Percentage Composition, Empirical Formula and Molecular Formula

Why Percentage Composition Matters

You cannot see the atoms in an unknown powder, but you can burn it, dissolve it and weigh the products. Each experiment asks what fraction of the mass is carbon, hydrogen, oxygen. That fraction, per hundred, is the percentage composition. It links the balance to the formula.

Key Point (Definition): The mass per cent of an element in a compound is the mass of that element in one mole of the compound, divided by the molar mass, times 100: $$\text{Mass \% of an element} = \frac{\text{mass of that element in one mole of the compound}}{\text{molar mass of the compound}} \times 100$$

1. It is a mass ratio, not an atom ratio. Water has two H atoms per O atom, yet hydrogen is only about 11% of its mass.
2. The percentages must add to 100. Use this as a check.

First example: water

One mole of $\mathrm{H_2O}$ contains 2 mol of H atoms and 1 mol of O atoms.

• Molar mass $= 2 \times 1.008 + 16.00 = 18.02\ \mathrm{g\ mol^{-1}}$
• Mass of hydrogen in one mole $= 2 \times 1.008 = 2.016$ g
• Mass of oxygen in one mole $= 16.00$ g

$$\text{Mass \% of H} = \frac{2.016}{18.02} \times 100 = 11.18\%$$

$$\text{Mass \% of O} = \frac{16.00}{18.02} \times 100 = 88.79\%$$

Check: $11.18 + 88.79 = 99.97 \approx 100$; the shortfall is rounding. In 100 g of water, 11.18 g is hydrogen and 88.79 g is oxygen.

What the data is used for

Pure $\mathrm{CaCO_3}$ must be 40.04% calcium by mass. A sample measuring 38.0% is impure, and $38.0/40.04$ gives its purity.

[Board] "Define mass per cent and calculate it for hydrogen in water" is a standard 2-mark question: formula, substitution with units, answer to two decimals.

Percentage Composition from a Formula

Formula to percentage composition is the easier direction:

1. Count the atoms of each element in one formula unit.
2. Find the molar mass (atoms $\times$ atomic mass, summed).
3. Mass of each element in one mole $=$ atoms $\times$ atomic mass.
4. Divide by the molar mass, multiply by 100.
5. Check that the percentages add to 100.

Second example: ethanol

$\mathrm{C_2H_5OH}$ contains 6 hydrogen atoms (five in $\mathrm{C_2H_5}$, one in $\mathrm{OH}$), not five.

$$(2 \times 12.01) + (6 \times 1.008) + 16.00 = 24.02 + 6.048 + 16.00 = 46.068\ \mathrm{g\ mol^{-1}}$$

Reading the figure

In water, oxygen dominates (88.79%) because one O atom outweighs two H atoms eight to one. In ethanol, carbon leads (52.14%): two carbons, one oxygen. Percentage composition is a fingerprint of the compound, Proust's law of definite proportions in numbers.

Mass per cent versus atom per cent

Questions mean mass per cent unless they say otherwise.

The commonest error is miscounting atoms in a condensed formula. $\mathrm{CH_3COOH}$ has 2 C, 4 H, 2 O. $\mathrm{(NH_4)_2SO_4}$ has 2 N, 8 H, 1 S, 4 O. $\mathrm{CuSO_4 \cdot 5H_2O}$ has 1 Cu, 1 S, 9 O, 10 H. Expand the formula first.

[NEET] "Which has the highest percentage of nitrogen?" (urea, ammonium nitrate, ammonium sulphate, ammonium chloride): urea, $\mathrm{CO(NH_2)_2}$, at 46.7%.

Empirical Formula versus Molecular Formula

Now from percentages back to a formula. Percentages can give only the ratio of atoms.

Key Point (Definition): An empirical formula gives the simplest whole-number ratio of the atoms in a compound. A molecular formula gives the exact number of atoms of each type in one molecule.

Mass divided by atomic mass gives moles, so composition gives the atom ratio. But 1 : 2 : 1 fits $\mathrm{CH_2O}$, $\mathrm{C_2H_4O_2}$ and $\mathrm{C_6H_{12}O_6}$ equally: all three are 40.0% C, 6.7% H, 53.3% O. To fix the molecule you need the molar mass.

