Mendeleev's periodic table — based on atomic mass: Worked examples

This page trains you to use Mendeleev's logic on real cases. We never guess: we line elements up by mass, watch the chemistry repeat, and read valency off oxide/hydride formulas. First, the map of every kind of problem this topic can throw.

Prerequisite ideas we lean on: Valency and oxide formulas, Atomic number and mass number, Isotopes, and the parent Mendeleev's periodic table.

The scenario matrix

Before any numbers, let's define the words we will reuse, so no symbol appears unearned.

Here is every case class this chapter can test. Each worked example below is tagged with the cell it fills.

Cell	Case class	What makes it tricky
A	Order-by-mass, spot the repeat	pure sequencing, no traps
B	Valency from an oxide formula	rising 1→7 across a period
C	Valency from a hydride formula	falling 4→1 on the right
D	Predict a gap element (eka-)	interpolate between neighbours
E	Correct a wrong atomic mass	trust pattern over data
F	Anomalous pair (heavier before lighter)	mass order violated on purpose
G	Degenerate/zero case — isotopes & noble gases	mass can't decide; $Z$ must
H	Word problem / exam twist	mixed reasoning, hidden trap

The 8 examples below hit all 8 cells.

Example 1 — Cell A: line them up, catch the repeat

Order by mass. $Li (7), Be (9), F (19), Na (23), Mg (24), Al (27)$ . Why this step? Mass is the only universal ruler Mendeleev had; sequencing first exposes the rhythm.
Recall Li's personality. Li is a soft, violently-reacting metal, valency $1$ , oxide $Li_{2} O$ . Why this step? We need a fingerprint to match against.
Scan forward for the return. Be(2), F(1 but a gas non-metal)... at Na(23) we again get a soft violent metal, oxide $Na_{2} O$ , valency $1$ . Why this step? Detecting the return is how Mendeleev knew where to wrap the row.

Answer: Na is Li's twin. They sit in the same group.

Example 2 — Cell B: valency from an oxide

Apply the oxide rule $valency = \frac{2 y}{x}$ with $x = 2, y = 5$ : $\frac{2 \cdot 5}{2} = 5$ . Why this step? The oxide formula is a hard chemical fact; it pins valency directly.
Map valency to group. Across a period oxides run $R_{2} O, RO, R_{2} O_{3}, RO_{2}, R_{2} O_{5}$ → groups I, II, III, IV, V. Why this step? The oxide series is the group axis.

Answer: Group V, valency 5 (e.g. nitrogen, phosphorus).

Example 3 — Cell C: valency from a hydride (the falling side)

Figure — Mendeleev's periodic table — based on atomic mass

Read hydride valency. In $YH_{3}$ , three H (each valency 1) bond to one Y ⇒ valency $= 3$ . Why this step? Hydride formula gives valency directly, just like oxides.
Place on the falling ladder. $RH_{4} (4), R H_{3} (3), RH_{2} (2), RH (1)$ correspond to groups IV, V, VI, VII. Why this step? On the right, hydride valency falls; this ladder is the group fingerprint. Look at the figure — the red step is $RH_{3}$ .
Cross-check with oxide. Group V's oxide is $R_{2} O_{5}$ (oxide valency 5); note $5 + 3 = 8$ . Why this step? Oxide + hydride valency = 8 on the right side — a self-consistency check.

Answer: hydride valency 3, Group V (nitrogen family).

Example 4 — Cell D: predict a gap (eka-element)

Interpolate mass from neighbours. The gap sits between Ca(40) and Ti(48); estimate the midpoint region: $\frac{40 + 48}{2} = 44$ . Mendeleev predicted ~44. Why this step? Within the local sequence, mass grows steadily; the average of the bracket is the natural guess.
Copy the group's oxide type. Group III oxide is $R_{2} O_{3}$ , so eka-aluminium's oxide is $Ea_{2} O_{3}$ , valency 3. Why this step? Same group ⇒ same valency pattern — that is the whole predictive power.

