Determining Relative Molecular Mass from a Vapour · Chemistry 5th Year OL

The idea: count the moles of vapour

You can weigh a liquid easily. What you don't know is how many moles that mass contains. If you knew the moles, you would get the molar mass straight away:

$$M_r = \frac{\text{mass in grams}}{\text{moles}}$$

Here $M_r$ is the relative molecular mass, which has the same number as the molar mass in $\text{g mol}^{-1}$.

For a gas you can find the moles from the volume. Avogadro's law says that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. At room temperature and pressure (RTP), one mole of any gas occupies about $24\ \text{L}$. At STP (273 K and 101.325 kPa) it occupies $22.4\ \text{L}$.

So the plan is:

Weigh a small mass of a volatile liquid. This is a liquid that evaporates easily and has a low boiling point.
Turn all of it into vapour and measure the volume.
Work out the moles of vapour.
Divide the mass by the moles.

Practice★☆☆

At room temperature and pressure, one mole of any gas occupies approximately $24\ \text{L}$. A sample of a gas has a volume of $0.060\ \text{L}$ at room temperature and…

At room temperature and pressure, one mole of any gas occupies approximately $24\ \text{L}$. A sample of a gas has a volume of $0.060\ \text{L}$ at room temperature and pressure. Calculate the number of moles.

Solve

Practice★☆☆

A volatile liquid is vaporised in a graduated gas syringe. State the measurements needed to calculate its relative molecular mass, and name the equation used.

Solve

Using PV = nRT with the right units

At room temperature and pressure the "24 L per mole" shortcut works. In the laboratory the oven is hot, so you need the ideal gas equation:

$$PV = nRT$$

Rearranged for moles, and then for the relative molecular mass (with $n = m/M_r$):

$$n = \frac{PV}{RT} \qquad\qquad M_r = \frac{mRT}{PV}$$

The equation only works in SI units:

Quantity	Symbol	Unit you must use	Common conversion
Pressure	$P$	pascals (Pa)	$1\ \text{kPa} = 1{,}000\ \text{Pa}$
Volume	$V$	cubic metres ($\text{m}^3$)	$1\ \text{L} = 10^{-3}\ \text{m}^3$, $1\ \text{mL} = 10^{-6}\ \text{m}^3$
Temperature	$T$	kelvin (K)	$T/\text{K} = \theta/^\circ\text{C} + 273$
Amount	$n$	mol

The gas constant is $R = 8.31\ \text{J K}^{-1}\ \text{mol}^{-1}$.

Why kelvin? The volume of a gas is proportional to the temperature measured from absolute zero, not from the freezing point of water. At $0\ ^\circ\text{C}$ a gas does not have zero volume. Extrapolating the volume of an ideal gas back down the temperature scale gives zero volume at $0\ \text{K}$ (a real gas would liquefy before it got there).

Practice★☆☆

Convert each temperature to kelvin: a) $20\ ^\circ\text{C}$ b) $100\ ^\circ\text{C}$ c) $-10\ ^\circ\text{C}$

Convert each temperature to kelvin:

a) $20\ ^\circ\text{C}$
b) $100\ ^\circ\text{C}$
c) $-10\ ^\circ\text{C}$

Solve

Worked calculation

Finding the relative molecular mass from experimental data

A student injects $0.230\ \text{g}$ of a volatile liquid into a gas syringe in an oven at $100\ ^\circ\text{C}$. The vapour occupies $120\ \text{mL}$ at $1.00\times10^{5}\ \text{Pa}$. Calculate $M_r$.

Convert to SI units. $T = 100 + 273 = 373\ \text{K}$. $V = 120\ \text{mL} = 120\times10^{-6} = 1.20\times10^{-4}\ \text{m}^3$. $P = 1.00\times10^{5}\ \text{Pa}$.

Find the moles of vapour.

$$n = \frac{PV}{RT} = \frac{(1.00\times10^{5})(1.20\times10^{-4})}{8.31 \times 373} = \frac{12.0}{3{,}099.6} = 3.87\times10^{-3}\ \text{mol}$$

Divide mass by moles.

$$M_r = \frac{0.230}{3.87\times10^{-3}} = 59.4$$

Interpret. A molecule of $\text{C}_3\text{H}_8\text{O}$ has $M_r = 3(12.0)+8(1.0)+16.0 = 60.0$, so the liquid could be propan-1-ol or propan-2-ol. The experiment gives an approximate value and you match it to a formula.

Your turn: calculate the percentage error between the experimental value $59.4$ and the true value $60.0$.

$$\%\ \text{error} = \frac{|59.4 - 60.0|}{60.0}\times 100 = 1.0\%$$

Practice★★☆

A student vaporises $0.152\ \text{g}$ of a volatile liquid. The vapour occupies $60.0\ \text{mL}$ at $373\ \text{K}$ and $1.01\times10^{5}\ \text{Pa}$. Using…

A student vaporises $0.152\ \text{g}$ of a volatile liquid. The vapour occupies $60.0\ \text{mL}$ at $373\ \text{K}$ and $1.01\times10^{5}\ \text{Pa}$. Using $R = 8.31\ \text{J K}^{-1}\text{mol}^{-1}$, calculate:

a) the number of moles of vapour
b) the relative molecular mass of the liquid

Solve

Practice★★☆

A sample of propanone vapour of mass $0.116\ \text{g}$ occupies $50.0\ \text{mL}$ at $373\ \text{K}$ and $1.00\times10^{5}\ \text{Pa}$. a) Calculate the relative…

A sample of propanone vapour of mass $0.116\ \text{g}$ occupies $50.0\ \text{mL}$ at $373\ \text{K}$ and $1.00\times10^{5}\ \text{Pa}$.

a) Calculate the relative molecular mass.
b) Compare it with the value calculated from the formula $\mathrm{C_3H_6O}$ $(A_r$: C = 12.0, H = 1.0, O = 16.0$)$ and calculate the percentage error.