The relation between the two formulae

$$\text{Molecular formula} = n \times (\text{Empirical formula}), \qquad n = \frac{\text{Molar mass}}{\text{Empirical formula mass}}$$

The empirical formula mass is the sum of atomic masses in the empirical formula. $n$ counts empirical units per molecule, so it is a whole number. Round 2.02 to 2; 2.5 means an arithmetic slip.

Families that share an empirical formula

When the two formulae are the same

If the subscripts have no common factor other than 1, the molecular formula is the empirical formula and $n = 1$: $\mathrm{H_2O}$, $\mathrm{CO_2}$, $\mathrm{CH_4}$, $\mathrm{NH_3}$, $\mathrm{H_2SO_4}$, $\mathrm{CH_2O}$. Hydrogen peroxide $\mathrm{H_2O_2}$, benzene $\mathrm{C_6H_6}$ and glucose $\mathrm{C_6H_{12}O_6}$ are not.

Key Point: Ionic compounds such as $\mathrm{NaCl}$, $\mathrm{Na_2SO_4}$ or $\mathrm{Fe_2O_3}$ have no discrete molecules, so their formula is always the empirical formula (formula unit). Exercises that ask for the "molecular formula" of iron oxide expect $\mathrm{Fe_2O_3}$ with $n = 1$.

[JEE Main] "Which pair has the same empirical formula?" Look for the same simplest ratio: benzene and acetylene, glucose and acetic acid, $\mathrm{NO_2}$ and $\mathrm{N_2O_4}$. And "empirical formula $\mathrm{CH_2}$, molar mass 42": $n = 42/14 = 3$, so $\mathrm{C_3H_6}$.

The Five-Step Recipe

Key Point (The recipe):
Step 1 Take 100 g of the compound, so mass per cent becomes grams.
Step 2 Divide each mass by the element's atomic mass to get moles.
Step 3 Divide every mole value by the smallest to get the simplest ratio; if it is not whole, multiply through by a small integer.
Step 4 Write the empirical formula with those whole numbers as subscripts.
Step 5 Find the empirical formula mass, compute $n =$ molar mass / empirical formula mass, and multiply the empirical formula by $n$.

A worked problem

A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. Find its empirical and molecular formulas.

Step 1. In 100 g: 4.07 g H, 24.27 g C, 71.65 g Cl. They add to 99.99, so there is no fourth element.

Step 2.

$$\text{Moles of H} = \frac{4.07\ \mathrm{g}}{1.008\ \mathrm{g\ mol^{-1}}} = 4.04$$

$$\text{Moles of C} = \frac{24.27\ \mathrm{g}}{12.01\ \mathrm{g\ mol^{-1}}} = 2.021$$

$$\text{Moles of Cl} = \frac{71.65\ \mathrm{g}}{35.453\ \mathrm{g\ mol^{-1}}} = 2.021$$

Step 3. Divide by the smallest, 2.021.

Step 4. H : C : Cl $=$ 2 : 1 : 1, so the empirical formula is $\mathrm{CH_2Cl}$.

Step 5. Empirical formula mass:

$$12.01 + (2 \times 1.008) + 35.453 = 49.48\ \mathrm{g}$$

$$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{98.96\ \mathrm{g}}{49.48\ \mathrm{g}} = 2$$

Molecular formula: $\mathrm{(CH_2Cl)_2} = \mathrm{C_2H_4Cl_2}$.

Answer: Empirical formula $\mathrm{CH_2Cl}$; molecular formula $\mathrm{C_2H_4Cl_2}$ (1,2- or 1,1-dichloroethane; composition cannot distinguish isomers).

When Step 3 does not give whole numbers

A ratio like 1 : 1.5 means 2 : 3. Never round 1.5 to 2 or 1.33 to 1. Multiply the set by the smallest integer that clears the fraction:

Round only small deviations: 1.98 is 2, 1.03 is 1, but 1.5 stays 1.5.

[JEE Main] Rounded atomic masses (H = 1, C = 12, O = 16, Cl = 35.5) are fine in Steps 2 and 3 (here H 4.07, C 2.02, Cl 2.02, still 2 : 1 : 1). Use precise values in Step 5 when the molar mass has three or four significant figures.

Where the Molar Mass Comes From, and Combustion Analysis

Problems supply the molar mass and the mass percentages in disguised forms.

