+

Synthesis/Combination

Decomposition

Single-Replacement Activity Series

Double-Replacement Solubility Rules

CombustionO2

+ +

hydrocarbon CO2 H2OO2

Metals

LithiumPotassium

CalciumSodium

MagnesiumAluminum

ZincChromium

IronNickel

TinLead

Hydrogen*CopperMercury

SilverPlatinum

Gold

Halogens

FluorineChlorineBromine

Iodine

Decreasing Activity

3.28 a) What is the mass of 2.50 x 10-2 mol of MgCl2?

FW(MgCl2) = Mg + 2 Cl = 24.305 g/mol + 2(35.453 g/mol)

FW(MgCl2) = 95.211 g/mol

Mass = (2.50 x 10-2 mol)(95.211 g/mol) = 2.38 g

b) How many moles of NH4Cl are there in 75.6 g of this substance?

FW(NH4Cl) = 53.491 g/mol

Moles = mass/FW = 76.5 g/(53.491 g/mol) = 1.43 mol

FW = Formula Weight gfm = gram formula massMW = Molar Weight

These are all equivalent terms for the molar mass of a substance:

c) How many molecules are there in 0.0772 mol HCHO2?

#m’cules = (6.022 x 1023)(0.0772 mol) = 4.65 x 1022 m’cules

d) How many nitrate ions, NO3, are there in 4.88 x 103

mol Al(NO3)3?

#f.u. = (6.022 x 1023)(4.88 x 103 mol) = 2.939 x 1021 f.u.

#ions = (2.939 x 1021 f.u.)(3 NO3/f.u.) = 8.82 x 1021 ions

Remember that ionic compounds don’t exist as discrete molecules, but are instead an extended lattice of ions. Hence, they are described in terms of the number of formula units (f.u.’s) that are present. In one f.u. of aluminum nitrate, there are three nitrate ions.

Perchlorate

Perbromate

Periodate

ClO4

BrO4

IO4

Nitrate NO3

Chlorate

Bromate

Iodate

ClO3

BrO3

IO3

Sulfate

Carbonate

SO42

CO32

Nitrite NO2

Chlorite

Bromite

Iodite

ClO2

BrO2

IO2

Sulfite SO32

Hypochlorite

Hypobromite

Hypoiodite

ClO

BrO

IO

PhosphateArsenate

PO43-

AsO43-

Polyatomic ions that you need to know are:

The ‘ides’

Peroxide O22

Hydroxide OH

Cyanide CN

Acetate C2H3O2

OR CH3COO

The halogenates

Permanganate MnO4

Dichromate Cr2O72

Chromate CrO42

Form colored solutions

The only cationAmmonium NH4

+

BrO4, +7 SO4

2, +6

BrO3, +5

BrO2, +3

BrO, +1SO3

2, +4S2O3

2, +6

ClO4, +7

ClO3, +5

ClO2, +3

ClO, +1

IO4, +7

IO3, +5

IO2, +3

IO, +1NO3

, +5

NO2, +3

PO43, +5

PO33, +3

NH4+, -3

CrO42, +6

Cr2O72, +6

CO32, +4

C2H3O2

Combustion Analysis3.48 (a) The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen.  Combustion of 2.78 mg of ethyl butyrate produces 6.32 mg of CO2 and 2.58 mg of H2O. What is the empirical formula of the compound (element order is C H O)?

CxHyOz + O2

CO2 + H2O2.78 mg 6.32 mg 2.58 mg

Step 1: Determine the number of moles of C and H atoms

FW(CO2) = 44.010 mg/mmol FW(H2O) = 18.015 mg/mmol

mmoles C = (6.32 mg CO2)/(44.010 mg/mmol) = 0.1436 mmol C

mmoles H2O = (2.58 mg H2O)/(18.015 mg/mmol) = 0.1432 mmol

mmoles H = (0.1432 mmoles H2O)(2 H atoms/H2O) = 0.2864 mmol H

Step 2: Convert the mmoles of C and H to mg of the two elements so can determine how much oxygen was in the compound.

mg C = (0.1436 mmol)(12.011 mg/mmol) = 1.7248 mg C

mg H = (0.2864 mmol)(1.00794 mg/mmol) = 0.2886 mg H

mg O = 2.78 mg – (1.7248 mg + 0.2886 mg) = 0.766 mg O

mmol O = (0.766 mg)/(15.9994 mg/mmol) = 0.0479 mmol

Step 3: Divide through by the smallest molar amount to determine the empirical formula.C: 0.1436 mmol/0.0479 mmol = 3

H: 0.2864 mmol/0.0479 mmol = 6

O: 0.0479 mmol/0.0479 mmol = 1

The empirical formula is C3H6O

Stoichiometry3.56 The fermentation of glucose (C6H12O6) produces ethyl alcohol (C2H5OH) and CO2.

(a) How many moles of CO2 are produced when 0.400 mol of  C6H12O6 reacts in this fashion?

C6H12O6 (aq)  2 C2H5OH (aq) + 2 CO2 (g)0.400 mol

#mol x 2x 2x

0.800 mol

(b) How many grams of  C6H12O6 are needed to form 7.50 g of C2H5OH?

mol C2H5OH = (7.50 g)/(46.069 g/mol) = 0.1628 mol

FW(C2H5OH) = 46.069 g/mol FW(C6H12O6) = 180.158 g/mol

0.1628 mol0.08140 mol

Mass glucose = (0.08140 mol)(180.158 g/mol) = 14.7 g

LR:

3.70 Aluminum hydroxide reacts with sulfuric acid as shown below.

(b) How many moles of Al2(SO4)3 form when 0.450 mol

Al(OH)3 and 0.550 mol H2SO4 are allowed to react?

Limiting Reactants/Theoretical Yield

2 Al(OH)3 (s) + 3 H2SO4 (aq)  Al2(SO4)3 (aq) + 6 H2O (l)0.450 mol

0.550 mol

2x 3x x

0.225 mol

0.183 mol

•The H2SO4 is the limiting reactant because it limits how much aluminum sulfate will be formed.

•The theoretical (calculated) yield for this reaction is 0.183 mol of aluminum sulfate and 1.10 mol of water.

•What is the percent yield for this reaction if only 0.750 mol of water are collected?

%yield = (0.750 mol)/(1.10 mol)*100 = 68.2%

6x