Module 5 · Chemical Formulas and ValencePractice · 化学式与化合价

A subscript after an element symbol gives the number of those atoms in one molecule: each $\ce{H2O}$ molecule has 2 hydrogen atoms. The coefficient 3 means 3 molecules, so there are 3 × 2 = 6 hydrogen atoms in all. An element is a kind of atom and isn’t counted, so “2 hydrogen elements” makes no sense. The atom ratio H : O is 2 : 1, but the mass ratio is (2 × 1) : 16 = 1 : 8.

In a compound, the valences add up to zero. K is +1 and O is −2; let Cr be x:

so x = +6.

$\ce{NH4NO3}$ is made of $\ce{NH4+}$ and $\ce{NO3-}$, so work them out separately. In a polyatomic group the valences add up to the group’s charge: in $\ce{NH4+}$, x + 4 × (+1) = +1, so N is −3; in $\ce{NO3-}$, y + 3 × (−2) = −1, so N is +5. Averaging over the whole formula gives +1 for both, which is wrong.

When writing a formula from valences, the valences must add up to zero. Al is +3 and O is −2, so it should be $\ce{Al2O3}$; Na is +1 and $\ce{SO4}$ is −2, so $\ce{Na2SO4}$; K is +1 and OH is −1, so $\ce{KOH}$. $\ce{CaCl2}$ (+2, −1) and $\ce{Fe2O3}$ (+3, −2) are correct.

First find R’s valence from $\ce{R2O3}$: 2x + 3 × (−2) = 0, so x = +3. With chlorine, each R (+3) needs 3 Cl (−1), giving $\ce{RCl3}$.

OH is −1, so two OH groups total −2 and the iron must be +2. Iron commonly forms +2 and +3 ions, so the name gives the charge in Roman numerals: iron(II) hydroxide. With +3 it would be $\ce{Fe(OH)3}$, iron(III) hydroxide; iron(II) oxide is FeO. Ionic compound names don’t use number prefixes, so “iron dihydroxide” is wrong.

Count the water of crystallization too: add the parts on both sides of the dot, and multiply H₂O by 5:

160 + 90 = 250.

Relative formula mass: 14 × 2 + 1 × 4 + 16 × 3 = 80. Each $\ce{NH4NO3}$ contains 2 N atoms, so the mass percent is $\dfrac{14 \times 2}{80}$ × 100% = 35%. Counting only one N gives 17.5%.

The relative formula mass of urea is 12 + 16 + (14 + 1 × 2) × 2 = 60, of which nitrogen accounts for 28.

The 2 outside the parentheses applies to the whole $\ce{NH2}$ group, so there are 2 N atoms.

Mass ratio = (atomic mass × number of atoms) for each element: (56 × 3) : (16 × 4) = 168 : 64 = 21 : 8. 3 : 4 is the ratio of atoms, 7 : 2 is the mass ratio in FeO, and 7 : 3 is the mass ratio in $\ce{Fe2O3}$.