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Dr. Chirie Sumanasekera

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1 Dr. Chirie Sumanasekera
Chapter 3: Stoichiometry 3.5 Conversion Factors from: Chemical formulas Balanced equations Concentration, volume & density AP Chemistry _ Notes Dr. Chirie Sumanasekera 10 /19/ 2017

2 3.5. Conversion factors Conversion factors from Chemical formulas A formula of an ionic compound is the empirical formula representing the simplest ratio of atoms (Ex: concentrated acids and bases such as, HCl, H2SO4, NaOH) A formula of a molecular compound indicates the numbers and types of atoms that make up a given molecule A chemical formula re[presents the ratio of atoms within the formula These formulas allow a chemist to form relationships within the formula as a whole and relationships between individual atoms in the formula

3 Ex: Formula of Sucrose (Table sugar) = C12H22O11
Moles Atoms 12 mol of C 12 atoms of C 11 mol of O 11 atoms of O 22 mol of H 1 molecule of C12H22O11 has 22 atoms of H 12 mol of C combining with 22 mol of H 12 atoms of C combining with 22 atoms of H 12 mol of C combining with 11 mol of O 12 atoms of C combining with 11 atoms of O 22 mol of H combining with 11 mol of O 22 atoms of H combining with 11 atoms of O The chemical relationships in C12H22O11 help us construct 6 different relationships each on a atomic and moles-basis for Stoichiometry-calculations. These are not TRUE-equalities and are only chemical-equalities.

4 In class Problems: 1. Determine the molar mass of the following compounds and round your answer to 2- decimal places. Cd(NO3)2 CH3(CH2)4Br (NH4)2SO4 (CH3CH2CH2)2O CuSO4 . 5H2O 2. How many different chemical-equivelances may be written for the following formulas? NaCl FeCl3 NiSO4 (NH4)3PO4 O3

5 3.5.2. Conversion from Balanced equations
The chemical formula only provides relationships between conversion factors within a single compound. However, a balanced chemical equations provides a group of relationships that can be used to generate additional conversion factors between compounds in that equation. EX. Write the balanced equation for the Combustion of benzene (C6H6): C6H6 + O  CO H2O 2C6H O2  CO H2O From this equation, on a mole-basis we get the following equivalences: 2 mol of C6H6 16 mol of O2 6 mol of H2O 12 mol of CO2 15 mol of O2 6 mol of H2O 12 mol of CO2 6 relationships 12 mol of CO2 12 mol of CO2

6 3.5.3. Concentration, volume & density conversion factors
I. MOLARITY (M) = the number of moles of solute dissolved in 1L (liter) of solute. Molarity is the most common concentration unit in chemistry. Ex. If a problem states the concentration of a solute as molar (0.250 M) NaOH, this value can be transformed in to conversion factors as follows: 0.250 M NaOH Inverse 0.250 mol NaOH 1L NaOH = mol NaOH 1000ml NaOH = 1000ml NaOH

7 3.5.3. Concentration, volume & density conversion factors
II. Volume = for a gas, 1 mol =22.4L of gas at STP [standard Temperature (0oC) and Pressure (1 atm)]. Ex. If a problem states that o.5 moles of N2 gas was produced in a reaction and the volume of gas has to be calculated, we have to find the volume at STP 1 mol of N2 gas = 22.4L of N2 gas at STP Volume of 0.5mol of N2 gas 0.5 mol N2 X L N2 1 mol N2 = L of N2 Remember also: (1) 1cm3 = 1 ml. So, 1000ml = 1000 cm3 (2) 1cm3 of H2O = 1 g of H2O. So, 1000 cm3 = 1000 cm3 of H2O = 1000g =1Kg

8 3.5.3. Concentration, volume & density conversion factors
III. Density = mass volume If you are given a density of a substance, along with either the mass or volume, you must be able to find the missing part of the above equation and then plug that into a stoichiometry calculation. Ex: If an organic liquid has a density of 0.741g cm-3 you can write the following conversion factors density Inverse 0.741g cm3

9 3.5.4. Conversion sequence of steps
Grams of substance A Grams of substance B Liters of solution A Liters of solution B Atomic mass Molar mass Atomic mass Molar mass Molarity Molarity chemical formula or Moles of substance A Moles of substance B 22.4L/mol Balanced chemical equation 22.4L/mol 6.02 x 1023 particles/mol 6.02 x 1023 particles/mol Liters of gas A at TSP Liters of gas B at TSP Units of substance A Units of substance B

