A. Naming Elements and Compounds 1. Polyatomic Molecular Elements
-
Upload
sophia-parsons -
Category
Documents
-
view
47 -
download
0
description
Transcript of A. Naming Elements and Compounds 1. Polyatomic Molecular Elements
A. Naming Elements and Compounds1. Polyatomic Molecular Elements
halogens, hydrogen, phosphorus, sulfur, oxygen, nitrogen
elements that exist in form instead of form
Chem 20 Review
molecular atomic
***SHPON (8, 2, 4, 2, 2)
2. Molecular Compounds
use prefixes to indicate the number of atoms in the molecule
there are many molecular compounds that still use their common names or IUPAC names (organic molecules)
eg) carbon dioxide =dinitrogen monoxide = CCl4 =
CO2
N2O carbon tetrachloride
eg) hydrogen peroxide, glucose, ammonia, sucrose
non-metals only
3. Acids
see data booklet and Nelson pg 553
Rules
1. hydrogen becomes acid
2. hydrogen becomes acid
3. hydrogen becomes acid
_____ide hydro____ic
_____ate _______ic
_____ite ______ous
Try These:
1. hydrogen iodide =
2. hydrogen phosphate =
3. hydrogen nitrite =
4. hydrogen sulphite =
hydroiodic acid
phosphoric acid
nitrous acid
sulphurous acid
4. Ionic Compounds
name in full then name the with or use
cation (positive ion) anion (negative ion)
eg) NaCl = Mg3(PO4)2 =
sodium chloride magnesium phosphate
look up the symbol for each ion then balance the charges using subscripts
the polyatomic ion name the “ide” ending
when writing formulas,
hydrated ionic compounds contain
indicated by where x is the
eg) CuSO45H2O = copper (II) sulphate pentahydrate
“xH2O”
water in their atomic structures
number of water molecules
B. Bonding
recall types of intermolecular forces
London Dispersion – occurs between all moleculesand is the attraction of the electrons in one molecule to the protons in another molecule
Hydrogen Bonding – occurs only when H is bonded to O, F or N
occurs only between polar moleculesDipole-Dipole – and is the electrostatic attraction of the polar ends of molecules
and is the attraction of the H to the O, F or N in another molecule
C. Chemical Reactions1. Classification of Reactions
clues to a chemical reaction:
conservation laws:
1. colour change
2. gas produced
3. energy change
4. precipitate formed
2. energy (1st Law of Thermo)
1. mass - number and kind of atom (balancing)
types of reactions:
1. composition/formation
2. decomposition
3. single replacement
4. double replacement
5. hydrocarbon combustion
6. other
2. Energy in Chemical Reactions
endothermic =
exothermic =
bond energy =
energy is absorbed
energy is released
energy released when bonds are formed or energy absorbed when bonds are broken
3. States of Matter
gives the states of elements at room temperature
all ionic compounds at room temperature
molecular compounds can be at room temperature
periodic table
(by themselves) are solid
solids, liquids or gases
acids are assumed to be
ionic compounds may be when mixed with water
aqueous
aqueous or solid
D. Significant Digits
represents the degree of
two different rules are used:
1. Addition/Subtraction:
2. Multiplication/Division:
add or subtract then round to the lowest number of decimal places
accuracy of using measured values
multiply or divide then round to the lowest number of sig digs
E. Solutions and Gases1. Preparation of Solutions
concentration is most commonly measured in
mol/L can also be expressed as
solubility =
solubility is measured in
mol/L
“molarity” or M
the concentration of solute in a saturated solution at a given temperature
g/100 mL
eg) 0.300 mol/L = 0.300 M
molar solubility is measured in
to determine the solubility of an ionic compound, check the solubility table in the Data Booklet
