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STOICHIOMETRY Review Questions: 1) How many H atoms are there in 45.0 g (NH4)2Cr2O7? Given: MMN = 14.01; MMH = 1.008; MMCr = 52.00; MMO = 16.00 A) 2.15×1023 atoms B) 6.02×1023 atoms C) 7.52×1023 atoms D) 8.60×1023 atoms 2) How many moles of calcium are required to produce 5.50 g of calcium phosphate? The molar mass of calcium phosphate is 310.18 g/mol A) 1.77×10–2 mol B) 5.32×10–2 mol C) 6.28×10–2 mol D) 4.07×10–2 mol 3) Combustion analysis of a 12.0–mg sample of ibuprofen produced 9.50 mg of water. What is the percent hydrogen in ibuprofen? A) 8.87% B) 17.7% C) 4.44% D) 79.2% Answers to the above questions: 1 d); 2 b); 3 a)

CHEMISTRY’S ACCOUNTING Reaction stoichiometry allows us to use balanced equations to convert/determine the following: Mass ↔ Moles ↔ Molecules ↔ Atoms What do balanced equations tell you? e.g., 3H2 + N2 → 2NH3 • • •

molecules? moles? masses?

•

atoms?

1

Example 1: What mass of I2 is produced if 13.1 g KI is reacted with excess KIO3 and HNO3 according to the following balanced equation? (Ans: 12.0 g) –1 –1 Given: MM(KI) = 166.0 g mol and MM(I2) = 253.8 g mol Rxn: 5KI(aq) + KIO3(aq) + 6HNO3(aq) → 6KNO3(aq) + 3I2(aq) + 3H2O(l) Strategy: 1) Convert mass of KI to moles of KI – using the molar mass of KI. 2) Convert moles of KI to moles of I2 – using the stoichiometry of the balanced equation. 3) Convert moles of I2 to mass of I2 – using the molar mass of I2.

Example 2:

PCl3 is prepared by a two step reaction: a. 4HCl(aq) + MnO2(s) → MnCl2(aq) + 2H2O(l) + Cl2(g) b. P4(s) + 6Cl2(g) → 4PCl3(s) How many moles of PCl3 will be prepared from 0.2743 moles HCl and excess MnO2 and P4? (Ans: 0.04572 moles)

Strategy: Convert moles of HCl to moles of PCl3 using the stoichiometry of the balanced equations: a. moles of HCl to moles of Cl2 (an intermediate which connects the two reactions) b. moles of Cl2 to PCl3

In general, unless we are TOLD that the other reagents are in EXCESS, we must NEVER assume it. This leads to the so–called LIMITING REAGENT (or REACTANT) problem. 2

Techniques for dealing with the Limited Reagent(or Reactant) problem are straightforward. (a) Convert all reagent MASSES to MOLES (b) To decide which of the reagents is LIMITING, simply divide the number of moles of each reagent by its coefficient in the BALANCED equation. (c) INSPECT the numbers so obtained; the smallest (or smaller if only two) value indicates the LIMITING REAGENT. The moles of products will depend ONLY on the moles of this reactant – all the others being in EXCESS. Example: Na2S2O3 is used in developing photographs. It can be synthesized by the following reaction: Na2CO3 + 2Na2S + 4SO2 → 3Na2S2O3 + CO2 Calculate the maximum mass of Na2S2O3 (MM =158.1 g/mole) that can be made by reacting the following: Reactants Mass (g) Molecular Mass(g/mol) Na2CO3 6.00 106.0 Na2S 5.00 78.0 SO2 4.00 64.1 The first step is to recognize that we have a Limiting Reagent problem. Next we have to decide which reagent is the Limiting Reagent. DECISION process:

mass → moles → number of times the reaction can proceed.

Reactants

MM (g/mol)

Mass

Na2CO3

106.0.