10 3.6. Problems

11 3.6.1 Calculations involving one substance
1. How many grams of FeCl3 (molar mass =162.2g/mol) must be weighed to have mol of FeCl3? 1mol of FeCl3 = g of FeCl3 162.2 g of FeCl3 1mol of FeCl3 ?g of FeCl3 = mol FeCl3 162.2 g of FeCl3 1mol of FeCl3 = 74.0 g FeCl3

12 3.6.1 Calculations involving one substance
2. A sample contains 24.6g of CaO. (Molar mass of CaO = 56.08g/mol) How many moles of CaO are in the sample? 1 mol CaO 56.08g CaO ? mol of CaO in 24.6 g = 24.6 g CaO = mol CaO

13 3.6.1 Calculations involving one substance
3. A solution has a molarity of mol/L MgBr2. How many moles of MgBr2 are in 0.4L of this solution? 0.658 M MgBr2 = mol MgBr2 1 L MgBr2 0.658 mol MgBr2 1 L MgBr2 ? mol of MgBr2 = L MgBr2 = mol MgBr2

14 3.6.1 Calculations involving one substance
4. The same solution (0.658 M) of MgBr2 in Q-3 must be used to obtain 0.5mol of MgBr2. How many milliliters of solution are needed? 1 L MgBr2 0.658 mol MgBr2 ? mL MgBr2 = mol MgBr2 1000 mL MgBr2 0.658 mol MgBr2 ? mL MgBr2 = mol MgBr2 = 760 ml MgBr2

15 3.6.1 Calculations involving one substance
5. How many grams of KCl (molar mass = 75.55g/mol) are there in 0.250L of a 0.3 molar solution of KCl? 0.3 mol KCl 1 L KCl 75.55g KCl 1 mol KCl ? g KCl = L KCl = 5.59 g KCl

16 3.6.1 Calculations involving one substance
6. How many moles of nitrogen are there in 6.50 mol of Ammonium phosphate, (NH4)3PO4? 3 mol of N in 1 mol of (NH4)3PO4 3 mol N 1 mol (NH4)3PO4 ? mol N = 6.50 mol (NH4)3PO4 = 19.5 mol of N

17 3.6.2 Calculations involving more than one substance
7. How many moles of water will be formed in the complete combustion of 2.50 mol of methane, CH4? CH4 + 2O2  CO2 + 2H2O 1 mol CH4  2 mol of H2O 2 mol H2O 1 mol CH4 ? mol H2O = 2.5 mol CH4 = 5.0 mol of H2O

18 3.6.2 Calculations involving more than one substance
8. Propane, C3H8, is a common heating and cooking fuel in rural areas of USA. It is also a fuel used in outdoor grills. (a) If 100 g of propane is burned with excess oxygen, how many grams of oxygen will be needed? (b) In addition, how many grams of carbon dioxide and water will be formed in the reaction? (a) Balanced equation: C3H8 + 5O2  3CO2 + 4H2O 1 mol C3H8  5 mol O2 C3H8 molar mass = 44.1 g O2 molar mass = 18 g 5 mol O2 1 mol C3H8 32 g O2 1 mol of O2 1 mol C3H8 44.1 g C3H8 ? g of O2 = 100 g C3H8 = 363 g O2

19 3.6.2 Calculations involving more than one substance
8. (b)In addition, how many grams of carbon dioxide and water will be formed in the reaction? 1 mol C3H8  3 mol CO2  4 mol H2O CO2 molar mass = g H2O molar mass = g Balanced equation: C3H8 + 5O2  3CO2 + 4H2O ? g of CO2 = 100 g C3H8 1 mol C3H8 44.1 g C3H8 3 mol CO2 1 mol C3H8 43.98 g CO2 1 mol CO2 = g of CO2 ? g of H2O = 100 g C3H8 1 mol C3H8 44.1 g C3H8 4 mol H2O 1 mol C3H8 18.02 g H2O 1 mol H2O = 163 g of H2O

20 3.6.2 Calculations involving more than one substance
9. A 45.0 ml Sample of molar FeCl3 is reacted with enough NaOH solution to precipitate all the iron as Fe(OH)3. How many grams of Fe(OH)3 will be precipitated?

