Presentation is loading. Please wait.

Presentation is loading. Please wait.

Stoichiometry Sherlock Holmes, in Sir Arthur Conan Doyles A Study in Scarlet In solving a problem of this sort, the grand thing is to be able to reason.

Similar presentations


Presentation on theme: "Stoichiometry Sherlock Holmes, in Sir Arthur Conan Doyles A Study in Scarlet In solving a problem of this sort, the grand thing is to be able to reason."— Presentation transcript:

1

2 Stoichiometry Sherlock Holmes, in Sir Arthur Conan Doyles A Study in Scarlet 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.

3 The Mole 1 dozen = 1 gross = 1 ream = 1 mole = x There are exactly 12 grams of carbon-12 in one mole of carbon-12.

4 Avogadros Number x is called Avogadros Number in honor of the Italian chemist Amadeo Avogadro ( ). Amadeo Avogadro I didnt discover it. Its just named after me!

5 Calculations with Moles: Converting moles to grams How many grams of lithium are in 3.50 moles of lithium? 3.50 mol Li = g Li 1 mol Li 6.94 g Li 45.1

6 Calculations with Moles: Converting grams to moles How many moles of lithium are in 18.2 grams of lithium? 18.2 g Li = mol Li 6.94 g Li 1 mol Li 2.62

7 Calculations with Moles: Converting grams to moles How many moles of lithium are in 18.2 grams of lithium? 18.2 g Li = mol Li 6.94 g Li 1 mol Li 2.62

8 Calculations with Moles: Using Avogadros Number How many atoms of lithium are in 3.50 moles of lithium? 3.50 mol Li = atoms Li 1 mol Li x atoms Li 2.11 x 10 24

9 Calculations with Moles: Using Avogadros Number How many atoms of lithium are in 18.2 g of lithium? 18.2 g Li = atoms Li 1 mol Li6.022 x atoms Li 1.58 x g Li1 mol Li (18.2)(6.022 x )/6.94

10 Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO g g + 3(16.00 g) = g

11 Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO 3. From previous slide: g g + 3(16.00 g) = g

12 Formulas molecular formula = (empirical formula) n [n = integer] molecular formula = (empirical formula) n [n = integer] molecular formula = C 6 H 6 = (CH) 6 molecular formula = C 6 H 6 = (CH) 6 empirical formula = CH empirical formula = CH 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.

13 Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

14 Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 OC 12 H 22 O 11

15 Empirical Formula Determination 1.Base calculation on 100 grams of compound. 2.Determine moles of each element in 100 grams of compound. 3.Divide each value of moles by the smallest of the values. 4.Multiply each number by an integer to obtain all whole numbers.

16 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?

17 Empirical Formula Determination (part 2) Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

18 Empirical Formula Determination (part 3) Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x Empirical formula: C3H5O2C3H5O2

19 Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C 3 H 5 O 2 3(12.01 g) + 5(1.01) + 2(16.00) = g

20 Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = g 2. Divide the molecular mass by the mass given by the emipirical formula. (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4

21 Review: Chemical Equations Chemical change involves a reorganization of the atoms in one or more substances. C 2 H 5 OH + 3O 2 2CO 2 + 3H 2 O reactantsproducts 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water When the equation is balanced it has quantitative significance:

22 Solving a Stoichiometry Problem 1.Balance the equation. 2.Convert masses to moles. 3.Determine which reactant is limiting. 4.Use moles of limiting reactant and mole ratios to find moles of desired product. 5.Convert from moles to grams.

23 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. Al+O2O2 Al 2 O 3 b. What are the reactants? a. Every reaction needs a yield sign! c. What are the products? d. What are the balanced coefficients? 43 2

24 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 + 3 O 2 2Al 2 O 3 = 6.50 g Al ? g Al 2 O 3 1 mol Al g Al 4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O g Al 2 O x 2 x ÷ ÷ 4 =12.3 g Al 2 O 3

25 Limiting Reactant The limiting reactant is the reactant The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed. that is consumed first, limiting the amounts of products formed. More of limiting reagents

26 % yield = Measured value x100 Calculated value A student reacts 6.50 grams of Aluminum with excess oxygen and measures 11.7 g of Al 2 O 3 is produced. What is the students % yield of aluminum oxide? 11.7 g x 100 = 95.1 % yield 12.3 g From previous calculation calculated value is 12.3 g of Al 2 O 3

27 6.50 grams of aluminum reacts with 2.50 g of Oxygen gas. How many grams of aluminum oxide are formed? Limiting Reagent Example Problem 4 Al + 3 O 2 2Al 2 O g Al 1 mol Al g Al = mol Al 2.50 g O 2 1 mol O g O 2 = mol O 2 Need Have 4Al = = O This 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.

28 = ? g Al 2 O 3 3 mol O 2 2 mol Al 2 O 3 1 mol Al 2 O g Al 2 O 3 4 Al + 3 O 2 2Al 2 O mol O 2 Aluminum is the excess reagent. How would you find the amount of excess aluminum present ?. 5.31


Download ppt "Stoichiometry Sherlock Holmes, in Sir Arthur Conan Doyles A Study in Scarlet In solving a problem of this sort, the grand thing is to be able to reason."

Similar presentations


Ads by Google