QUESTION IMAGE
Question
- write the chemical formula. you do not need to include the charges of the ions.
a. in the case that more than one of a certain polyatomic ion is necessary to balance the charge, the entire formula for the polyatomic ion(s) is enclosed in parentheses and the numerical subscript is placed outside the parentheses. this shows that the subscript applies to the entire polyatomic ion.
example
| name | cation | anion | chemical formula |
|---|
practice exercises
write the chemical formula for each of the following compounds.
- calcium nitrite
- potassium carbonate
- silver citrate
- aluminum arsenate
- cadmium sulfite
- rubidium phosphite
- strontium hydroxide
- beryllium sulfate
- zinc hydrogen sulfate
- potassium perchlorate
rules for writing formulas for ternary compounds with variable charges
- identify the cation part of the compound and determine its charge. add the roman numeral as a positive charge to the element symbol.
chemistry 2a - sa 3 writing chemical formulas
Step1: Identify cations and anions
For calcium nitrite, calcium ($Ca^{2 +}$) is the cation and nitrite ($NO_{2}^{-}$) is the anion.
Step2: Balance charges
To balance the +2 charge of calcium, we need 2 nitrite ions. So the formula is $Ca(NO_{2})_{2}$.
Step3: For potassium carbonate
Potassium ($K^{+}$) is the cation and carbonate ($CO_{3}^{2 -}$) is the anion. We need 2 potassium ions to balance the - 2 charge of carbonate, so the formula is $K_{2}CO_{3}$.
Step4: For silver citrate
Silver ($Ag^{+}$) is the cation and citrate ($C_{6}H_{5}O_{7}^{3 -}$) is the anion. We need 3 silver ions to balance the - 3 charge of citrate, so the formula is $Ag_{3}C_{6}H_{5}O_{7}$.
Step5: For aluminum arsenate
Aluminum ($Al^{3+}$) is the cation and arsenate ($AsO_{4}^{3 -}$) is the anion. The charges balance with a 1:1 ratio, so the formula is $AlAsO_{4}$.
Step6: For cadmium sulfite
Cadmium ($Cd^{2+}$) is the cation and sulfite ($SO_{3}^{2 -}$) is the anion. The charges balance with a 1:1 ratio, so the formula is $CdSO_{3}$.
Step7: For rubidium phosphite
Rubidium ($Rb^{+}$) is the cation and phosphite ($PO_{3}^{3 -}$) is the anion. We need 3 rubidium ions to balance the - 3 charge of phosphite, so the formula is $Rb_{3}PO_{3}$.
Step8: For strontium hydroxide
Strontium ($Sr^{2+}$) is the cation and hydroxide ($OH^{-}$) is the anion. We need 2 hydroxide ions to balance the + 2 charge of strontium, so the formula is $Sr(OH)_{2}$.
Step9: For beryllium sulfate
Beryllium ($Be^{2+}$) is the cation and sulfate ($SO_{4}^{2 -}$) is the anion. The charges balance with a 1:1 ratio, so the formula is $BeSO_{4}$.
Step10: For zinc hydrogen sulfate
Zinc ($Zn^{2+}$) is the cation and hydrogen sulfate ($HSO_{4}^{-}$) is the anion. We need 2 hydrogen - sulfate ions to balance the + 2 charge of zinc, so the formula is $Zn(HSO_{4})_{2}$.
Step11: For potassium perchlorate
Potassium ($K^{+}$) is the cation and perchlorate ($ClO_{4}^{-}$) is the anion. The charges balance with a 1:1 ratio, so the formula is $KClO_{4}$.
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- $Ca(NO_{2})_{2}$
- $K_{2}CO_{3}$
- $Ag_{3}C_{6}H_{5}O_{7}$
- $AlAsO_{4}$
- $CdSO_{3}$
- $Rb_{3}PO_{3}$
- $Sr(OH)_{2}$
- $BeSO_{4}$
- $Zn(HSO_{4})_{2}$
- $KClO_{4}$