Unit V: Chemical Quantities

Information in Chemical Equations As we have seen in the last unit, chemistry is about reactions Reactions are described by equations that give the identities of the reactants and products The amount of each is shown by the coefficients

Example: Making a sandwich – Ratio 2:3:1 Using the equation in a reaction permits us to determine the amounts of reactants needed give a particular amount of product The equation would look like this… → 1 The #’s are all coefficients that would be placed in front of the elements and or compounds

Mol-Mol Relationships We can use an equation to predict the mols of products that a given number of mols of reactants will yield Example: 2 H 2 O (l) → 2 H 2 (g) + O 2 (g) If we decompose 4 mol of water, how many mol of product do we get? 4 mol H 2 O mol H = 4 mol H 2

Using the same equation, suppose we were to decompose 5.8 mol of water Equation: 2 H 2 O (l) → 2 H 2 (g) + O 2 (g) The answer is to use mol ratios Examples: 2 mol H 2 O = 2 mol H 2 (1:1 ratio) 2 mol H 2 O = 1 mol O 2 (2:1 ratio) Set up a mol ratio: 1 mol O 2 = 2 mol H 2 O 5.8 mol H 2 O (1 mol O 2 = 2 mol H 2 O) = 2.9 mol O 2 Final Answer: 5.8 H 2 O (l) → 5.8 H 2 (g) O 2 (g)

mol – mol Conversions

Sample Problems: 1A. Iron (II) oxide reacts with oxygen to form iron (III) oxide Equation: 4 FeO (s) + O 2 (g) → 2 Fe 2 O 3 (s) 2.4 mol FeO → ? mol O mol FeO mol O = 6.0 x mol O 2

Equation: 4 FeO (s) + O 2 (g) → 2 Fe 2 O 3 (s) 1B. What number of moles of iron (III) oxide will be produced by reacting 9.2 mol of iron (II) oxide with excess oxygen? 9.2 mol FeO → ? Fe 2 O mol FeO mol Fe = 4.6 x 10 0 mol Fe 2 O 3

Mass Calculations We know that moles represent the number of molecules We can not count molecules directly We count by weighing Grams to grams stoichiometry

Consider the following problem: C 3 H 8 (g)+ 5 O 2 (g)→ 3 CO 2 (g) + 4 H 2 0 (g) 2A. What mass of oxygen will be required to react exactly with 54.1 g of propane? 54.1 g C 3 H 8 → ? g O 2 (grams → grams) ( ) / ( ) = Round to 3 SF’s = 196 → SSN = 1.96 x 10 2 Add units of measure = 54.1g C 3 H 8 mol C 3 H 8 mol O 2 g O x 10 2 g O 2

Consider the following problem: C 3 H 8 (g)+ 5 O 2 (g)→ 3 CO 2 (g) + 4 H 2 0 (g) 2B. What mass of water will be required to react exactly with 24.2 g of oxygen gas? 24.2 g O 2 → ? g H 2 O (grams → grams) ( ) / (32 5 1) = Round to 3 SF’s = 10.9 → SSN = 1.09 x 10 1 Add units of measure = 24.2g O 2 mol O 2 mol H 2 O g H 2 O x 10 1 g H 2 O

Consider the following problem: C 3 H 8 (g)+ 5 O 2 (g)→ 3 CO 2 (g) + 4 H 2 0 (g) 2C. How many L of CO 2 will be required to react exactly with 7.5 x atoms of O 2 gas? 7.5 x atoms O 2 → ? L CO 2 (4 steps) (7.5 x ) / (6.022 x ) = Round to 2 SF’s = 170 → SSN = 1.7 x 10 2 Add units of measure = 7.5 x atoms O 2 mol O 2 mol CO 2 L CO x x 10 2 L CO 2

Mass Calculations: Comparing Two Reactions Question: Which antacid can consume the most stomach acid, 1.00 g of NaHCO 3 or 1.00 g of Mg(OH) 2 ? Remember – moles are used to compare! Reactions: A. NaHCO 3 (s)+HCl (aq)→ NaCl(aq) + H 2 O(l) + CO 2 (g) B. Mg(OH) 2 (s) + 2 HCl (aq)→ 2 H 2 O (l) + MgCl 2 (aq) Problems: (3 steps: g → mol) A g NaHCO 3 → ? mol HCl A g Mg(OH) 2 → ? mol HCl

The Concept of Limiting Reactants Earlier in the unit, we discussed making sandwiches… 2 pieces of bread + 3 slices of meat + 1 slice of cheese → 1 sandwich Suppose you came to work and found the following: 20 slices of bread 24 slices of meat 12 slices of cheese

Example: Available Tools Sets of Tools Extra Tools

Calculations Involving a Limiting Reactant 3. What mass of water is required exactly to react exactly with 249 g of methane? 249 g CH 4 → ? g H 2 O (4 steps) Balanced Equation: CH 4 (g) + H 2 0 (g) → 3 H 2 (g) + CO (g) 249 g CH 4 mol CH 4 mol H 2 O g H 2 O

Problem: 249 g CH 4 → ? g H 2 O Equation: CH 4 (g) + H 2 0 (g) → 3 H 2 (g) + CO (g) Calculation:( ) / (16.05) = Round to 3 SF’s = 280. g H 2 O Normally we would place the answer in SSN Question: If 249 g of CH 4 were mixed with 300 g of water… Remember: 249 g CH 4 → 280 g H 2 O The CH 4 will be consumed before the H 2 O runs out The CH 4 is the limiting reactant H 2 O will be in excess

Percent Yield In the previous sections, we learned many aspects about chemical reactions Essentially, products stop running out when one reactant runs out The amount calculated in this way is called the theoretical yield It is the amount of product predicted from the amounts of reactant used

The actual yield of the product (which is actually obtained) is called the percent yield AY TY% Actual Yield (g) Theoretical Yield (g) What actually occurs during the experiment Based on Stoichiometry Percent Yield (g) Expressed as a decimal

Sample Problem 4: if 124 g of Iron is produced from 222 g of Iron (III) oxide, what is the A) Theoretical Yield, B) Percent Yield? Equation: Iron (III) oxide decomposes… 2 Fe 2 O 3 (s) → 4 Fe (s) + 3 O 2 (g) Actual Yield = 124 g Fe Have to convert 222 g Fe 2 O 3 to g Fe Grams to grams problem (4 steps) This will be the Theoretical Yield

( ) / ( ) = Round to 3 SF’s = 155 → SSN = 1.55 x 10 2 Add units of measure = % Yield = (AY / TY) 100 = (124 g Fe / 155 g Fe) 100 = 80% Yield 222 g Fe 2 O 3 mol Fe 2 O 3 mol Fe g Fe x 10 2 g Fe