Three ways a problem gives the molar mass

[JEE Main] $M = 2 \times \mathrm{VD}$ holds because VD is the gas density relative to $\mathrm{H_2}$ (molar mass 2) at the same temperature and pressure.

Combustion analysis

A weighed sample is burned in excess oxygen and the masses of $\mathrm{CO_2}$ and $\mathrm{H_2O}$ formed are measured.

• All the carbon ends up in $\mathrm{CO_2}$, and carbon is $\dfrac{12.01}{44.01}$ of its mass: $$\text{mass of C} = \frac{12.01}{44.01} \times m(\mathrm{CO_2}), \qquad \text{moles of C} = \text{moles of } \mathrm{CO_2} = \frac{m(\mathrm{CO_2})}{44.01}$$
• All the hydrogen ends up in $\mathrm{H_2O}$, hydrogen is $\dfrac{2.016}{18.02}$ of its mass, and each molecule carries two H atoms: $$\text{mass of H} = \frac{2.016}{18.02} \times m(\mathrm{H_2O}), \qquad \text{moles of H} = 2 \times \frac{m(\mathrm{H_2O})}{18.02}$$
• Oxygen cannot be read from the products (the supply was in excess), so it is found by difference: $$\text{mass of O} = \text{mass of sample} - \text{mass of C} - \text{mass of H}$$

Key Point: Moles of C $=$ moles of $\mathrm{CO_2}$; moles of H $= 2 \times$ moles of $\mathrm{H_2O}$. Going straight to moles skips the percentages.

The welding-gas problem in outline

A C, H fuel gas gives 3.38 g $\mathrm{CO_2}$ and 0.690 g $\mathrm{H_2O}$; 10.0 L at STP weighs 11.6 g. Moles of C $= 3.38/44.01 = 0.0768$; moles of H $= 2 \times 0.690/18.02 = 0.0766$; ratio 1 : 1, so $\mathrm{CH}$. $M = (11.6/10.0) \times 22.7 = 26.3\ \mathrm{g\ mol^{-1}}$, $n = 26.3/13.02 \approx 2$, molecular formula $\mathrm{C_2H_2}$ (acetylene). Full working in Question 12.

"A hydrocarbon has 92.3% carbon; find its empirical formula": $92.3/12 = 7.69$, $7.7/1 = 7.7$, ratio 1 : 1, $\mathrm{CH}$. Fingerprints: 92.3% C is $\mathrm{CH}$; 85.7% is $\mathrm{CH_2}$; 80.0% is $\mathrm{CH_3}$; 75.0% is $\mathrm{CH_4}$.

Purity Checks and Common Mistakes

Composition as a purity test

A pure compound has a fixed mass per cent of each element (Proust's law). An impure sample has less of the element per gram, if the impurity contains none of it:

$$\text{Purity (\%)} = \frac{\text{measured mass \% of the element in the sample}}{\text{theoretical mass \% of the element in the pure compound}} \times 100$$

Equivalent form: mass of pure compound in the sample $= \dfrac{\text{mass of element found}}{\text{mass fraction of the element in the pure compound}}$.

Pure $\mathrm{CaCO_3}$ is 40.04% Ca. Limestone with 38.0% Ca is $\dfrac{38.0}{40.04} \times 100 = 94.9\%$ calcium carbonate, if the impurity (silica or clay) has no calcium.

Extraction problems are the same calculation: copper from 100 g of $\mathrm{CuSO_4}$ is $63.5/159.6 \times 100 = 39.8$ g.

Checklist

Four traps

1. Stopping at the empirical formula. If a molar mass is given, the molecular formula is wanted.
2. Forgetting the factor of 2 for hydrogen in combustion analysis: one $\mathrm{H_2O}$ has two H atoms.
3. Reading mass ratio as atom ratio. 24.27 g C and 71.65 g Cl look like 1 : 3 but are 1 : 1 in atoms. Convert to moles first.
4. Sloppy significant figures. Keep three or four figures through the working; rounding 2.021 to 2 early can hide a 1 : 1.5 ratio.

Key Point: Composition gives ratio; ratio plus molar mass gives molecule.

