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Writing Formulas of Ionic Compounds. By Sharon Mace, West Stanly HS, NC. Binary Ionic Compounds:. Binary Ionic Compounds:. Have only two types of elements. Are made of monatomic ions only. The name always ends in “-ide ”. The first ion is always positive. - PowerPoint PPT Presentation
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WRITING FORMULAS OF IONIC COMPOUNDS
By Sharon Mace, West Stanly HS, NC
Binary Ionic Compounds:
Binary Ionic Compounds: Have only two types of elements. Are made of monatomic ions only. The name always ends in “-ide”. The first ion is always positive. The second ion is always negative. They DO NOT have an overall charge. They NEVER have ( )’s. (Parentheses are
only used with more than one of a polyatomic ion, never with a monatomic ion.)
Example:Sodium Chloride
Example:Sodium Chloride
Na+1 Cl-1
Example:Sodium Chloride
Na+1 Cl-1
It takes one of each ion to make a neutral compound.
Example:Sodium Chloride
Na+1 Cl-1
It takes one of each ion to make a neutral compound.
NaClNaCl
OR…Use the criss-cross method:
Example:Sodium Chloride
Na+1 Cl-1
Example:Sodium Chloride
Na+1 Cl-1
Take the 1 from the Na and put it as the invisible subscript on Cl.
Example:Sodium Chloride
Na+1 Cl-1
Take the 1 from the Na and put it as the invisible subscript on Cl.
Take the 1 from Cl and put it as the invisible subscript on Na.
Example:Sodium Chloride
Na+1 Cl-1
Take the 1 from the Na and put it as the invisible subscript on Cl.
Take the 1 from Cl and put it as the invisible subscript on Na.
NaClNaCl
Example:Magnesium Chloride
Example:Magnesium ChlorideMg+2 Cl-1
Example:Magnesium ChlorideMg+2 Cl-1
It takes one magnesium and two chlorides to make a neutral compound.
Example:Magnesium ChlorideMg+2 Cl-1
It takes one magnesium and two chlorides to make a neutral compound.
Mg+2 + Cl-1 + Cl-1 = No charge
Example:Magnesium ChlorideMg+2 Cl-1
It takes one magnesium and two chlorides to make a neutral compound.
Mg+2 + Cl-1 + Cl-1 = No charge
MgCl2MgCl2
OR…Use the criss-cross method:
Example:Magnesium ChlorideMg+2 Cl-1
Example:Magnesium ChlorideMg+2 Cl-1
Take the 2 from the Mg and put it as the subscript on Cl.
Example:Magnesium ChlorideMg+2 Cl-1
Take the 2 from the Mg and put it as the subscript on Cl.
MgCl2
Example:Magnesium ChlorideMg+2 Cl-1
Take the 2 from the Mg and put it as the subscript on Cl.
MgCl2
Take the 1 from Cl and put it as the invisible subscript on Mg.
Example:Magnesium ChlorideMg+2 Cl-1
Take the 2 from the Mg and put it as the subscript on Cl.
MgCl2
Take the 1 from Cl and put it as the invisible subscript on Mg.
MgCl2MgCl2
Example:Aluminum Sulfide
Example:Aluminum Sulfide
Al+3 S-2
Example:Aluminum Sulfide
Al+3 S-2
It takes two aluminum ions and three sulfide ions to make a neutral
compound.
Example:Aluminum Sulfide
Al+3 S-2
It takes two aluminum ions and three sulfide ions to make a neutral
compound.
Al+3 + Al+3 + S-2 + S-2 + S-2 = No Charge
Example:Aluminum Sulfide
Al+3 S-2
It takes two aluminum ions and three sulfide ions to make a neutral
compound.
Al+3 + Al+3 + S-2 + S-2 + S-2 = No Charge
Al2S3
Al2S3
OR…Use the criss-cross method:
Example:Aluminum Sulfide
Al+3 S-2
Example:Aluminum Sulfide
Al+3 S-2
Take the 3 from the Al and put it as the subscript on S.
Example:Aluminum Sulfide
Al+3 S-2
Take the 3 from the Al and put it as the subscript on S.
AlS3
Example:Aluminum Sulfide
Al+3 S-2
Take the 3 from the Al and put it as the subscript on S.
AlS3
Take the 2 from S and put it as the subscript on Al.
Example:Aluminum Sulfide
Al+3 S-2
Take the 3 from the Al and put it as the subscript on S.
