Exercises — Empirical formula vs molecular formula — determination from % composition

A molecular formula must be a whole-number multiple of the empirical unit $C{H}_2$ — meaning both subscripts scale by the same integer $n$.

• $C{H}_3$: ratio $1:3$, not $1:2$. ✗
• $C_2{H}_4=(C{H}_2{)}_2$: $n=2$, both scaled by 2. ✓
• $C_3{H}_8$: ratio $3:8$, not $1:2$. ✗
• $C_2{H}_5$: ratio $2:5$, not $1:2$. ✗ Answer: $C_2{H}_4$.

$n=\frac{M_{\text{molar}}}{M_{\text{empirical}}}=\frac{18.02}{18.02}=1.$ When $n=1$, the empirical and molecular formulas are identical. Water is one of those cases (like $C{O}_2$, $NaCl$). This is why you can't assume they always differ.

Step 1 — Assume 100 g (so percent becomes grams): $80.0\text{g C}$, $20.0\text{g H}$. Step 2 — Grams → moles (divide by atomic mass): $\text{C}:\frac{80.0}{12.01}=6.661\text{mol},\text{H}:\frac{20.0}{1.008}=19.84\text{mol}.$ Step 3 — Divide by smallest ($6.661$): $\text{C}:\frac{6.661}{6.661}=1.00,\text{H}:\frac{19.84}{6.661}=2.98\approx 3.$ Answer: $C{H}_3$. (Here $2.98$ rounds to $3$ — within $0.1$ of an integer, so it's genuine rounding noise.)

Assume 100 g: $52.2\text{g C}$, $13.0\text{g H}$, $34.8\text{g O}$. Moles: $\text{C}:\frac{52.2}{12.01}=4.347,\text{H}:\frac{13.0}{1.008}=12.90,\text{O}:\frac{34.8}{16.00}=\mathrm{2.175.}$ Divide by smallest ($2.175$): $\text{C}:1.999\approx 2,\text{H}:5.93\approx 6,\text{O}:\mathrm{1.00.}$ Answer: $C_2{H}_{6}O$ (ethanol).

Assume 100 g: $69.6\text{g Mn}$, $30.4\text{g O}$. Moles: $\text{Mn}:\frac{69.6}{54.94}=1.267,\text{O}:\frac{30.4}{16.00}=\mathrm{1.900.}$ Divide by smallest ($1.267$): $\text{Mn}:1.00,\text{O}:\frac{1.900}{1.267}=\mathrm{1.500.}$ A clean $1.5$ is a real half, not rounding noise. Multiply both by 2: $\text{Mn}:2,\text{O}:3.$ Answer: $M{n}_2{O}_3$.

Moles (from 100 g): $\text{Co}:\frac{71.1}{58.93}=1.207,\text{O}:\frac{28.9}{16.00}=\mathrm{1.806.}$ Divide by smallest ($1.207$): $\text{Co}:1.00,\text{O}:\frac{1.806}{1.207}=1.497\approx \mathrm{1.50.}$ Again a half → multiply by 2: $C{o}_2{O}_3$. Answer: $C{o}_2{O}_3$.

Moles: $\text{C}:\frac{85.6}{12.01}=7.127,\text{H}:\frac{14.4}{1.008}=\mathrm{14.29.}$ Divide by smallest ($7.127$): $\text{C}:1.00$, $\text{H}:2.005\approx 2$. Empirical formula $=C{H}_2$, empirical mass $=12.01+2(1.008)=14.03$. Find $n$: $n=\frac{M_{\text{molar}}}{M_{\text{empirical}}}=\frac{56.0}{14.03}=3.99\approx 4.$ Molecular formula $=(C{H}_2{)}_4=C_4{H}_8$.

Step 1 — mass of C (all carbon in the sample ends up in $C{O}_2$; C is $12.01$ of the $44.01\text{g/mol}$ of $C{O}_2$): $m_{\text{C}}=0.733\times \frac{12.01}{44.01}=0.2000\text{g}.$ Step 2 — mass of H (each $H_{2}O$ of mass $18.02$ carries $2(1.008)=2.016$ of H): $m_{\text{H}}=0.300\times \frac{2.016}{18.02}=0.03356\text{g}.$ Step 3 — mass of O by difference (the sample was only C, H, O): $m_{\text{O}}=0.500-0.2000-0.03356=0.2664\text{g}.$ Step 4 — grams → moles: $\text{C}:\frac{0.2000}{12.01}=0.01665,\text{H}:\frac{0.03356}{1.008}=0.03330,\text{O}:\frac{0.2664}{16.00}=\mathrm{0.01665.}$ Step 5 — divide by smallest ($0.01665$): $\text{C}:1.00$, $\text{H}:2.00$, $\text{O}:1.00$. Empirical formula $=C{H}_{2}O$, empirical mass $=30.03$. Step 6 — $n$: $n=60.0/30.03=2.00\Rightarrow$ $C_2{H}_4{O}_2$ (acetic acid).

Moles (from 100 g): $\text{K}:\frac{26.57}{39.10}=0.6795,\text{Cr}:\frac{35.36}{52.00}=0.6800,\text{O}:\frac{38.07}{16.00}=\mathrm{2.379.}$ Divide by smallest ($0.6795$): $\text{K}:1.00,\text{Cr}:1.001\approx 1,\text{O}:3.501\approx \mathrm{3.5.}$ A clean half on O → multiply all by 2: $\text{K}:2$, $\text{Cr}:2$, $\text{O}:7$. Answer: $K_{2}C{r}_2{O}_7$ (potassium dichromate).

This is the same "moles → ratio" logic, but the ratio we want is water molecules per formula unit. Moles of $MgS{O}_4$: $\frac{0.976}{120.4}=0.008106\text{mol}.$ Moles of $H_{2}O$ driven off: $\frac{1.024}{18.02}=0.05683\text{mol}.$ Ratio (water : salt): $x=\frac{0.05683}{0.008106}=7.01\approx 7.$ Answer: ${\text{MgSO}}_4\cdot 7{\text{H}}_2\text{O}$ (Epsom salt).

Moles (from 100 g): $\text{N}:\frac{30.4}{14.01}=2.170,\text{O}:\frac{69.6}{16.00}=\mathrm{4.350.}$ Divide by smallest ($2.170$): $\text{N}:1.00$, $\text{O}:2.005\approx 2$. Empirical $=N{O}_2$, empirical mass $=14.01+2(16.00)=46.01$. $n$: $n=92.0/46.01=2.00\Rightarrow$ $N_2{O}_4$ (dinitrogen tetroxide). Subtlety: $N{O}_2$ also exists as a real molecule ($n=1$, molar mass $46$). So the same empirical formula corresponds to two different molecules — exactly why percent composition alone is not enough. Only the measured molar mass ($92$) tells us we have $N_2{O}_4$, not $N{O}_2$. This is the Law of Definite Proportions giving a fixed ratio, while molar mass fixes the actual size.

$N{O}_2$ from $N_2{O}_4$ given identical % composition? The measured molar mass. Both have empirical formula $N{O}_2$ (mass $46$); $n=M/46$ selects between them ($n=1\to N{O}_2$, $n=2\to N_2{O}_4$).

Because H atoms are ~12× lighter, so $13\text{g H}=12.9\text{mol}$ beats $52\text{g C}=4.35\text{mol}$. Atom count follows moles, not grams.