mol/L
Steps for Solution Preparation
1. Calculate the mass of the solute required to achieve a specific concentration and volume.
2. Measure mass.
3. Dissolve the solute in half of the volume of solvent.
4. Transfer solution to a volumetric flask.
5. Bring solution up to final volume and mix by inverting.
2. Dilution of Solutions
when a solution is diluted, only the amount of is increased
the remains constant
solvent (usually water)
number of moles
ViCi = VfCf
3. Types of Solutes
nonelectrolytes:
electrolytes:
dissociation of solutes in water:
1. electrolytes:
2. non-electrolytes:
substances that dissolve to yield solutions that do not conduct electricity
substances that dissolve to yield solutions that conduct electricity
eg) molecular compounds
eg) ionic compounds and acids
Na2CO3(s)
C6H12O6(s)
2Na+(aq) + CO3
2-(aq)
C6H12O6(aq)
4. Determining Ionic Concentrations
you can use the concentration of a solute to determine the ion concentrations once it is dissolved in water (dissociated)
Steps
1. Write a balanced dissociation equation.
2. Write down concentration given.
3. Determine the ion concentrations using the mole ratio.
Example
What is the concentration of each ion in a 0.23 mol/L solution of aluminum sulphate?
1 Al2(SO4)3(s)
C = 0.23 mol/L C = 0.23 mol/L C = 0.23 mol/L x 2 1
x 3 1
= 0.46 mol/L = 0.69 mol/L
2 Al3+(aq) + 3 SO4
2-(aq)
5. Non Ionic vs. Net Ionic Equations
net ionic reactions are used to show
write the non-ionic reaction, the total ionic reaction and the net ionic reaction
only the reacting ions…spectator ions (non-reacting) are omitted
Example
What is the net ionic reaction for the reaction of bromine and sodium iodide ?
Non Ionic:
Total Ionic:
Net Ionic:
+ Br–(aq)
+ 2Br–(aq)
Br2(l) + NaI(aq) I2(s) + NaBr(aq)2 2
Br2(l) + Na+(aq) + I–
(aq) I2(s) + Na+(aq)2 2 2 2
Br2(l) + 2I–(aq) I2(s)
6. Gases
Ideal Gas Law:
STP =
SATP =
conversion to Kelvin =
PV = nRT
273.15 K, 101.325 kPa, 22.4 L/mol
298.15 K, 100 kPa, 24.8 L/mol
273.15 + xC
F. Stoichiometry
Steps
1. Write a balanced chemical (or net ionic) equation.
2. Write down the given information.
3. Find moles of given using n = m, C = n , PV = nRT M V
4. Find moles of wanted using the mole ratio: wanted/given.
5. Find answer to question.
Example 1
If 5.00 g of sodium reacts with excess chlorine gas, how much sodium chloride is produced?
2Na(s) + Cl2(g) 2 NaCl(aq)
m = 5.00 gM = 22.99 g/mol
x gM = 58.44 g/mol
n = m M = 5.00g 22.99 g/mol = 0.217… mol
n = 0.217… mol x 2 2
m = nM = (0.217… mol)(58.44 g/mol) = 12.7 g
Example 2
Liquid bromine is added to 500 mL of a solution containing 0.150 mol/L iodide ion. What mass of iodine will be produced?
2I-(aq) + Br2(l) 2Br-
(aq) + I2(s)
V = 500 mL = 0.500 LC = 0.150 mol/L
x gM = 253.80 g/mol
n = CV = (0.150 mol/L)(0.500 L) = 0.0750 mol
n = 0.0750 mol x 1 2
= 0.0375 mol
m = nM = (0.0375mol)(253.80 g/mol) = 9.52 g
Example 3
If 5.0 g of sodium reacts with 5.0 g of chlorine, how much sodium chloride is produced?
2Na(s) + Cl2(g) 2NaCl(aq)
x g
M = 58.44 g/moln = 0.0705 mol x 2
1 = 0.141…mol
m = nM = (0.141..mol)(58.44 g/mol) = 8.2 g
m = 5.0 gM = 22.99 g/mol
m = 5.0 gM = 70.90 g/mol
n = m M= 5.0 g 22.99 g/mol= 0.217…mol
n = m M= 5.0 g 70.90 g/mol= 0.0705…mol
2 = 0.108…mol
1 = 0.0705…molCl2(g) is limiting