Answer: mass ≈ 44, oxide $Ea_{2} O_{3}$ , valency 3. This gap was filled by scandium (real mass 45.0, oxide $Sc_{2} O_{3}$ ). ✓

Example 5 — Cell E: correct a wrong mass

State the relation. $atomic mass = equivalent weight \times valency$ . Why this step? Equivalent weight is what the weighings actually give; mass depends on the assumed valency.
Wrong assumption (valency 3): $4.5 \times 3 = 13.5 \approx 14$ — the old, contradictory value. Why this step? Shows how a wrong valency inflated the mass.
Right assumption (valency 2, from oxide $BeO$ ): $4.5 \times 2 = 9.0$ . Why this step? $BeO$ gives valency $\frac{2 \cdot 1}{1} = 2$ ; trusting chemistry fixes the mass.

Answer: corrected mass ≈ 9, placing Be above Mg in Group II. Mendeleev trusted the pattern over the data — and was right.

Example 6 — Cell F: an anomalous pair (mass order broken on purpose)

Compare masses. $A (Ar) = 39.9 > A (K) = 39.1$ . Pure mass order says K first. Why this step? Establishes the apparent contradiction.
Compare chemistry. Ar is inert (noble gas, valency 0); K is a violent metal, valency 1 like Na and Li. Why this step? Chemistry must decide the column — Ar belongs with noble gases, K with alkali metals.
Place by property, not mass. Mendeleev put Ar first so it stacks under the noble gases and K under the alkali metals, inverting the mass order deliberately. Why this step? Grouping twins is the point; a small mass swap is worth it.
Name the true ruler. Atomic number $Z$ : Ar has $Z = 18$ , K has $Z = 19$ . Ordering by $Z$ (Moseley 1913) puts Ar(18) before K(19) — no inversion. Why this step? The anomaly vanishes under $Z$ ; see Modern Periodic Law — based on atomic number.

Answer: property beats mass; $Z = 18 < 19$ restores natural order. In the figure the red arrow shows the mass order (backwards) vs the $Z$ order (forward).

Example 7 — Cell G: the degenerate case (isotopes & noble gases)

Count distinct masses. Two masses, 35 and 37, would demand two separate positions. Why this step? Mass-ordering treats different masses as different places.
Count distinct chemistry. Both isotopes react identically (valency 1, $HCl$ , $Cl_{2} O_{7}$ ) — they are the same element. Why this step? Same chemistry ⇒ must be one box; the mass rule contradicts reality.
Resolve with $Z$ . Both have $Z = 17$ ; ordering by atomic number gives them one shared position. Why this step? This is exactly the defect that atomic number cures.

Answer: mass says 2 boxes, chemistry says 1 — a genuine defect fixed by $Z = 17$ .

Example 8 — Cell H: exam-style word twist

Identify by position. "Eka" = "one" = one row below. One below silicon ⇒ eka-silicon, later found as germanium. Why this step? The eka- naming maps directly to position: eka-silicon = the box under Si.
Check oxide type. Group IV oxide is $RO_{2}$ ; $QO_{2}$ matches ⇒ valency $\frac{2 \cdot 2}{1} = 4$ . ✓ Why this step? Confirms Q really is Group IV before trusting the density.
Compute measured density. $ρ = \frac{mass}{volume} = \frac{27.0}{4.9} \approx 5.51 g/cm^{3}$ . Why this step? Density is mass per volume — the physical fingerprint Mendeleev predicted (~5.5).

Answer: (a) eka-silicon = germanium; (b) $5.51 \approx 5.5$ g/cm³ — the prediction matches beautifully.

Connections

Modern Periodic Law — based on atomic number — how $Z$ erases every anomaly above.
Atomic number and mass number — the real ordering variable behind Cells F, G.
Isotopes — Cell G's degenerate case.
Valency and oxide formulas — engine behind Cells B, C, D.
Newlands' Law of Octaves, Döbereiner's Triads — earlier repeat-detection attempts.