6.00

Na2S

78.0

5.00

SO2

64.1

4.00

Limiting Reagent:

Moles

Rxn Coefficient

# of Rxns Possible

Excess Reagents:

Now finish off the problem:

(Answer: Because SO2 is the limiting reagent, only 0.740 g Na2S2O3 is formed.) 3

ALTERNATE METHOD: Reactants

MM (g/mol)

Mass

Na2CO3

106.0.

6.00

Na2S

78.0

5.00

SO2

64.1

4.00

Moles

Mole Ratio

Moles of Na2S2O3 Produced

3 mol Na2S2O3 requires 1 mol Na2CO3

REACTION YIELDS – 4 types to be aware of: 1. Theoretical yield: what we PREDICT or CALCULATE we will obtain. 2. Actual yield: what we obtain by doing the EXPERIMENT.

3. Percentage yield: % yield =

4. FRACTIONAL yield =

Actual Yield × 100% Theoretica l Yield

Actual Yield Theoretica l Yield

Chapter 4: Chemical Reactions Types of Chemical Reactions (refer to page 133 in text) 1. Precipitation – mixing of two ionic substances produces a solid ionic substance (a precipitate). e.g., compare the following two reactions: BaCl2(aq) + K2SO4(aq) → 2KCl(aq) + BaSO4(s) BaCl2(aq) + KNO3(aq) → no reaction 2. Acid–Base Reactions – an “acid” reacts with a “base”. Such reactions involve proton transfer. 3. Redox Reactions (Oxidation and Reduction) – reactions that involve the transfer of electrons. (This topic will be dealt with thoroughly in CHEM*1050 when we look at electrochemistry.) 4

AQUEOUS SOLUTIONS There are 3 types of solutions: 1. STRONG electrolyte solution – solute exists entirely as ions. e.g., 2. WEAK electrolyte solution – solute is only partially ionized. e.g., 3. NON–electrolyte sol’n – has no ions at all. e.g., Can you predict the electrical conductivity if each solution?

SOLUBILITY and PRECIPITATION REACTIONS Based on scientific observations, the follow rules allow us to do qualitative predictions: •

If a substance has a solubility < 0.01 mol/L, it is considered insoluble.

•

If a substance has a solubility > 0.1 mol/L, it is considered soluble.

•

If a substance has a solubility of 0.1 – 0.01 mol/L, it is considered sparingly soluble. (Note in passing the number of times the letter a occurs in the word soluble!)

SOLUBILITY RULES (Refer to Table 4.1 p. 128 1. All compounds of group IA elements (i.e., 2. All ammonium (i.e.,

in text or Expt#2 p. 2-6 in lab manual)

) are soluble.

) salts are soluble.

3. All NO3–, ClO3–, ClO4–, acetate (i.e.,

), salts are soluble.

4. Most Cl–, Br– and I– salts are soluble EXCEPT Ag+, Pb2+, and Hg(I) [i.e., INSOLUBLE as well as Hg(II) bromide and Hg(II) iodide (e.g.,

] salts are ).

5. Most SO42– salts are soluble EXCEPT Ag2SO4, PbSO4, Hg2SO4, as well as CaSO4, SrSO4, BaSO4. 6. The following form insoluble salts: • Carbonates (CO32–) • Phosphates (PO43–) • Oxalates (C2O42–) • Chromates (CrO42–) EXCEPT those of Group 1 elements and NH4+ are soluble – Rules 1 and 2. 7. Hydroxides (OH–), Oxides (O2–) and Sulphides (S2–) are insoluble, EXCEPT those of Group 1 (Rule 1) and salts from Group 2 elements (e.g., Ca2+, Sr2+ and Ba2+).

NOTE: You need to KNOW the above rules for the midterm! 5

Example 1: Use the solubility rules to predict the solubility of the following compounds:

a) AgBr b) Ag2S c) Ag2CO3 d) AgNO3 e) AgClO4 f) AgI g) KNO3 h) CuCl2 i) Hg2Br2 j) BaS k) SrCO3 m) Pb(ClO4)2

Molecular Equations – a chemical equation in which the reactants and products are written as if they were molecular substances, even though they may actually exist in solution as ions.