21 3.6.2 Calculations involving more than one substance
(b)In addition, how many grams of carbon dioxide and water will be formed in the reaction? 1 mol C3H8  3 mol CO2  4 mol H2O CO2 molar mass = g H2O molar mass = g Balanced equation: C3H8 + 5O2  3CO2 + 4H2O ? g of CO2 = 100 g C3H8 1 mol C3H8 44.1 g C3H8 3 mol CO2 1 mol C3H8 43.98 g CO2 1 mol CO2 = 29.0 g of CO2 ? g of H2O = 100 g C3H8 1 mol C3H8 44.1 g C3H8 4 mol H2O 1 mol C3H8 18.02 g H2O 1 mol H2O = 29.0 g of CO2

22 3.6.3. Limiting reactant calculations
When chemicals are mixed together under the right conditions, chemical reactions occur. The reaction will stop when one of the reactants is completely used up. Then, the other reactant(s) will be in excess. The amount of the limiting reactant determines how much of the other reactants are used up and how much product is formed. Product Excess reactant Limiting reactant * In the AP exam you will need to know how to draw diagrams like this one!

23 3.6.3 Calculations involving Limiting –Reactants
1. For the reaction below, determine the limiting reactant if 100g of FeCl3 is reacted with 50 g of H2S. 2 FeCl3(aq) + 3H2S (g) Fe 2S3 (s)+ 6HCl (aq) FeCl3= 162.2g HCl = 36.45g H2S = 34.08g Fe2S3 =207.9g Step-1: Molar masses: Step-2: Convert the amount given for one reactant, in to the amount of another reactant needed: 1 mol H2S 34.08 g H2S 2 mol FeCl3 3 mol H2S 162.2 g FeCl3 1 mol of FeCl3 ? g of FeCl3 = 50 g H2S = 159 g FeCl3 Step-3: Analyze. 159g of FeCl3 is needed to react with 50g of H2S. And we are given only 100g of FeCl3. So, the limiting reactant is: FeCl3.

24 3.6.4 Calculations Theoretical Yield
Theoretical yield refers to the maximum amount of product formed in a reaction based on the amounts of reactants used. It is Theoretical as no experiment is done to determine the yield. In practice, the experimental yield may differ from the theoretical yield as all the reactants may not combine to form the product(s). Or, because the reactants produce different products than that was expected. Products can also be lost due to poor purification techniques. 2FeCl3(aq) + 3H2S (g) 1Fe 2S3 (s)+ 6HCl (aq) 1. What is the theoretical yield of Fe2S3 that can be obtained from 100g of FeCl3 and 50 g of H2S?. FeCl3= 162.2g HCl = 36.45g H2S = 34.08g Fe2S3 =207.9g Step-1: Molar masses: 1 mol FeCl3 162.2g FeCl3 1 mol Fe2S3 2 mol FeCl3 207.9 g Fe2S3 1 mol of Fe2S3 ? g of Fe2S3 = 100g FeCl3 = 64.1g Fe2S3

25 Theoretical Yield (continued.) 10-26-17
2FeCl3(aq) + 3H2S (g) 1Fe 2S3 (s)+ 6HCl (aq) 2. When 100g of FeCl3 is reacted with 50 g of H2S when the reaction is complete? FeCl3= 162.2g HCl = 36.45g H2S = 34.08g Fe2S3 =207.9g Molar masses: * We know from the previous problem that FeCl3 Is the limiting reactant, then all other reactants are in excess! To determine how much H2S is left over, first calculate the grams of H2S that is left over: 1 mol FeCl3 162.2g FeCl3 3 mol H2S 2 mol FeCl3 34.08 g H2S 1 mol of H2S ? g of H2S = 100g FeCl3 = 31.5g H2S Starting grams of H2S = 50 g Remaining grams = 50 – 31.5 g = 18.5g

26 2FeCl3(aq) + 3H2S (g) 1Fe 2S3 (s)+ 6HCl (aq)
Theoretical Yield 2FeCl3(aq) + 3H2S (g) 1Fe 2S3 (s)+ 6HCl (aq) 3. When 100g of FeCl3 is reacted with 50 g of H2S how many grams of HCl are formed? FeCl3= 162.2g HCl = 36.45g H2S = 34.08g Fe2S3 =207.9g Molar masses: * To find the amount of a product made, you NEED to know the limiting reactant, first. Snice we already know it is FeCl3 we can proceed: 1 mol FeCl3 162.2g FeCl3 6 mol HCl 2 mol FeCl3 36.45 g HCl 1 mol of HCl ? g of HCl = 100g FeCl3 = g HCl

27 All stoichiometry calculations obey the law of mass conservation.
Tip: All stoichiometry calculations obey the law of mass conservation.

