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.

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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

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.

I didn’t discover it. Its just named after me!

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

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

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

Calculations with Moles: Using Avogadro’s Number
How many atoms of lithium are in 3.50 moles of lithium? 3.50 mol Li 6.022 x 1023 atoms Li = atoms Li 2.11 x 1024 1 mol Li

Calculations with Moles: Using Avogadro’s Number
How many atoms of lithium are in 18.2 g of lithium? 18.2 g Li 1 mol Li 6.022 x 1023 atoms Li 6.94 g Li 1 mol Li (18.2)(6.022 x 1023)/6.94 = atoms Li 1.58 x 1024

Calculating Formula Mass
Calculate the formula mass of magnesium carbonate, MgCO3. 24.31 g g + 3(16.00 g) = 84.32 g

Calculating Percentage Composition
Calculate the percentage composition of magnesium carbonate, MgCO3. From previous slide: 24.31 g g + 3(16.00 g) = g 100.00

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)6 empirical formula = CH

Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaCl MgCl2 Al2(SO4)3 K2CO3

Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2O C6H12O6 C12H22O11 Empirical: H2O CH2O C12H22O11

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.

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?

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

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 2 x 2 x 2 3 5 2 Empirical formula: C3H5O2

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 C3H5O2 3(12.01 g) + 5(1.01) + 2(16.00) = g

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

Review: Chemical Equations
Chemical change involves a reorganization of the atoms in one or more substances. C2H5OH + 3O2 ® 2CO H2O reactants products When the equation is balanced it has quantitative significance: 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

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.

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. 4 Al + 3 O2 2 Al2O3 a. Every reaction needs a yield sign! b. What are the reactants? c. What are the products? d. What are the balanced coefficients?

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  2Al2O3 6.50 g Al 1 mol Al 2 mol Al2O3 g Al2O3 = ? g Al2O3 26.98 g Al 4 mol Al 1 mol Al2O3 6.50 x 2 x ÷ ÷ 4 = 12.3 g Al2O3

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

% 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 Al2O3 11.7 g x = 95.1 % yield 12.3 g

4 Al + 3 O2  2Al2O3 Limiting Reagent Example Problem Need Have
6.50 grams of aluminum reacts with 2.50 g of Oxygen gas. How many grams of aluminum oxide are formed? 4 Al O2  2Al2O3 Need Have 4Al = = 3.08 3 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. 6.50 g Al 1 mol Al = mol Al 26.98 g Al 2.50 g O2 1 mol O2 = mol O2 31.99 g O2

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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