[Board] Typical 3-mark question: "A compound contains 40.0% C, 6.7% H and 53.3% O and its molar mass is 60 g mol$^{-1}$. Find its empirical and molecular formula." $\mathrm{CH_2O}$, then $n = 60/30 = 2$, so $\mathrm{C_2H_4O_2}$ (acetic acid). Show all five steps for step marks.

Solved Examples

Question 1: Mass per cent of hydrogen and oxygen in water

Calculate the mass per cent of hydrogen and oxygen in water, $\mathrm{H_2O}$. Atomic masses: H = 1.008, O = 16.00.

Answer: Molar mass $= 2 \times 1.008 + 16.00 = 18.02\ \mathrm{g\ mol^{-1}}$. One mole holds $2 \times 1.008 = 2.016$ g of H and 16.00 g of O.

$$\text{Mass \% of H} = \frac{2.016}{18.02} \times 100 = 11.18\%, \qquad \text{Mass \% of O} = \frac{16.00}{18.02} \times 100 = 88.79\%$$

Check: $11.18 + 88.79 = 99.97\%$, fine within rounding.

Ans: Hydrogen 11.18%, oxygen 88.79%.

Question 2: Percentage composition of ethanol

Find the mass per cent of carbon, hydrogen and oxygen in ethanol, $\mathrm{C_2H_5OH}$.

Answer: Atoms: 2 C, 6 H (5 + 1), 1 O. Molar mass $= (2 \times 12.01) + (6 \times 1.008) + 16.00 = 24.02 + 6.048 + 16.00 = 46.068\ \mathrm{g\ mol^{-1}}$.

Carbon: $\dfrac{24.02}{46.068} \times 100 = 52.14\%$. Hydrogen: $\dfrac{6.048}{46.068} \times 100 = 13.13\%$. Oxygen: $\dfrac{16.00}{46.068} \times 100 = 34.73\%$.

Ans: C 52.14%, H 13.13%, O 34.73%.

Watch out: Count the hydroxyl hydrogen; $\mathrm{C_2H_5OH}$ has 6 H, not 5.

Question 3: Mass per cent of elements in sodium sulphate

Calculate the mass per cent of the different elements present in sodium sulphate, $\mathrm{Na_2SO_4}$. (Na = 23.0, S = 32.1, O = 16.00)

Answer: Molar mass $= 2 \times 23.0 + 32.1 + 4 \times 16.00 = 46.0 + 32.1 + 64.0 = 142.1\ \mathrm{g\ mol^{-1}}$.

Sodium: $\dfrac{46.0}{142.1} \times 100 = 32.37\%$. Sulphur: $\dfrac{32.1}{142.1} \times 100 = 22.59\%$. Oxygen: $\dfrac{64.0}{142.1} \times 100 = 45.04\%$.

Check: $32.37 + 22.59 + 45.04 = 100.00\%$.

Ans: Na 32.37%, S 22.59%, O 45.04% (an answer key using other atomic masses gives 32.38%, 22.57%, 45.05%; rounding only).

Question 4: Copper from copper sulphate

How much copper can be obtained from 100 g of copper sulphate, $\mathrm{CuSO_4}$? (Cu = 63.5, S = 32.1, O = 16.00)

Answer: This is the mass per cent of Cu in $\mathrm{CuSO_4}$. Molar mass $= 63.5 + 32.1 + 4 \times 16.00 = 159.6\ \mathrm{g\ mol^{-1}}$; 159.6 g contains 63.5 g of Cu. For 100 g:

$$\text{Cu obtainable} = \frac{63.5}{159.6} \times 100\ \mathrm{g} = 39.8\ \mathrm{g}$$

Ans: About 39.8 g of copper (39.81 g with S = 32.0).

Question 5: The reverse problem, and a family resemblance

Calculate the percentage composition of glucose, $\mathrm{C_6H_{12}O_6}$, and compare it with that of acetic acid, $\mathrm{CH_3COOH}$. What do you conclude?

Answer: Glucose: molar mass $= 6 \times 12.01 + 12 \times 1.008 + 6 \times 16.00 = 72.06 + 12.096 + 96.00 = 180.16\ \mathrm{g\ mol^{-1}}$. C $= 72.06/180.16 \times 100 = 40.00\%$; H $= 12.096/180.16 \times 100 = 6.71\%$; O $= 96.00/180.16 \times 100 = 53.29\%$.

