Rules for Writing Chemical Formulae - Ninth Grade Science
The chemical formula is the shorthand universal script of chemistry. Just as mathematical equations express complex geometric relationships through concise algebraic symbols, a chemical formula conveys the exact elemental identity, atomic proportions, and charge balance of a chemical compound in a compact string of letters and subscripts. When a chemist writes $ ext{Al}_2( ext{SO}_4)_3$, that single formula communicates that the compound is aluminum sulfate, that it is constructed from aluminum cations and polyatomic sulfate anions, and that aluminum and sulfate combine in a strict two-to-three ratio to achieve electrical neutrality. Writing chemical formulae accurately is an indispensable literacy in physical science. In this chapter, you will master the step-by-step rules for writing chemical formulae, learn the criss-cross valency method, analyze the use of brackets with polyatomic ions, and practice naming and formulating ionic and molecular compounds.
The Meaning of a Chemical Formula
The chemical formula of a compound is a symbolic representation of its chemical composition, indicating the elements present in the compound and the exact numerical ratio of atoms of each element that combine to form that compound.
Anatomy of a Chemical Formula: Al_2(SO_4)_3
Element Symbol: Aluminum Element Symbol: Sulfur
| |
v v
Al _2 ( S O _4 ) _3
^ ^ ^ ^
| | | |
Subscript: Subscript: Subscript: Multiplier Subscript:
2 Aluminum Atoms 1 Sulfur 4 Oxygen 3 entire Sulfate groups!
Atom Atoms (3 * 1 = 3 Sulfur atoms)
(3 * 4 = 12 Oxygen atoms)
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| The Fundamental Rule of Chemical Formulae |
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| ELECTRICAL NEUTRALITY: In any stable chemical compound, the total |
| positive charge of the cations must exactly balance the total negative |
| charge of the anions: |
| |
| Total Positive Charge + Total Negative Charge = 0 |
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The Four Fundamental Rules for Writing Chemical Formulae
To construct any chemical formula systematically, adhere to four foundational IUPAC rules:
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| The Four Rules for Writing Chemical Formulae |
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| Rule 1: Balance the Valencies or Charges so net compound charge is zero. |
| Rule 2: In binary compounds of metal and non-metal, write the METAL |
| symbol first on the left, followed by the NON-METAL on the right.|
| Rule 3: When writing subscripts, do NOT write the numeral 1. |
| Rule 4: Enclose a POLYATOMIC ION in brackets if its subscript exceeds 1. |
+--------------------------------------------------------------------------+
Rule 1: Metal First, Non-Metal Second
When a compound is composed of a metal and a non-metal, the metal is written first on the left, and the non-metal is written second on the right:
- In calcium oxide ($ ext{CaO}$), calcium (metal) is written first, oxygen (non-metal) second.
- In sodium chloride ($ ext{NaCl}$), sodium (metal) is written first, chlorine (non-metal) second.
- In copper(II) oxide ($ ext{CuO}$), copper (metal) is written first, oxygen (non-metal) second.
- In iron(II) sulfide ($ ext{FeS}$), iron (metal) is written first, sulfur (non-metal) second.
Rule 2: The Criss-Cross Valency Method
The most intuitive and foolproof technique for determining the subscripts of a chemical formula is the criss-cross method:
1. Write the chemical symbols of the constituent elements or polyatomic ions side by side, placing the positive ion (cation) on the left and the negative ion (anion) on the right.
2. Write the valency (or ionic charge magnitude) of each element directly beneath its corresponding symbol.
3. Criss-cross the valencies: transfer the valency number of the first element to become the subscript of the second element, and transfer the valency number of the second element to become the subscript of the first element.
4. Simplify the ratio: if the subscripts share a common divisor, divide by the greatest common divisor to express the formula in its simplest whole-number ratio.
5. Omit the number 1: if a subscript is 1, omit it from the final formula.
The Criss-Cross Method Illustrated
Example A: Magnesium Chloride Example B: Aluminum Oxide
Symbols: Mg Cl Symbols: Al O
Valencies: 2 1 Valencies: 3 2
\ / \ /
\ / \ /
v v v v
Formula: Mg_1 Cl_2 Formula: Al_2 O_3
Final: MgCl_2 Final: Al_2O_3
Rule 3: Simplification to Smallest Integer Ratios
If criss-crossing yields subscripts that share a common factor, divide by that factor to obtain the empirical formula:
Example: Calcium Oxide Example: Carbon Dioxide
Symbols: Ca O Symbols: C O
Valencies: 2 2 Valencies: 4 2
\ / \ /
v v v v
Formula: Ca_2 O_2 Formula: C_2 O_4
Divide by 2: Ca_1 O_1 Divide by 2: C_1 O_2
Final: CaO Final: CO_2
In calcium oxide, criss-crossing yields $ ext{Ca}_2 ext{O}_2$. Dividing both subscripts by 2 yields $ ext{CaO}$. In carbon dioxide, criss-crossing yields $ ext{C}_2 ext{O}_4$. Dividing both subscripts by 2 yields $ ext{CO}_2$.
