Ionic Compound Naming Race | Science 9th Grade - Free Printable

To solve this "Compound Naming Race," we need to write the chemical formula for each named compound. Here is the step-by-step logic for finding the formulas:

General Rules:
1. Identify the ions: Break the name into the positive ion (cation) and the negative ion (anion).
2. Find the charges:
* For metals with Roman numerals (like Copper (II)), the numeral tells you the positive charge.
* For main group metals (like Sodium, Lithium), use the periodic table group number (Group 1 is +1, Group 2 is +2, Aluminum is +3).
* For polyatomic ions (like acetate, hydroxide, nitrate), you need to know their specific charges (e.g., Nitrate is $NO_3^-$).
* For non-metals ending in "-ide" (like oxide, sulfide), the charge is usually determined by how many electrons they need to reach 8 (Group 16 is -2, Group 17 is -1, Group 15 is -3).
3. Balance the charges: The total positive charge must equal the total negative charge so the compound is neutral. Use subscripts (small numbers at the bottom right) to show how many of each ion are needed. If you need more than one polyatomic ion, put it in parentheses.

Step-by-Step Solutions:

1) copper (II) acetate: Copper (II) is $Cu^{2+}$. Acetate is $C_2H_3O_2^-$ (or $CH_3COO^-$). You need two acetates to balance one copper.
Formula: $Cu(C_2H_3O_2)_2$

2) sodium hydroxide: Sodium is $Na^+$. Hydroxide is $OH^-$. Charges are +1 and -1, so they balance 1:1.
Formula: $NaOH$

3) lithium oxide: Lithium is $Li^+$. Oxide is $O^{2-}$. You need two lithiums (+2) to balance one oxide (-2).
Formula: $Li_2O$

4) cobalt (III) carbonate: Cobalt (III) is $Co^{3+}$. Carbonate is $CO_3^{2-}$. To balance +3 and -2, find the least common multiple (6). You need two cobalts ($2 \times +3 = +6$) and three carbonates ($3 \times -2 = -6$).
Formula: $Co_2(CO_3)_3$

5) aluminum sulfide: Aluminum is $Al^{3+}$. Sulfide is $S^{2-}$. Similar to #4, you need two aluminums and three sulfides to balance charges (+6 and -6).
Formula: $Al_2S_3$

6) ammonium cyanide: Ammonium is $NH_4^+$. Cyanide is $CN^-$. Charges are +1 and -1.
Formula: $NH_4CN$

7) iron (III) phosphide: Iron (III) is $Fe^{3+}$. Phosphide is $P^{3-}$ (Group 15). Charges are +3 and -3. They balance 1:1.
Formula: $FeP$

8) vanadium (V) phosphate: Vanadium (V) is $V^{5+}$. Phosphate is $PO_4^{3-}$. Least common multiple of 5 and 3 is 15. You need three vanadiums ($3 \times 5 = 15$) and five phosphates ($5 \times 3 = 15$).
Formula: $V_3(PO_4)_5$

9) sodium permanganate: Sodium is $Na^+$. Permanganate is $MnO_4^-$. Charges are +1 and -1.
Formula: $NaMnO_4$

10) manganese (III) fluoride: Manganese (III) is $Mn^{3+}$. Fluoride is $F^-$ (Group 17). You need three fluorides to balance one manganese.
Formula: $MnF_3$

11) beryllium nitrate: Beryllium is $Be^{2+}$ (Group 2). Nitrate is $NO_3^-$. You need two nitrates to balance one beryllium.
Formula: $Be(NO_3)_2$

12) nickel (III) sulfite: Nickel (III) is $Ni^{3+}$. Sulfite is $SO_3^{2-}$. Least common multiple of 3 and 2 is 6. You need two nickels and three sulfites.
Formula: $Ni_2(SO_3)_3$

13) potassium oxide: Potassium is $K^+$ (Group 1). Oxide is $O^{2-}$. You need two potassiums to balance one oxide.
Formula: $K_2O$

14) silver bromide: Silver is typically $Ag^+$. Bromide is $Br^-$ (Group 17). Charges balance 1:1.
Formula: $AgBr$

15) zinc phosphate: Zinc is typically $Zn^{2+}$. Phosphate is $PO_4^{3-}$. Least common multiple of 2 and 3 is 6. You need three zincs and two phosphates.
Formula: $Zn_3(PO_4)_2$

16) copper (II) bicarbonate: Copper (II) is $Cu^{2+}$. Bicarbonate is $HCO_3^-$. You need two bicarbonates to balance one copper.
Formula: $Cu(HCO_3)_2$

17) nickel (II) selenide: Nickel (II) is $Ni^{2+}$. Selenide is $Se^{2-}$ (Group 16, same as oxygen/sulfur). Charges are +2 and -2. They balance 1:1.
Formula: $NiSe$

18) manganese (IV) carbonate: Manganese (IV) is $Mn^{4+}$. Carbonate is $CO_3^{2-}$. You can simplify the ratio 4:2 to 2:1. You need one manganese and two carbonates.
Formula: $Mn(CO_3)_2$

19) lead (IV) nitride: Lead (IV) is $Pb^{4+}$. Nitride is $N^{3-}$ (Group 15). Least common multiple of 4 and 3 is 12. You need three leads ($3 \times 4 = 12$) and four nitrides ($4 \times 3 = 12$).
Formula: $Pb_3N_4$

20) tin (II) hydroxide: Tin (II) is $Sn^{2+}$. Hydroxide is $OH^-$. You need two hydroxides to balance one tin.
Formula: $Sn(OH)_2$

Final Answer:
1) $Cu(C_2H_3O_2)_2$
2) $NaOH$
3) $Li_2O$
4) $Co_2(CO_3)_3$
5) $Al_2S_3$
6) $NH_4CN$
7) $FeP$
8) $V_3(PO_4)_5$
9) $NaMnO_4$
10) $MnF_3$
11) $Be(NO_3)_2$
12) $Ni_2(SO_3)_3$
13) $K_2O$
14) $AgBr$
15) $Zn_3(PO_4)_2$
16) $Cu(HCO_3)_2$
17) $NiSe$
18) $Mn(CO_3)_2$
19) $Pb_3N_4$
20) $Sn(OH)_2$