Question

this means that it takes two - iron (iii) ions (total charge 6+) to balance out the charge of 3 sulfate ions (total charge 6-). 4. write the chemical formula. to show that there are multiple polyatomic ions, enclose that section of the formula in parentheses. write the subscript showing the number of atoms after the close parenthesis. example name cation anion chemical formula iron (iii) sulfate fe³⁺ (so₄)²⁻ fe₂(so₄)₃ practice exercises write the chemical formula for each of the following compounds. 1. antimony (iii) chlorate __________ 6. cobalt (ii) arsenate ________ 2. tantalum (v) sulfite ________ 7. platinum (iv) borate ________ 3. manganese (ii) arsenite ________ 8. copper (ii) oxalate ________ 4. rhenium (vi) perchlorate ________ 9. gold (i) iodate ________ 5. iron (iii) sulfate ________ 10. gallium (iii) tartrate __________ chemistry 2a - 6a 3 writing chemical formulas

Explanation:

Step1: Determine ion charges

For each compound, identify the charge of the cation and the anion. For example, in Antimony (III) chlorate, the antimony (III) ion ($Sb^{3 + }$) has a charge of 3+ and the chlorate ion ($ClO_{3}^{-}$) has a charge of 1-.

Step2: Balance charges

To balance the charges, find the least - common multiple of the absolute values of the charges of the cation and anion. Then, use subscripts to make the total positive charge equal the total negative charge. For example, for $Sb^{3 + }$ and $ClO_{3}^{-}$, we need 3 chlorate ions to balance the 3+ charge of the antimony ion, so the formula is $Sb(ClO_{3})_{3}$.

Step3: Write formulas for all compounds

1. Antimony (III) chlorate: The cation is $Sb^{3+}$ and the anion is $ClO_{3}^{-}$. The formula is $Sb(ClO_{3})_{3}$.
2. Tantalum (V) sulfite: The cation is $Ta^{5 + }$ and the anion is $SO_{3}^{2-}$. The least - common multiple of 5 and 2 is 10. So we need 2 $Ta^{5+}$ ions and 5 $SO_{3}^{2-}$ ions, and the formula is $Ta_{2}(SO_{3})_{5}$.
3. Manganese (II) arsenite: The cation is $Mn^{2+}$ and the anion is $AsO_{2}^{-}$. The formula is $Mn(AsO_{2})_{2}$.
4. Rhenium (VI) perchlorate: The cation is $Re^{6+}$ and the anion is $ClO_{4}^{-}$. The formula is $Re(ClO_{4})_{6}$.
5. Iron (III) sulfate: Given in the example as $Fe_{2}(SO_{4})_{3}$.
6. Cobalt (II) arsenate: The cation is $Co^{2+}$ and the anion is $AsO_{4}^{3 - }$. The least - common multiple of 2 and 3 is 6. So the formula is $Co_{3}(AsO_{4})_{2}$.
7. Platinum (IV) borate: The cation is $Pt^{4+}$ and the anion is $BO_{3}^{3 - }$. The least - common multiple of 4 and 3 is 12. So the formula is $Pt_{3}(BO_{3})_{4}$.
8. Copper (II) oxalate: The cation is $Cu^{2+}$ and the anion is $C_{2}O_{4}^{2-}$. The formula is $CuC_{2}O_{4}$.
9. Gold (I) iodate: The cation is $Au^{+}$ and the anion is $IO_{3}^{-}$. The formula is $AuIO_{3}$.
10. Gallium (III) tartrate: The cation is $Ga^{3+}$ and the anion is $C_{4}H_{4}O_{6}^{2 - }$. The least - common multiple of 3 and 2 is 6. So the formula is $Ga_{2}(C_{4}H_{4}O_{6})_{3}$.

Answer:

1. $Sb(ClO_{3})_{3}$
2. $Ta_{2}(SO_{3})_{5}$
3. $Mn(AsO_{2})_{2}$
4. $Re(ClO_{4})_{6}$
5. $Fe_{2}(SO_{4})_{3}$
6. $Co_{3}(AsO_{4})_{2}$
7. $Pt_{3}(BO_{3})_{4}$
8. $CuC_{2}O_{4}$
9. $AuIO_{3}$
10. $Ga_{2}(C_{4}H_{4}O_{6})_{3}$