A. Binary Compounds of Metal and Nonmetal
This compound is composed of metal and nonmetal. If you want to learn how to name this compound just follow the simple rule below:
Rule: Name of metal + Stem name of nonmetal + -ide ending
Example:
NaCl = Na + Cl
= Sodium + Chlor (stem name of chlorine) + ide
= Sodium Chloride
Some more examples:
KCl = Potassium chloride
MgO = Magnesium oxide
Li2S = Lithium sulfide
MgCl2 = Magnesium chloride
BeO = Berrylium oxide
B. Binary Compounds of Nonmetal and Nonmetal
This compound is composed of two nonmetals. Now in naming this compounds prefixes are used.
Prefixes
1 - mono
2 - di
3 - tri
4 - tetra
5 - pent
6 - hex
7 - hept
8 - oct
9 - nona
10 - dec
Rule: Prefix of first element except mono + Name of first nonmetal + Prefix of second nonmetal + stem name of second nonmetal + ide
Example:
CO = C + O
= Carbon (1 carbon no prefix since mono) + Mono (1 oxygen ) + ox (stem name of oxygen) + ide
= Carbon monoxide
Other examples:
CO2 - Carbon dioxide
NO3 - Nitrogen trioxide
N2O4 - Dinitrogen tetraoxide
CCl4 - Carbon tetrachloride
P2O4 - Diphosphorus tetraoxide
C. Binary Acids
This compound is composed of hydrogen bonded to a nonmetal
Rule : Hydro (constant prefix) + stem name of nonmetal + -ic acid
Example :
HCl = H + Cl
= Hydro + chlor (stem name of chlorine) + ic acid
= Hydrochloric acid
Other examples;
H2S - Hydrosulfuric acid ( sulfur is exempted to the rule, full name is used)
HF - Hydroflouric acid
HBr - Hydrobromic acid
D. Binary compounds of metal with variable oxidation number and nonmetal
This compound is composed of metal having two or more oxidation number paired with a nonmetal.
There are two rules to follow, one is by the use of IUPAC system of naming and the latin system of naming.
IUPAC System:
Rule : Name of metal + Roman Numeral showing the oxidation used in metal element + Stem name of nonmetal + ide
Example :
FeCl2 = Fe (+2 oxidation used) + Cl (-1 oxidation used)
= Iron (II) (II - indicates the oxidation number used of Fe) + Chlor (Stem name of chlorine + ide
= Iron (II) chloride
Other examples:
FeCl3 - Iron (III) chloride
Fe2O3 - Iron (III) oxide
FeO - Iron (II) oxide
Latin System:
Rule : Stem name of the latin name of metal ( but there are some exemption like Hg) + ic ( is attached when the oxidation number used of the metal is higher) or -ous (is attached when the oxidation number used of the metal is lower) + stem name of nonmetal + ide
Example:
FeCl2 = Fe (+2 oxidation used) + Cl (-1 oxidation used)
= Ferr ( Stem name of the latin name of Fe which is Ferrum) + ous ( ous is the suffix used for lower oxidation number + chlor (stem name of chlorine) + ide
= Ferrous chloride
Other examples:
FeCl3 - Ferric chloride
Fe2O3 - Ferric oxide
FeO - Ferrous oxide
E. Ternary Ionic Compounds
This compound is composed of metal attached to a polyatomic anion.
Rule: Name of metal ( use roman numeral of the oxidation number when metal has variable oxidation number + name of the polyatomic anion.
Example :
NaOH = Na + OH
= Sodium (the name of the metal) + hydroxide ( name of the polyatomic anion)
= Sodium hydroxide
Other Examples:
CaSO4 - Calcium Sulfate
Cu3PO4 - Copper (I) phosphate
Cu3(PO3)2 - Copper (II) phosphite
F. Ternary Acids (Oxyacids)
Ternary acids are acids composed of hydrogen element paired to polyatomic anion. This is also called oxyacids, acids containing oxygen.
Rule : Stem name of the element which is not hydrogen and oxygen (but there are some exemption, other element use full name like sulfur, carbon) + ous acid ( this is used when the ending of the polyatomic anion is ite) or ic acid (when the ending name of the polyatomic anion is ate).
Example
H2SO4 = H + SO4 (sulfate, name of polyatomic anion)
= Sulfur + ic acid ( ic acid is used because sulfate ends in ate)
Other examples:
H3PO3 - Phosphorous acid
H3PO4 - Phosphoric acid
H2SO3 - Sulfurous acid
Try your knowledge by answering Pactice Test. Click NAMING OF COMPOUNDS WORKSHEET
Friday, June 3, 2016
Thursday, June 2, 2016
Concentration of Solution Sample Problems
A. Percent by Mass Sample Problems
1. What is percent by mass concentration of 12.5 g of NaCl dissolved in 200 g of water?
Identify the given: mass solute = 12.5 g NaCl
mass solvent = 200 g water
The problem is asking the percentage by mass of NaCl given the mass of solute and the mass of solution. You have to take note that the mass of the solution is equal to the mass of solute + the mass of solvent.