Complete Ionic Equations – a chemical equation in which strong electrolytes are written as separate ions in solution. Net Ionic Equation – an ionic equation from which spectator ions have been cancelled

Example 2: Predict whether or not a precipitate would result from the following reactions. Give relevant net–ionic equations.

Hg2(ClO4)2 + NaBr →

Fe(ClO4)3 + NaOH → 6

Pb(NO3)2 + H2SO4 →

CaCl2 + Na2CO3 →

ACID–BASE REACTIONS Definitions: ARRHENIUS ¾ Acid supplies _____________ ¾ Base supplies _____________

There are ONLY 6 STRONG ACIDS: Acids Hydrochloric HCl Hydrobromic HBr hydroiodic HI nitric HNO3 Perchloric HClO4 sulphuric* H2SO4

Anions chloride Cl– bromide Br – iodide I– nitrate NO3– perchlorate ClO4– hydrogen sulphate HSO4–

There are ONLY 8 STRONG BASES (Alkali Metal and Alkaline Earth Hydroxides): LiOH NaOH KOH

Ca(OH)2

RbOH

Sr(OH)2

CsOH

….. Ba(OH)2

Note: Strong bases can also be formed by the reaction of alkali metal and alkaline earth oxides with water,

e.g.,

Na2O(s) + H2O(l) → 2NaOH(aq) CaO(s) + H2O(l) → Ca(OH)2(aq)

. 7

Neutralization: reaction of an acid and a base to yield a salt and water. Typical acid–base neutralization reaction: HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) acid

base

Complete Ionic:

water

H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) → H2O(l) + Na+(aq) + Cl–(aq) H+(aq) + OH–(aq) → H2O(l)

Net ionic equation (NIE):

Example 1:

"salt"

HNO3(aq) + Ba(OH)2 (aq) →

Balanced equation: Complete Ionic:

Net ionic equation: Example 2:

HF + KOH →

Balanced equation: Complete Ionic:

Net ionic equation:

Brønsted–Lowry Definitions: Acid is a PROTON ______________ Base is a PROTON ____________________ Examples of Weak Acids:

HF(aq) + H2O(l) ↔ H3O+(aq) + F–(aq)

CH3COOH(aq) + H2O(l) ↔ H3O+(aq) + CH3COO–(aq)

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Examples of Weak Bases:

NH3(aq) + H2O(l) ↔ NH4+(aq) + OH–(aq) CH3NH2(aq) + H2O(l) ↔ CH3NH3+(aq) + OH–(aq)

Neutralization reactions:

CH3COOH(aq) + NaOH(aq) ↔ H2O(l) + CH3COONa(aq) NH3(aq) + HBr(aq) ↔ NH4Br(aq)

Acid–Base Reactions with Gas Formation (pg. 142) Certain salts (such as carbonates, sulphites and sulphides) react with acids to form a gas. e.g., Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2CO3(aq) BUT

H2CO3(aq) → H2O(l) + CO2(g)

∴ Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)

OR

NIE: CO32– (aq) + 2H+(aq) → H2O(l) + CO2(g)

MOLARITY (section 4.7, pg. 155) Many reactions take place in solution.

We need to define concentration units.

Molarity (M) =

amount of solute n volume of solution = V

Units: MOLES PER LITRE (mol/L or mol L–1) (Also can use mol/dm3 because 1 L = 1 dm3) Main thing to note: Volume (L) × conc (moles/L) = moles

mass For gases:

↔

PV = nRT

moles OR

↔

molecules

n = PV/RT

where R = 8.314 J/(K·mol) = 8.314 kPa·dm3/(K·mol) = 0.0821 atmL/(K·mol) (Refer to Chapter 5 for further details.) 9

DILUTING SOLUTIONS (section 4.8, pg. 157) Dilution is much easier and cheaper to do than weighing! Types of situations:

(a)

Provide a precise recipe to dilute a solution of known concentration to make a more dilute solution with a precise concentration.