Rule 4: Brackets Around Polyatomic Ions
When a compound contains a polyatomic ion and the required subscript for that ion is greater than 1, the entire formula of the polyatomic ion must be enclosed in parentheses (brackets) before writing the subscript outside:
Example A: Calcium Hydroxide Example B: Sodium Hydroxide
Symbols: Ca OH Symbols: Na OH
Valencies: 2 1 Valencies: 1 1
\ / \ /
v v v v
Formula: Ca(OH)_2 Formula: NaOH
(Subscript is 2: BRACKETS REQUIRED!) (Subscript is 1: NO BRACKETS!)
In calcium hydroxide, the subscript 2 applies to the entire hydroxide group ($ ext{OH}$), meaning the compound contains two oxygen atoms and two hydrogen atoms for every calcium atom; writing $ ext{CaOH}_2$ without brackets is chemically incorrect because it implies one calcium, one oxygen, and two hydrogens. However, if the subscript for a polyatomic ion is 1, brackets are omitted: sodium hydroxide is written simply as $ ext{NaOH}$, not $ ext{Na(OH)}$.
Step-by-Step Worked Examples of Formula Writing
+-----------------------+-------------------+-----------------+--------------------+
| Compound Name | Ions Involved | Criss-Cross | Final Formula |
+-----------------------+-------------------+-----------------+--------------------+
| Hydrogen Sulfide | H (1), S (2) | H_2 S_1 | H_2S |
| Carbon Tetrachloride | C (4), Cl (1) | C_1 Cl_4 | CCl_4 |
| Magnesium Oxide | Mg (2), O (2) | Mg_2 O_2 -> 1:1 | MgO |
| Sodium Nitrate | Na (1), NO_3 (1) | Na_1 (NO_3)_1 | NaNO_3 |
| Calcium Nitrate | Ca (2), NO_3 (1) | Ca_1 (NO_3)_2 | Ca(NO_3)_2 |
| Aluminum Sulfate | Al (3), SO_4 (2) | Al_2 (SO_4)_3 | Al_2(SO_4)_3 |
| Ammonium Sulfate | NH_4 (1), SO_4 (2)| (NH_4)_2 SO_4 | (NH_4)_2SO_4 |
| Calcium Carbonate | Ca (2), CO_3 (2) | Ca_2(CO_3)_2->1 | CaCO_3 |
| Ammonium Phosphate | NH_4 (1), PO_4 (3)| (NH_4)_3 PO_4 | (NH_4)_3PO_4 |
| Iron(III) Hydroxide | Fe (3), OH (1) | Fe_1 (OH)_3 | Fe(OH)_3 |
+-----------------------+-------------------+-----------------+--------------------+
Detailed Walkthrough: Aluminum Sulfate
Construct the chemical formula for aluminum sulfate:
1. Write the symbols: Aluminum is $ ext{Al}$; the polyatomic sulfate ion is $ ext{SO}_4$.
2. Write their valencies beneath: Aluminum cation has a valency of 3 ($ ext{Al}^{3+}$); sulfate anion has a valency of 2 ($ ext{SO}_4^{2-}$).
3. Criss-cross the numbers: The 2 moves to aluminum ($ ext{Al}_2$), and the 3 moves to sulfate ($( ext{SO}_4)_3$).
4. Apply brackets: Because sulfate is a polyatomic ion and its subscript is 3 (greater than 1), enclose sulfate in brackets: $ ext{Al}_2( ext{SO}_4)_3$.
5. Check charge neutrality: Two $ ext{Al}^{3+}$ ions contribute $2 imes (+3) = +6$; three $ ext{SO}_4^{2-}$ ions contribute $3 imes (-2) = -6$. The net charge is $+6 + (-6) = 0$. The formula is balanced.
Chapter Exercises
Exercise 1: Using the criss-cross valency method, write the correct chemical formula for each of the following eight chemical compounds, showing all intermediate valency steps: sodium oxide, aluminum chloride, calcium fluoride, magnesium hydroxide, potassium carbonate, ammonium nitrate, copper(II) sulfate, and iron(III) oxide.