Formula:
Solution:
C. Molarity Problems
Calculation of Molarity:
2. What mass of solute is present in each of the following solutions?
Calculating for the mass of solute, needs the molar mass of Ca(OH)2
b. Given : volume of solution = 2.5 L
Molarity = 1 M of NaCl solution
Before the mass can be calculated the mole of solute should be calculated first and the molar mass of the substance
Molar Mass NaCl = 1 (23) + 1 (35)
= 23 + 35
= 58 g/mol
Calculating the mole of solute:
Calculating mass of substance:
1. What is percent by mass concentration of 12.5 g of NaCl dissolved in 200 g of water?
Identify the given: mass solute = 12.5 g NaCl
mass solvent = 200 g water
The problem is asking the percentage by mass of NaCl given the mass of solute and the mass of solution. You have to take note that the mass of the solution is equal to the mass of solute + the mass of solvent.
Formula:
Solution:
2. Calculate the percentage by mass of 30 g of KCl in 250 g of solution.
Find the given: mass of solute = 30 g KCl
mass of solution = 250 g of solution
Here the problem is asking to calculate the percentage by mass given the mass of the solute and the mass of the solution.
Formula:
Solution:
3. What is the mass of the CuSO4 present in 15% by mass 316 g of CuSO4 solution?
Find the given: Percent by mass = 15 %
mass of solution = 316 g of solution
The problem asked to calculate the mass of CuSO4, which is the solute of the solution.
Formula :
B. Percent by Volume Sample Problems
1. A solution of hydrochloric acid is produced by dissolving 15 mL of HCl in enough water to make 240 mL of solution. What is the percentage composition of the solution?
Find the given : volume of solute = 15 mL
volume of solution = 240 mL
This problem is asking the percentage by volume of hydrochloric acid given the volume of solute and the volume of solution.
Formula :
Solution:
2. Calculate the volume of alcohol present in 250 mL 70% alcohol solution and 40 % alcohol solution.
Find the given: volume of alcohol solution = 250 mL
Percentage by Volume = 70 % and 40 %
The problem is asking the volume of alcohol present in 70 % and 40% alcohol solution having 250 total volume of solution. Two solutions are needed in this problem.
Formula:
Solution:
For 70% alcohol solution
For 40% alcohol solution
3. The average adult human body contains about 5 L of blood. Only 0.72% consists of leukocytes (white blood cells). What volume of leukocyte cells is present in the body of a small child, with only 2.5 L of blood?
The given in this problem is the percent of white blood cells = 0.72% and the total volume of blood of a child blood = 2.5 L.
The problem is asking to calculate the volume of leukocytes or the white blood cells.
Formula:
Solution:
C. Molarity Problems
1. What is the molar concentration of the following solutions:
a. 0.20 mol of NaCl in 0.25 L of solution.
b. 5 g of sugar, C12H22O11 in 20 mL of solution.
c. 3 g of KCl in 80 mL solution.
Formulas needed:
Solutions:
a. Given : mole solute = 0.20 mol
volume of solution = 0.25 L of solution
Unit for Molarity can be mol/L or simply M.
b. Given: mass of the substance = 5 g
volume of solution = 20 mL or 0.02 L
In this problem moles of solute should be calculated first, and needs the molar mass of sugar
Molar mass of sugar = 12 (Atomic Mass of Carbon) + 22 (Atomic Mass of Hydrogen) + 11(Atomic Mass of Oxygen)
Molar Mass of Sugar = 12(12) +22(1) + 11(16)
= 144 + 22 + 176
= 342 g/mol
How to calculate mole of solute?
c. Given: mass of the substance = 3 g of KCl
volume of solution = 80 mL solution or 0.08 L
Molar mass of KCl = 1(atomic mass of K) + 1(Atomic Mass)
= 1(39) + 1(35)
= 39 + 35
= 74 g/mol
Calculation for mole of solute
Calculation for molarity:
a. 400 mL of 0.05 M of Calcium hydroxide, (Ca(OH)2) solution.
b. 2.5 L of 1.00 M of sodium chloride, (NaCl) solution
Solution:
a. Given : Volume of solution = 400 mL or 0.4 L
molarity = 0.05 M or mol/L
In this problem we need to calculate first the mole of the solute before we can calculate the mass.
Calculating the moles of solute, formula is just rearrange:
Calculating for the mass of solute, needs the molar mass of Ca(OH)2
Molar Mass of Ca(OH)2 = 1(Atomic Mass Ca) + 2 (Atomic Mass O) + 2(atomic Mass H)
= 1(40g) + 2(16g) + 2(1g)
= 40g + 32g + 2g
= 74 g/mol
Calculating mass of the substance:
b. Given : volume of solution = 2.5 L
Molarity = 1 M of NaCl solution
Before the mass can be calculated the mole of solute should be calculated first and the molar mass of the substance
Molar Mass NaCl = 1 (23) + 1 (35)
= 23 + 35
= 58 g/mol
Calculating the mole of solute:
Calculating mass of substance:
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