(b)

Given the dilution information (or recipe), calculate the exact concentration of molecules or ions present in the dilute solution.

Example for case (a):

Supply a fool–proof recipe for making 5.000 L of 0.01234 M NaOH solution. You have access to a solution of NaOH in water which was made by adding 45.67 g NaOH to a 2.000 L volumetric flask and making it up to the 2.000 L mark with water. [Ans = 108.1 mL] THINK: 1) How many moles of NaOH do you need in the 5.000 L sample? 2) What volume of the concentrated solution is needed to obtain the correct amount of moles of NaOH?

Recipe: To a 5.000 L volumetric flask, add ________ mL of the concentrated NaOH solution and enough water to bring the volume up to the mark on the volumetric flask. Seal and shake well. Example for case (b):

What are the molar concentrations of the strontium and nitrate ions in a solution prepared by adding 25.00 mL of 0.2346 M strontium nitrate to 275.0 mL of pure water? Ans: [Sr2+] = 0.01955 M; [NO3–] = 0.03910 M THINK:

1) How many moles of strontium nitrate were transferred? 2) What is the new volume? 3) What is the new concentration of the dilute solution? 4) What is the concentration of the ions present in that solution? 10

Solution Stoichiometry Example 1: What volume (in mL) of 10.0 M HCl solution is needed to prepare 2.00 g Cl2 based on the

following reaction and assuming excess MnO2: 4 HCl(aq) + MnO2(s) → MnCl2(aq) + Cl2(g) + 2H2O(l) (a) If the reaction is 100% efficient? [Ans: 11.3 mL] (b) If the reaction is only 31% efficient? [Ans: 36 mL]

Example 2: In a titration, 25.00 mL of barium hydroxide solution were placed in the flask and the concentration of HCl added was 0.1234 mol L–1. The volume of HCl needed to react completely (decolourise the phenolphthalein indicator solution) was 24.68 mL. Calculate the concentration of the barium hydroxide solution. [Ans: = 6.091×10–2 mol L–1]

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Example 3: Assuming that calcium carbonate precipitates completely, calculate the mass of calcium carbonate formed by the reaction of 200.0 mL of 0.01246 mol L–1 CaCl2 solution with 50.00 mL of 0.07223 mol L–1 Na2CO3 solution, according to the equation: CaCl2(aq) + Na2CO3(aq) → CaCO3(s) + 2 NaCl(aq) [Ans: 0.2494 g] (Question: Is this a limiting reactant problem? How does one tell?)

Example 4: A 100.0 mL sample of drinking water containing chlorine (Cl2) is treated with excess KI in acidic solution, and the liberated iodine (I2) is titrated against 2.468 mmol L–1 Na2S2O3 solution. An amount of 10.23 mL of Na2S2O3 was needed to react with all the I2. Calculate the concentration of chlorine in the water, in moles per litre. [Ans: 1.262 x 10–4 mol L–1] – – Given: Cl2(aq) + 2I (aq) → 2Cl (aq) + I2(aq) – – – I2(aq) + 2S2O32 (aq) → 2I (aq) + S4O62 (aq)

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BACK TITRATION Example: A 0.1609 g sample of limestone (impure calcium carbonate) is dissolved in 50.00 mL of 0.07543 mol L–1 hydrochloric acid. The HCl left over after the reaction is titrated against 0.02222 mol L– 1 NaOH solution, of which 26.89 mL are needed to reach the end point. Calculate the percent of CaCO3 in the limestone sample. [Ans: 98.76%] Strategy: 1. Write balance equations for the reactions that take place. 2. Determine the total amount of HCl added and the amount left over [= n(NaOH)] 3. Calculate the number of moles of CaCO3 present in the sample. 4. Determine the mass of CaCO3 and the percent of CaCO3 in the sample.

Further Practice Questions: 1.

The number of H atoms in 45.0 g (NH4)3AsO3 is?

[Ans: 1.84×1024]

2.

