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Solving Problems: A Chemistry Handbook
Chemistry: Matter and Change
67
SOLVING PROBLEMS:
A CHEMISTRY HANDBOOK
CHAPTER
7
Example Problem 7-3
Determining the Formula for an Ionic Compound
Determine the correct chemical formula for the ionic compound
formed from strontium (Sr) and bromine (Br).
Strontium is an element in group 2 of the periodic table. Bromine is
in group 17. The formulas for their ions are therefore Sr 2ϩ and
Br Ϫ , respectively. Sr must lose two electrons to form its ion, and Br
must gain one electron. The total number of electrons lost must
equal the total number of electrons gained so that overall charge is
zero. Subscripts must be chosen so that this is the case. For
electrical neutrality, there must be two Br for each Sr. The correct
formula is thus SrBr 2 , where the subscript 1 is understood after the
Sr. This formula can be verified by multiplying the subscripts by the
ion charges and summing the result, as follows.
(1 ϫ 2ϩ) ϩ (2 ϫ 1Ϫ) ϭ 0
Practice Problems
Write the correct formula for the ionic compound formed
between atoms of each of the following pairs of elements.
11. sodium (Na) and sulfur (S)
12. magnesium (Mg) and nitrogen (N)
13. potassium (K) and phosphorus (P)
14. barium (Ba) and fluorine (F)
15. aluminum (Al) and nitrogen (N)
Polyatomic ions An ion that contains more than one atom
is called a polyatomic ion. The charge on such an ion applies to
the entire group of atoms. The writing of chemical formulas for
compounds containing such ions follows the same rules as for
compounds containing only monatomic ions. The overall charge
must be zero. Subscripts within the formula for a polyatomic ion
must not be changed during formula writing. If there is more than
one such ion in a formula unit, parentheses are written around
the formula of the ion and a subscript is written after the final
parenthesis.
▲
Solving Problems: A Chemistry Handbook
Chemistry: Matter and Change
67
SOLVING PROBLEMS:
A CHEMISTRY HANDBOOK
CHAPTER
7
Example Problem 7-3
Determining the Formula for an Ionic Compound
Determine the correct chemical formula for the ionic compound
formed from strontium (Sr) and bromine (Br).
Strontium is an element in group 2 of the periodic table. Bromine is
in group 17. The formulas for their ions are therefore Sr 2ϩ and
Br Ϫ , respectively. Sr must lose two electrons to form its ion, and Br
must gain one electron. The total number of electrons lost must
equal the total number of electrons gained so that overall charge is
zero. Subscripts must be chosen so that this is the case. For
electrical neutrality, there must be two Br for each Sr. The correct
formula is thus SrBr 2 , where the subscript 1 is understood after the
Sr. This formula can be verified by multiplying the subscripts by the
ion charges and summing the result, as follows.
(1 ϫ 2ϩ) ϩ (2 ϫ 1Ϫ) ϭ 0
Practice Problems
Write the correct formula for the ionic compound formed
between atoms of each of the following pairs of elements.
11. sodium (Na) and sulfur (S)
12. magnesium (Mg) and nitrogen (N)
13. potassium (K) and phosphorus (P)
14. barium (Ba) and fluorine (F)
15. aluminum (Al) and nitrogen (N)
Polyatomic ions An ion that contains more than one atom
is called a polyatomic ion. The charge on such an ion applies to
the entire group of atoms. The writing of chemical formulas for
compounds containing such ions follows the same rules as for
compounds containing only monatomic ions. The overall charge
must be zero. Subscripts within the formula for a polyatomic ion
must not be changed during formula writing. If there is more than
one such ion in a formula unit, parentheses are written around
the formula of the ion and a subscript is written after the final
parenthesis.
▲