Exercise 2: A student writes the chemical formula for calcium phosphate as $ ext{CaPO}_4$ and magnesium hydroxide as $ ext{MgOH}_2$. Identify the chemical errors in both written formulae, explain why they violate electrical neutrality, and provide the correct IUPAC chemical formulae.
Exercise 3: Write the chemical names corresponding to each of the following chemical formulae: $ ext{KNO}_3$, $ ext{ZnSO}_4$, $ ext{FeCl}_2$, $ ext{FeCl}_3$, $ ext{NaHCO}_3$, and $( ext{NH}_4)_2 ext{CO}_3$. Explain why Roman numerals are required in the names of iron compounds.
Solutions and Explanations
Solution 1: Applying the criss-cross method step by step: Sodium Oxide: Symbols $ ext{Na}$ (valency 1) and $ ext{O}$ (valency 2). Criss-cross gives $ ext{Na}_2 ext{O}_1 \implies ext{Na}_2 ext{O}$. Aluminum Chloride: Symbols $ ext{Al}$ (valency 3) and $ ext{Cl}$ (valency 1). Criss-cross gives $ ext{Al}_1 ext{Cl}_3 \implies ext{AlCl}_3$. Calcium Fluoride: Symbols $ ext{Ca}$ (valency 2) and $ ext{F}$ (valency 1). Criss-cross gives $ ext{Ca}_1 ext{F}_2 \implies ext{CaF}_2$. Magnesium Hydroxide: Symbols $ ext{Mg}$ (valency 2) and $ ext{OH}$ (valency 1). Criss-cross gives $ ext{Mg}_1( ext{OH})_2 \implies ext{Mg(OH)}_2$. Potassium Carbonate: Symbols $ ext{K}$ (valency 1) and $ ext{CO}_3$ (valency 2). Criss-cross gives $ ext{K}_2( ext{CO}_3)_1 \implies ext{K}_2 ext{CO}_3$. Ammonium Nitrate: Symbols $ ext{NH}_4$ (valency 1) and $ ext{NO}_3$ (valency 1). Criss-cross gives $( ext{NH}_4)_1( ext{NO}_3)_1 \implies ext{NH}_4 ext{NO}_3$. Copper(II) Sulfate: Symbols $ ext{Cu}$ (valency 2) and $ ext{SO}_4$ (valency 2). Criss-cross gives $ ext{Cu}_2( ext{SO}_4)_2$, simplifying by 2 yields $ ext{CuSO}_4$. Iron(III) Oxide: Symbols $ ext{Fe}$ (valency 3) and $ ext{O}$ (valency 2). Criss-cross gives $ ext{Fe}_2 ext{O}_3$.
Solution 2: In the student's formula $ ext{CaPO}_4$, the valencies are unbalanced: calcium has a valency of 2 ($ ext{Ca}^{2+}$), while the phosphate ion has a valency of 3 ($ ext{PO}_4^{3-}$). A $1:1$ ratio produces a net charge of $+2 - 3 = -1$, violating electrical neutrality. Criss-crossing gives three calcium ions balancing two phosphate ions, yielding the correct formula $ ext{Ca}_3( ext{PO}_4)_2$ ($3 imes (+2) = +6$ and $2 imes (-3) = -6$). In the student's formula $ ext{MgOH}_2$, the student omitted brackets around the polyatomic hydroxide ion. Writing $ ext{MgOH}_2$ states that there is one magnesium, one oxygen, and two hydrogens, resulting in an unneutral charge of $+2 - 2 - 1 + 2 = +1$. The subscript 2 must apply to both oxygen and hydrogen, requiring brackets: $ ext{Mg(OH)}_2$.
Solution 3: The systematic chemical names are: $ ext{KNO}_3$ is potassium nitrate; $ ext{ZnSO}_4$ is zinc sulfate; $ ext{FeCl}_2$ is iron(II) chloride (ferrous chloride); $ ext{FeCl}_3$ is iron(III) chloride (ferric chloride); $ ext{NaHCO}_3$ is sodium hydrogen carbonate (sodium bicarbonate); and $( ext{NH}_4)_2 ext{CO}_3$ is ammonium carbonate. Roman numerals (Stock notation) are required in the names of iron compounds because iron exhibits variable valency, existing as either the divalent iron(II) ion ($ ext{Fe}^{2+}$) or the trivalent iron(III) ion ($ ext{Fe}^{3+}$). Without Roman numerals, the name "iron chloride" is ambiguous, as it could refer to either green $ ext{FeCl}_2$ or brown $ ext{FeCl}_3$.