Question

ec5 classwork ionic compound practice
write the ions and then the formula for each named compound: the first one is done for you as an example.
hints:

1. use the criss - cross pattern for charges that dont match, to make subscripts that tell how many.
2. remember that the names of transition elements with multiple +charges possible have roman numerals in the name.
3. remember that ionic compounds have only 2 ions and therefore 2 names, a first and a last. look for polyatomic ions if there are more than 2 elements present. (on the back of your periodic table)

aluminum chlorate
ions: al⁺³ (clo₃)⁻¹
formula: al₁(clo₃)₃
ammonium phosphate
ions: (nh₄)⁺¹ (po₄)⁻³
formula: (nh₄)₃(po₄)
iron (ii) oxide
copper (ii) hydroxide
magnesium sulfate
ammonium chloride
silver acetate
barium phosphate
sodium carbonate
potassium sulfide
lead (ii) phosphate
strontium chloride

Explanation:

Step1: Identify ions for silver acetate

Silver has a +1 charge ($Ag^{+}$) and acetate is a poly - atomic ion with a - 1 charge ($C_{2}H_{3}O_{2}^{-}$).

Step2: Write the formula for silver acetate

Using the criss - cross rule (since charges are equal in magnitude), the formula is $AgC_{2}H_{3}O_{2}$.

Step3: Identify ions for barium phosphate

Barium has a +2 charge ($Ba^{2 + }$) and phosphate is a poly - atomic ion with a - 3 charge ($PO_{4}^{3-}$).

Step4: Write the formula for barium phosphate

Using the criss - cross rule, we get $Ba_{3}(PO_{4})_{2}$.

Step5: Identify ions for iron (II) oxide

Iron (II) has a +2 charge ($Fe^{2+}$) and oxide has a - 2 charge ($O^{2-}$).

Step6: Write the formula for iron (II) oxide

Using the criss - cross rule (simplifying as the charges are equal in magnitude), the formula is $FeO$.

Step7: Identify ions for sodium carbonate

Sodium has a +1 charge ($Na^{+}$) and carbonate is a poly - atomic ion with a - 2 charge ($CO_{3}^{2-}$).

Step8: Write the formula for sodium carbonate

Using the criss - cross rule, the formula is $Na_{2}CO_{3}$.

Step9: Identify ions for copper (II) hydroxide

Copper (II) has a +2 charge ($Cu^{2+}$) and hydroxide is a poly - atomic ion with a - 1 charge ($OH^{-}$).

Step10: Write the formula for copper (II) hydroxide

Using the criss - cross rule, the formula is $Cu(OH)_{2}$.

Step11: Identify ions for potassium sulfide

Potassium has a +1 charge ($K^{+}$) and sulfide has a - 2 charge ($S^{2-}$).

Step12: Write the formula for potassium sulfide

Using the criss - cross rule, the formula is $K_{2}S$.

Step13: Identify ions for magnesium sulfate

Magnesium has a +2 charge ($Mg^{2+}$) and sulfate is a poly - atomic ion with a - 2 charge ($SO_{4}^{2-}$).

Step14: Write the formula for magnesium sulfate

Using the criss - cross rule (simplifying as the charges are equal in magnitude), the formula is $MgSO_{4}$.

Step15: Identify ions for lead (II) phosphate

Lead (II) has a +2 charge ($Pb^{2+}$) and phosphate is a poly - atomic ion with a - 3 charge ($PO_{4}^{3-}$).

Step16: Write the formula for lead (II) phosphate

Using the criss - cross rule, the formula is $Pb_{3}(PO_{4})_{2}$.

Step17: Identify ions for ammonium chloride

Ammonium has a +1 charge ($NH_{4}^{+}$) and chloride has a - 1 charge ($Cl^{-}$).

Step18: Write the formula for ammonium chloride

Using the criss - cross rule (since charges are equal in magnitude), the formula is $NH_{4}Cl$.

Step19: Identify ions for strontium chloride

Strontium has a +2 charge ($Sr^{2+}$) and chloride has a - 1 charge ($Cl^{-}$).

Step20: Write the formula for strontium chloride

Using the criss - cross rule, the formula is $SrCl_{2}$.

Answer:

Silver acetate:
Ions: $Ag^{+}, C_{2}H_{3}O_{2}^{-}$
Formula: $AgC_{2}H_{3}O_{2}$
Barium phosphate:
Ions: $Ba^{2+}, PO_{4}^{3-}$
Formula: $Ba_{3}(PO_{4})_{2}$
Iron (II) oxide:
Ions: $Fe^{2+}, O^{2-}$
Formula: $FeO$
Sodium carbonate:
Ions: $Na^{+}, CO_{3}^{2-}$
Formula: $Na_{2}CO_{3}$
Copper (II) hydroxide:
Ions: $Cu^{2+}, OH^{-}$
Formula: $Cu(OH)_{2}$
Potassium sulfide:
Ions: $K^{+}, S^{2-}$
Formula: $K_{2}S$
Magnesium sulfate:
Ions: $Mg^{2+}, SO_{4}^{2-}$
Formula: $MgSO_{4}$
Lead (II) phosphate:
Ions: $Pb^{2+}, PO_{4}^{3-}$
Formula: $Pb_{3}(PO_{4})_{2}$
Ammonium chloride:
Ions: $NH_{4}^{+}, Cl^{-}$
Formula: $NH_{4}Cl$
Strontium chloride:
Ions: $Sr^{2+}, Cl^{-}$
Formula: $SrCl_{2}$