AlS3
Take the 2 from S and put it as the subscript on Al.
Al2S3Al2S3
Example:Magnesium Oxide
Example:Magnesium Oxide
Mg+2 O-2
Example:Magnesium Oxide
Mg+2 O-2
It only takes ONE of each ion to make a neutral compound.
Example:Magnesium Oxide
Mg+2 O-2
It only takes ONE of each ion to make a neutral compound.
Mg+2 + O-2 = No charge
Example:Magnesium Oxide
Mg+2 O-2
It only takes ONE of each ion to make a neutral compound.
Mg+2 + O-2 = No charge
MgCl2MgO
OR…Use the criss-cross method:
Example:Magnesium Oxide
Mg+2 O-2
First, you must reduce the charge ratio!
A +2 to -2 ratio cancels to +1 to -1:
Example:Magnesium Oxide
Mg+2 O-2
First, you must reduce the charge ratio!
A +2 to -2 ratio cancels to +1 to -1:
Mg+1 O-1
NOW you can criss-cross:
Example:Magnesium Oxide
Mg+2 O-2
First, you must reduce the charge ratio!
A +2 to -2 ratio cancels to +1 to -1:
Mg+1 O-1
NOW you can criss-cross:MgO
Now you try it:QUESTION ANSWER
1. Potassium selenide
2. Aluminum
phosphide
3. Beryllium carbide
________
________
________
Watch out for canceling!
Now you try it:QUESTION ANSWER
1. Potassium selenide
2. Aluminum
phosphide
3. Beryllium carbide
K2S
AlP
Be2C
Adding Transition Metal Ions:
When the name of a compound contains a transition metal ion, the
Roman numeral or the classical name refers to the charge on the ion
BEFORE you figure out the formula of the compound.
Example:Manganese (II) Bromide
ORManganous Bromide:
Example:Manganese (II) Bromide
ORManganous Bromide:
Mn+2 Br-1
Example:Manganese (II) Bromide
ORManganous Bromide:
Mn+2 Br-1
MnBr2
Example:Ferric Nitride
ORIron (III) Nitride:
Example:Ferric Nitride
ORIron (III) Nitride:
Fe+3 N-3
Watch out for canceling!
Example:Ferric Nitride
ORIron (III) Nitride:
Fe+3 N-3
FeN
Watch out for canceling!
Example:Ferric Nitride
ORIron (III) Nitride:
Fe+3 N-3
FeNWatch out for canceling!
Question: Does the Roman numeral in the name have anything to do
with the subscript of the final answer?
Example:Ferric Nitride
ORIron (III) Nitride:
Fe+3 N-3
FeNWatch out for canceling!
Answer: NOT REALLY!
Now you try it:QUESTION ANSWER
1. Stannous fluoride
2. Chromium (III)
sulfide
3. Plumbic oxide
________
________
________
Watch out for canceling!
Now you try it:QUESTION ANSWER
1. Stannous fluoride
2. Chromium (III)
sulfide
3. Plumbic oxide
SnF2
Cr2S3
PbO2
Ternary Ionic Compounds:
Ternary Ionic Compounds: Have 3 or more different elements. Are still made of ONLY 2 DIFFERENT IONS! Contain at least one polyatomic ion. The first ion is always positive. The second ion is always negative. They DO NOT have an overall charge. The first ion is usually the first element (unless
you have ammonium, NH4+1).
All the rest of the elements are in the second ion. Only have ( )’s around MORE THAN ONE of a
POLYATOMIC ion!
Example:Aluminum Sulfate
Example:Aluminum Sulfate
Al+3 SO4-2
Example:Aluminum Sulfate
Al+3 SO4-2
It takes two aluminum ions and three sulfate ions to make a neutral compound.
Al+3 + Al+3 + SO4-2 + SO4
-2 + SO4-2 = No
Charge
Example:Aluminum Sulfate
Al+3 SO4-2
It takes two aluminum ions and three sulfate ions to make a neutral compound.
Al+3 + Al+3 + SO4-2 + SO4
-2 + SO4-2 = No
ChargeAl2(SO4)3
Example:Aluminum Sulfate
Al+3 SO4-2
It takes two aluminum ions and three sulfate ions to make a neutral compound.
