many moles of lithium are in 18.2 grams oflithium?18.2 g Li1 mol Li= mol Li2.626.94 g Li
6 Calculations with Moles: Converting grams to moles How many moles of lithium are in 18.2 grams oflithium?18.2 g Li1 mol Li= mol Li2.626.94 g Li
7 Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 3.50 moles oflithium?3.50 mol Li6.022 x 1023 atoms Li= atoms Li2.11 x 10241 mol Li
8 Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 18.2 g oflithium?18.2 g Li1 mol Li6.022 x 1023 atoms Li6.94 g Li1 mol Li(18.2)(6.022 x 1023)/6.94= atoms Li1.58 x 1024
9 Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO3.24.31 g g + 3(16.00 g) =84.32 g
10 Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO3.From previous slide:24.31 g g + 3(16.00 g) = g100.00
11 Formulas molecular formula = (empirical formula)n [n = integer] Empirical formula: the lowest whole number ratio of atoms in a compound.Molecular formula: the true number of atoms of each element in the formula of a compound.molecular formula = (empirical formula)n [n = integer]molecular formula = C6H6 = (CH)6empirical formula = CH
12 Formulas (continued)Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio).Examples:NaClMgCl2Al2(SO4)3K2CO3
13 Formulas (continued)Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio).Molecular:H2OC6H12O6C12H22O11Empirical:H2OCH2OC12H22O11
14 Empirical Formula Determination Base calculation on 100 grams of compound.Determine moles of each element in 100 grams of compound.Divide each value of moles by the smallest of the values.Multiply each number by an integer to obtain all whole numbers.
15 Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?
16 Empirical Formula Determination (part 2) Divide each value of moles by the smallest of the values.Carbon:Hydrogen:Oxygen:
17 Empirical Formula Determination (part 3) Multiply each number by an integer to obtain all whole numbers.Carbon: 1.50Hydrogen: 2.50Oxygen: 1.00x 2x 2x 2352Empirical formula:C3H5O2
18 Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?1. Find the formula mass of C3H5O23(12.01 g) + 5(1.01) + 2(16.00) = g
19 Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?2. Divide the molecular mass by the mass given by the emipirical formula.3(12.01 g) + 5(1.01) + 2(16.00) = g(C3H5O2) x 2 =C6H10O4
20 Review: Chemical Equations Chemical change involves a reorganization ofthe atoms in one or more substances.C2H5OH + 3O2 ® 2CO H2OreactantsproductsWhen the equation is balanced it has quantitative significance:1 mole of ethanol reacts with 3 moles of oxygento produce 2 moles of carbon dioxide and 3 moles of water
21 Solving a Stoichiometry Problem Balance the equation.Convert masses to moles.Determine which reactant is limiting.Use moles of limiting reactant and mole ratios to find moles of desired product.Convert from moles to grams.
22 Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.1. Identify reactants and products and write the balanced equation.4Al+3O22Al2O3a. Every reaction needs a yield sign!b. What are the reactants?c. What are the products?d. What are the balanced coefficients?
23 Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?4 Al O2 2Al2O36.50 g Al1 mol Al2 mol Al2O3g Al2O3=? g Al2O326.98 g Al4 mol Al1 mol Al2O36.50 x 2 x ÷ ÷ 4 =12.3 g Al2O3
24 Limiting Reactant The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed.More of limiting reagents
25 % yield = Measured value x100 Calculated value A student reacts 6.50 grams of Aluminum with excess oxygen and measures 11.7 g of Al2O3 is produced. What is the students % yield of aluminum oxide?From previous calculation calculated value is 12.3 g of Al2O311.7 g x = 95.1 % yield12.3 g
26 4 Al + 3 O2 2Al2O3 Limiting Reagent Example Problem Need Have 6.50 grams of aluminum reacts with 2.50 g ofOxygen gas. How many grams of aluminum oxide are formed?4 Al O2 2Al2O3Need Have4Al = = 3.083 OThis shows we have excess Al and Oxygen is the limiting reagent. We use the moles of oxygen (limiting reagent) to find the mass of aluminum oxide that can be formed.6.50 g Al1 mol Al= mol Al26.98 g Al2.50 g O21 mol O2= mol O231.99 g O2
27 4 Al + 3 O2 2Al2O3 = 0.078 mol O2 2 mol Al2O3 101.96 g Al2O3 5.31 Aluminum is the excess reagent.How would you find the amount of excess aluminum present ?.
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