The number of molecules of in 454 mg DDT (C14H9Cl5) is?

[Ans: 7.71×1020]

3.

A sample containing Ba, Cl and O weighs 0.850 g. On heating it loses all its oxygen and the residue weighs 582 mg. A reaction of ALL this residue with an excess of silver nitrate solution yields 0.800 g AgCl. What is the empirical formula of the sample? Can you suggest a possible name for the compound? [Ans: Barium chlorate]

4.

How would you prepare 75.0 mL of 0.350 M KCl solution from 0.575 M KCl(aq)?

5.

The volume of 0.300 M HCl solution which would be required to react completely with 24.6 g of sodium carbonate is? [Ans: 1.55 L]

[Ans: 45.7 mL]

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6.

Consider the reactions:

Fe2(CO3)3 + 6HCl → 2FeCl3 + 3CO2 + 3H2O CO 2 + Ca(OH)2 → CaCO3 + H2O The mass in grams of CaCO3 produced if 2.90 g Fe2(CO3)3 reacts with excess HCl and the resultant CO2 is bubbled through a Ca(OH)2 solution is? [Ans: 2.98 g]

7. A compound referred to by its mnemonic NDMA is thought to be carcinogenic, and has been detected in trace amounts in some water supplies in a nearby community. Analysis shows that the compound has 32.43% C, 8.163% H, 37.81% N and 21.60% O. The approximate molar mass for NDMA is 7×101. Determine: (a) the simplest molecular formula, (b) the true molecular formula, and [Ans a): C2H6N2O; c) 74.07 g/mol] (c) the precise molar mass for NDMA. 8. Quicklime, CaO(s), is prepared from limestone by roasting it. The reaction is: CaCO3(s) → CaO(s) + CO2(g). What mass (in t) of quicklime can be prepared from 1.000 Mg of limestone that contains 94.6% CaCO3? [Ans: 0.530 t ] 9*.

A 7.78 g sample of a mixture of NaCl and NaHCO3 is heated to 300°C. The NaHCO3 decomposes as shown, but the NaCl is stable (i.e., does not react): 2NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) After decomposition, the sample weighs 5.93 g. Calculate the percent by mass of NaCl in the original mixture. *[Ans: This question is much more challenging and involves solving simultaneous equations; 35.2% ]

10.

Nitric acid is produced commercially by the Ostwald process, represented by the following reactions: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) 2NO(g) + O2(g) → 2NO2(g) 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) How many moles of NH3 are needed to prepare 4 mole of HNO3? [Ans: 6 mol]

11.

Potassium perchlorate, KClO4, can be prepared by the following sequence of reactions: Cl2 + 2KOH → KCl + KClO + H2O 3KClO → 2KCl + KClO3 4KClO3 → 3KClO4 + KCl a) How many moles of Cl2 are needed to prepare 2 mole of KClO4? b) What mass of Cl2 (in g) is needed to prepare 1.00 g of KClO4 by this sequence of reactions?

[Ans: 8 mol] [Ans: 2.05 g ]

12.

The copper in a 2.00 g mineral sample was determined quantitatively by the following reactions: 2Cu2+ + 4I– → 2CuI + I2 I2 + 2S2O32– → S4O62– + 2I– If 31.4 mL of 0.0500 M Na2S2O3 solution was required for the titration, calculate the percent by mass of Cu in the mineral. [ Ans: 4.99%]

13.

Give net ionic equations for the following reactions: AgNO3(aq) + Na2S(aq) → HClO3(aq) + KOH(aq) → AgClO4(aq) + (NH4)2SO4(aq) →

All in One Problem: When aluminum metal is added to sulphuric acid, hydrogen gas and aluminum sulphate is produced.

What volume of hydrogen gas will evolve if we use 15.00 mL of a 0.250M solution of sulphuric acid and add 0.0712 g of aluminum? Assume that this reaction is only 90.0% efficient and that the experiment was performed at 1.00 atm and 25.0°C. (MMAl = 26.98 g/mole) 14...