2.5 · Nomenclature (Focus on English Names)Practice · 命名法（英文命名为重点）

Tin commonly forms both +2 and +4 ions, so the name gives the charge in Roman numerals. Each F is −1, so two total −2 and the tin is +2: tin(II) fluoride. The numeral is the ion’s charge, not an atom count; ionic compounds don’t take prefixes like di-; and stannic is the old name for +4 tin (the old name for +2 is stannous, which is why toothpaste labels often say stannous fluoride).

(II) means $\ce{Pb^2+}$, and nitrate is $\ce{NO3-}$. Balancing the charges takes 2 nitrates, and a polyatomic ion that appears more than once goes in parentheses: $\ce{Pb(NO3)2}$. $\ce{Pb(NO3)4}$ misreads what the (II) means.

Ammonium, $\ce{NH4+}$, is 1+ and carbonate, $\ce{CO3^2-}$, is 2−, so two ammonium ions are needed, in parentheses: $\ce{(NH4)2CO3}$. $\ce{NH4HCO3}$ is ammonium hydrogen carbonate, a different compound used as a fertilizer.

$\ce{HSO4-}$ is sulfate, $\ce{SO4^2-}$, with one $\ce{H+}$ added, so “hydrogen” goes in front: hydrogen sulfate. $\ce{KHSO4}$ is potassium hydrogen sulfate. Potassium sulfate is $\ce{K2SO4}$; sulfite is $\ce{SO3^2-}$ (one fewer O); sulfide is the monatomic $\ce{S^2-}$.

First write the formula: (III) means $\ce{Fe^3+}$, and nitrate is $\ce{NO3-}$, so 3 nitrates balance the charge: $\ce{Fe(NO3)3}$. The 3 outside the parentheses multiplies the whole $\ce{NO3}$: 3 × 3 = 9 O atoms.

A molecular compound of two nonmetals uses prefixes for the number of atoms: di- = 2, tri- = 3, and the second element ends in -ide (oxide). So it’s dinitrogen trioxide. Roman numerals are only for ionic compounds with metals; “nitrogen trioxide” would mean $\ce{NO3}$; dinitrogen oxide is $\ce{N2O}$.

No prefix on the first element means one Si (mono- is dropped there), and tetra- means 4 Cl: $\ce{SiCl4}$. A molecular name states the atom counts directly; you don’t need charges to work them out.

Removing 2 $\ce{H+}$ leaves $\ce{CrO4^2-}$, chromate. For oxyacids, an -ate ion gives an -ic acid and an -ite ion gives an -ous acid, so $\ce{H2CrO4}$ is chromic acid. The hydro-…-ic pattern is only for acids without oxygen (like HCl), and acids aren’t named “… -ate acid.”

Hydro-…-ic acids contain no oxygen and come from -ide ions: iodide ($\ce{I-}$) → hydroiodic acid, which is $\ce{HI}$ in water. $\ce{I-}$ is only 1−, so it pairs with one $\ce{H+}$, not two. $\ce{HIO3}$ and $\ce{HIO}$ are oxyacids (iodic acid and hypoiodous acid), named the same way as $\ce{HClO3}$ and $\ce{HClO}$.

Water lost: 2.87 − 1.61 = 1.26 g. Compare amounts using mass ÷ formula mass: water 1.26 ÷ 18.02 = 0.0699, $\ce{ZnSO4}$ 1.61 ÷ 161.44 = 0.00997, so x = 0.0699 ÷ 0.00997 ≈ 7.01, which rounds to 7: $\ce{ZnSO4.7H2O}$, zinc sulfate heptahydrate.