Al+3 + Al+3 + SO4-2 + SO4
-2 + SO4-2 = No
ChargeAl2(SO4)3
There must be ( )‘s around SO4 because you have more than one of this polyatomic ion. You cannot use ( )’s around Al because it is
a monatomic ion.
OR…Use the criss-cross method:
Example:Aluminum SulfateAl+3 SO4
-2
Example:Aluminum SulfateAl+3 SO4
-2
Take the 3 from the Al and put it as the subscript on SO4. You must use ( )’s because you have more than one of
this polyatomic ion.
Example:Aluminum SulfateAl+3 SO4
-2
Take the 3 from the Al and put it as the subscript on SO4. You must use ( )’s because you have more than one of
this polyatomic ion.
Al(SO4)3
Example:Aluminum SulfateAl+3 SO4
-2
Take the 3 from the Al and put it as the subscript on SO4. You must use ( )’s because you have more than one of
this polyatomic ion.
Al(SO4)3
Then take the 2 from SO4 and put it as the subscript on Al. You cannot use
( )’s on a monatomic ion.
Example:Aluminum SulfateAl+3 SO4
-2
Take the 3 from the Al and put it as the subscript on SO4. You must use ( )’s because you have more than one of
this polyatomic ion.
Al(SO4)3
Then take the 2 from SO4 and put it as the subscript on Al. You cannot use
( )’s on a monatomic ion.Al2(SO4)3
Example:Cupric Sulfate/Copper (II) Sulfate
Example:Cupric Sulfate/Copper (II) Sulfate
Cu+2 SO4-2
Example:Cupric Sulfate/Copper (II) Sulfate
Cu+2 SO4-2
It takes only ONE copper (II) ion and ONE sulfate ion to make a neutral compound.
Cu+2 + SO4-2 = No Charge
Example:Cupric Sulfate/Copper (II) Sulfate
Cu+2 SO4-2
It takes only ONE copper (II) ion and ONE sulfate ion to make a neutral compound.
Cu+2 + SO4-2 = No ChargeCuSO4
There cannot be ( )‘s around SO4 because you have only one of this polyatomic ion.
You cannot use ( )’s around Cu because it is a monatomic ion.
OR…Use the criss-cross method:
Example:Cupric Sulfate/Copper (II) Sulfate
Cu+2 SO4-2
Example:Cupric Sulfate/Copper (II) Sulfate
Cu+2 SO4-2
First cancel the charges:
Example:Cupric Sulfate/Copper (II) Sulfate
Cu+2 SO4-2
First cancel the charges: Cu+1 SO4
-1
Example:Cupric Sulfate/Copper (II) Sulfate
Cu+2 SO4-2
First cancel the charges: Cu+1 SO4
-1
NOW you can criss-cross:
Example:Cupric Sulfate/Copper (II) Sulfate
Cu+2 SO4-2
First cancel the charges: Cu+1 SO4
-1
NOW you can criss-cross:
You cannot use ( )’s around SO4 because you have only one of this
polyatomic ion.You cannot use ( )’s around Cu because
it is a monatomic ion.
CuSO4
Now you try it:QUESTION ANSWER
1. Lithium carbonate
2. Mercury (II) nitrate
3. Beryllium hydroxide
________
________
________
Now you try it:QUESTION ANSWER
1. Lithium carbonate
2. Mercury (II) nitrate
3. Beryllium hydroxide
Li2CO3
Hg(NO3)2
Be(OH)2Note: In #3, the hydroxide ion must have ( )’s
around it because there is more than one of this polyatomic ion. The 2 must be outside the ( )’s because it is not a part of the hydroxide ion; it is
telling how many hydroxides there are.
Now you try it:QUESTION ANSWER
1. Ammonium oxalate
2. Sodium hydrogen
carbonate (Hint: there
are only 2 ions here.)
3. Lead (IV) cyanide
________
________
________
Now you try it:QUESTION ANSWER
1. Ammonium oxalate
2. Sodium hydrogen
carbonate (Hint: there
are only 2 ions here.)
3. Lead (IV) cyanide
(NH4)2 C2O4
NaHCO3
Pb(CN)4For #1, the C2O4 ion cannot have ( )’s because there is only one of this polyatomic ion.
For #2, the ions are Na+1 (sodium) and HCO3-1 (hydrogen
carbonate).For #3, CN must have ( )’s because there is more than one of this polyatomic ion. The 4 must be outside the ( )’s because it is not
part of the ion.