Acetic acid: $\mathrm{CH_3COOH} = \mathrm{C_2H_4O_2}$, molar mass $= 24.02 + 4.032 + 32.00 = 60.05\ \mathrm{g\ mol^{-1}}$. C $= 24.02/60.05 \times 100 = 40.00\%$; H $= 4.032/60.05 \times 100 = 6.71\%$; O $= 32.00/60.05 \times 100 = 53.29\%$.

Identical to two decimal places. Both reduce to $\mathrm{CH_2O}$: glucose is $(\mathrm{CH_2O})_6$, acetic acid $(\mathrm{CH_2O})_2$.

Ans: Both are 40.00% C, 6.71% H, 53.29% O; composition alone cannot separate compounds with the same empirical formula.

Question 6: From percentages to $\mathrm{C_2H_4Cl_2}$

A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. Find its empirical and molecular formulae.

Answer: In 100 g: 4.07 g H, 24.27 g C, 71.65 g Cl. Moles: H $= 4.07/1.008 = 4.04$; C $= 24.27/12.01 = 2.021$; Cl $= 71.65/35.453 = 2.021$.

Dividing by 2.021: H $= 2.00$, C $= 1.00$, Cl $= 1.00$, so the empirical formula is $\mathrm{CH_2Cl}$.

Empirical formula mass $= 12.01 + 2 \times 1.008 + 35.453 = 49.48$ g; $n = 98.96/49.48 = 2$; molecular formula $(\mathrm{CH_2Cl})_2 = \mathrm{C_2H_4Cl_2}$.

Ans: Empirical formula $\mathrm{CH_2Cl}$; molecular formula $\mathrm{C_2H_4Cl_2}$.

Question 7: Oxide of iron

An oxide of iron contains 69.9% iron and 30.1% oxygen by mass. (i) Determine its empirical formula. (ii) If its molar mass is 159.7 g mol$^{-1}$, determine its molecular formula. (Fe = 55.85, O = 16.00)

Answer: In 100 g: 69.9 g Fe, 30.1 g O. Moles: Fe $= 69.9/55.85 = 1.252$; O $= 30.1/16.00 = 1.881$.

Dividing by 1.252: Fe $= 1.00$; O $= 1.881/1.252 = 1.50$. Not whole, so I multiply both by 2: Fe : O $= 2 : 3$, $\mathrm{Fe_2O_3}$.

Empirical formula mass $= 2 \times 55.85 + 3 \times 16.00 = 159.7$ g; $n = 159.7/159.7 = 1$; molecular formula also $\mathrm{Fe_2O_3}$.

Ans: (i) $\mathrm{Fe_2O_3}$; (ii) $\mathrm{Fe_2O_3}$ ($n = 1$).

Watch out: Multiply 1.5 by 2, never round it; rounding gives the non-existent $\mathrm{FeO_2}$.

Question 8: A ratio that needs clearing

A green oxide of chromium contains 68.4% chromium and 31.6% oxygen. Find its empirical formula. (Cr = 52.0, O = 16.00)

Answer: Moles in 100 g: Cr $= 68.4/52.0 = 1.315$; O $= 31.6/16.00 = 1.975$.

Dividing by 1.315: Cr $= 1.00$; O $= 1.975/1.315 = 1.50$. Times 2: Cr : O $= 2 : 3$.

Reverse check: $\mathrm{Cr_2O_3}$ has molar mass $104.0 + 48.0 = 152.0$; mass % Cr $= 104.0/152.0 \times 100 = 68.4\%$, matching the data.

Ans: $\mathrm{Cr_2O_3}$ (chromium(III) oxide).

Question 9: Hydrocarbon from percentage and molar mass

A gaseous hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass and has a molar mass of 56 g mol$^{-1}$. Find its empirical and molecular formulae.

Answer: Moles in 100 g: C $= 85.7/12.01 = 7.14$; H $= 14.3/1.008 = 14.19$.

Dividing by 7.14: C $= 1.00$; H $= 14.19/7.14 = 1.99 \approx 2$. Empirical formula $\mathrm{CH_2}$, mass $= 12.01 + 2.016 = 14.03$ g. $n = 56/14.03 = 3.99 \approx 4$, so the molecular formula is $(\mathrm{CH_2})_4 = \mathrm{C_4H_8}$.