28 Dr. Chirie Sumanasekera
Chapter 3: Stoichiometry Theoretical yield: Titrations AP Chemistry _ Notes Dr. Chirie Sumanasekera 10 /30/ 2017

29 Theoretical Yield (10-30-17)
4. Silver tarnishes in Air because of a complex reaction with oxygen and hydrogen sulfide, H2S, in the air. The equation for this reaction: What is the Theoretical yield of Silver sulfide, Ag2S that can be produced from a mixture of g of silver, 1.50 L of oxygen at STP, and 65.0ml of M H2S? 4 Ag+ O H2S  2 Ag2S H2O Ag = g Ag2S = g O2 = 34.08g H2O = 207.9g Molar masses: H2S = 34.08g Steps for solving problem: Find the limiting reactant: ? L of O2 that reacts with 0.2g Ag ?ml of H2S that reacts with 0.2g Ag ? g of Ag2S using the limiting reactant 3 ways for solving problem: How much Ag2S is formed with the g of Ag used in the rxn much Ag2S is formed with the ml of 0.35mol H2S used in the rxn How much Ag2S is formed with the L of O2 used in the rxn

30 So, both O2 and H2S are in excess and Ag is the limiting agent
Theoretical Yield 4 Ag + O H2S  2 Ag2S H2O Ag = g Ag2S = g O2 = 34.08g H2O = 207.9g Molar masses: H2S = 34.08g 1. ? L of O2 that reacts with 0.2g Ag 1 mol Ag 107.9 g Ag 1 mol O2 4 mol Ag 22.4 L O2 1 mol O2 ? L of O2 = 0.2g Ag = g O2 (we have L) 2. ? ml of H2S that reacts with 0.2g Ag 1 mol Ag 107.9 g Ag 2 mol H2S 4 mol Ag 100 ml H2S 0.35 mol H2S ? ml of H2S = 0.2g Ag = ml H2S (we have 65 ml) So, both O2 and H2S are in excess and Ag is the limiting agent

31 Theoretical Yield 1 mol Ag 107.9 g Ag 2 mol Ag2S 4 mol Ag 247.8 g Ag2S
4 Ag + O H2S  2 Ag2S H2O Ag = g Ag2S = g O2 = 34.08g H2O = 207.9g Molar masses: H2S = 34.08g 3. ? g of Ag2S that reacts with 0.2g Ag 1 mol Ag 107.9 g Ag 2 mol Ag2S 4 mol Ag 247.8 g Ag2S 1 mol Ag2S ? g of Ag2S = 0.2g Ag = g Ag2S

32 Titrations A titration is a technique where a solution of known concentration is used to determine the concentration of an unknown solution Typically, the titrant (the know solution) is added from a buret to a known quantity of the analyte (the unknown solution) is placed in a beaker until the reaction is complete The buret can be adjusted with the stopcock to gradually drop controlled amounts of the titrant into the beaker to detect the exact end point of the reaction. An indicator is added to the beaker so that a color change occurs when the reaction is complete Knowing the volume of titrant added allows the determination of the concentration of the unknown (Total amount– amount left in buret)

33 Titrations

34 5 Fe3+ + MnO4- + 8H+  Mn2+ + 5 Fe3+ + 4H2O
Titrations 5 Fe MnO H+  Mn Fe H2O It takes ml of M KMnO4 solution (0.240 M MnO4-) to titrate an unknown amount of Fe2+ to its end point. How many grams of Fe2+ are in the sample? Fe = g Molar masses: ? g of Fe3+ that reacts with ml of M MnO4- 0.24 mol MnO4- 1000 ml MnO4- 5 mol Fe3+ 1 mol MnO4- 55.85 g Fe3+ 1 mol Fe3+ ? g of Fe3+ = ml MnO4- = g Fe3+

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