Ans: Empirical formula $\mathrm{CH_2}$, molecular formula $\mathrm{C_4H_8}$ (butene or cyclobutane).

Question 10: A purity check from percentage composition

A sample of limestone is analysed and found to contain 38.0% calcium by mass. Assuming the impurities contain no calcium, find the percentage purity of the limestone as $\mathrm{CaCO_3}$. (Ca = 40.08, C = 12.01, O = 16.00)

Answer: Pure $\mathrm{CaCO_3}$: molar mass $= 40.08 + 12.01 + 3 \times 16.00 = 100.09$; mass % Ca $= 40.08/100.09 \times 100 = 40.04\%$. The sample has only 38.0% Ca:

$$\text{Purity} = \frac{38.0}{40.04} \times 100 = 94.9\%$$

Cross-check: 38.0 g Ca belongs to $38.0/0.4004 = 94.9$ g of $\mathrm{CaCO_3}$.

Ans: The limestone is about 94.9% $\mathrm{CaCO_3}$.

Question 11: Combustion analysis of a compound containing oxygen

1.80 g of an organic compound containing only carbon, hydrogen and oxygen is burnt completely in excess oxygen to give 2.64 g of $\mathrm{CO_2}$ and 1.08 g of $\mathrm{H_2O}$. The molar mass of the compound is 180 g mol$^{-1}$. Find its empirical and molecular formulae.

Answer: Carbon from $\mathrm{CO_2}$: mass of C $= \dfrac{12.01}{44.01} \times 2.64 = 0.720$ g; moles of C $= 0.720/12.01 = 0.0600$ mol.

Hydrogen from $\mathrm{H_2O}$: mass of H $= \dfrac{2.016}{18.02} \times 1.08 = 0.121$ g; moles of H $= 0.121/1.008 = 0.120$ mol.

Oxygen by difference: mass of O $= 1.80 - 0.720 - 0.121 = 0.959$ g; moles of O $= 0.959/16.00 = 0.0599$ mol.

Dividing by 0.0599: C $= 1.00$, H $= 2.00$, O $= 1.00$, so $\mathrm{CH_2O}$ (mass 30.03). $n = 180/30.03 = 5.99 \approx 6$; molecular formula $\mathrm{C_6H_{12}O_6}$.

Ans: Empirical formula $\mathrm{CH_2O}$; molecular formula $\mathrm{C_6H_{12}O_6}$ (glucose).

Watch out: Oxygen cannot be read from the products (the supply was in excess); subtract C and H from the sample mass.

Question 12: The welding gas (Textbook Exercise 1.34)

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) the empirical formula, (ii) the molar mass of the gas, and (iii) the molecular formula.

Answer: All the carbon is in $\mathrm{CO_2}$: moles of C $= 3.38/44.01 = 0.0768$ mol (mass $= 3.38 \times 12.01/44.01 = 0.922$ g). Each $\mathrm{H_2O}$ carries two H: moles of H $= 2 \times 0.690/18.02 = 0.0766$ mol (mass $= 0.690 \times 2.016/18.02 = 0.0772$ g). C : H $= 0.0768 : 0.0766 = 1 : 1$, so the empirical formula is $\mathrm{CH}$.

Molar mass: 10.0 L at STP weighs 11.6 g and one mole occupies 22.7 L at STP (1 bar):

$$M = \frac{11.6\ \mathrm{g}}{10.0\ \mathrm{L}} \times 22.7\ \mathrm{L\ mol^{-1}} = 26.3\ \mathrm{g\ mol^{-1}}$$

(With 22.4 L, $M = 26.0\ \mathrm{g\ mol^{-1}}$.) Empirical formula mass of $\mathrm{CH} = 12.01 + 1.008 = 13.02$; $n = 26.3/13.02 = 2.02 \approx 2$; molecular formula $(\mathrm{CH})_2 = \mathrm{C_2H_2}$.

Ans: (i) $\mathrm{CH}$; (ii) about 26 g mol$^{-1}$ (26.3 with 22.7 L, 26.0 with 22.4 L); (iii) $\mathrm{C_2H_2}$, acetylene (ethyne).