Solve

The experiment: getting a reliable result

In the laboratory the method is:

Fill a gas syringe and its holder so it is clean and the plunger moves freely. Record the starting volume (if any).
Heat the oven or gas-syringe holder to a steady temperature well above the boiling point of the liquid, and record it.
Weigh a small injecting syringe with the liquid. Inject a few drops through the septum, then reweigh. The difference is the mass of liquid.
Wait until the plunger stops moving, then read the volume of vapour. Record the atmospheric pressure from a barometer.
Calculate $n$ and then $M_r$.

Precautions and why they matter

Precaution	Why
Oven well above the boiling point	Ensures all the liquid is a vapour. If some stays liquid, the volume is too small.
Inject only a small mass	The vapour must fit inside the syringe and all of it must vaporise.
Wait for a steady reading at a constant temperature	The volume changes with temperature, so reading while it is still changing gives the wrong $V$ for the stated $T$.
Plunger lubricated and free to move; no leaks	Friction keeps the pressure in the syringe from equalling atmospheric pressure, and leaks lose vapour.
Mass found by weighing the syringe before and after	Avoids the loss of drops on the needle or septum.
Work in a fume cupboard, goggles on, no naked flames near the liquid	Many volatile liquids are flammable and the vapour is harmful.

Practice★★☆

In an experiment a small mass of volatile liquid is injected into a gas syringe in a heated oven. State two precautions that should be taken to ensure the result is…

In an experiment a small mass of volatile liquid is injected into a gas syringe in a heated oven. State two precautions that should be taken to ensure the result is reliable, and explain why each one matters.

Solve

Practice★★★

In an experiment to find the relative molecular mass of a liquid, a student failed to allow for the fact that a small portion of the liquid had not fully vaporised. a)…

In an experiment to find the relative molecular mass of a liquid, a student failed to allow for the fact that a small portion of the liquid had not fully vaporised.

a) State, with reasoning, whether the calculated relative molecular mass would be too high or too low.
b) The student also read the syringe volume while the oven was cooling slightly. State the effect of this error on the result.
c) Suggest one improvement to the procedure.

Solve

From $M_r$ to a formula, and how ideal is the vapour?

The experiment gives only an approximate $M_r$. To identify the liquid you combine it with its percentage composition. This is the method you used in Strand 1.4:

Divide each percentage by the relative atomic mass to get moles in $100\ \text{g}$.
Divide by the smallest value to get the simplest ratio. This gives the empirical formula.
Compare the empirical formula mass with $M_r$. The molecular formula is the empirical formula multiplied by $M_r/M_{r}(\text{empirical})$, rounded to a whole number.

Formula from composition and $M_r$

A liquid contains $52.2\%$ C, $13.0\%$ H and $34.8\%$ O by mass, and its experimental $M_r$ is $46.5$. Find its molecular formula.

Moles in 100 g. C: $52.2/12.0 = 4.35$. H: $13.0/1.0 = 13.0$. O: $34.8/16.0 = 2.175$.

Simplest ratio. Divide by $2.175$: C : H : O = $2 : 6 : 1$. The empirical formula is $\text{C}_2\text{H}_6\text{O}$.

Compare masses. The empirical formula mass is $2(12.0)+6(1.0)+16.0 = 46.0$. The ratio $46.5/46.0 \approx 1$, so the molecular formula is $\text{C}_2\text{H}_6\text{O}$ (ethanol).

Your turn: if the experimental $M_r$ had been $92$, what molecular formula would you get? Does it make sense?

$92/46 = 2$ would give $\text{C}_4\text{H}_{12}\text{O}_2$, but this molecule cannot exist: four carbon atoms can hold at most $2(4)+2 = 10$ hydrogens (as in $\text{C}_n\text{H}_{2n+2}$). $\text{C}_2\text{H}_6\text{O}$ is already saturated, so the only possible molecular formula is $\text{C}_2\text{H}_6\text{O}$. An experimental $M_r$ of $92$ would mean the measurement (or the composition data) is wrong.

Is the vapour really ideal?

An ideal gas has molecules with negligible volume and no attractive forces between them. A real vapour that is only a little above its boiling point does not match this well, because its molecules attract each other (intermolecular forces such as dipole–dipole attractions or hydrogen bonding). Real gases behave most ideally at high temperature and low pressure. This is why the oven is set well above the boiling point.

Because of these attractions the measured volume is slightly smaller than an ideal gas would have. The calculated $M_r$ is therefore slightly too high, and the result is an approximation of the true value.

Back to the bottle. Weigh a small mass of the unknown liquid, vaporise it in the syringe, and use $M_r = mRT/PV$. Combine the result with its composition from analysis and you can deduce the molecular formula. You have identified the liquid without a label.

Summary

A volatile liquid is vaporised completely in a gas syringe in a hot oven. You need its mass (weighing the syringe before and after), the volume of vapour, the temperature and the pressure.
Convert to SI units: $T$ in K ($\theta + 273$), $V$ in $\text{m}^3$, $P$ in Pa.
Use $PV = nRT$ with $R = 8.31\ \text{J K}^{-1}\text{mol}^{-1}$ to find $n$, then $M_r = m/n = mRT/PV$.
Incomplete vaporisation, a falling temperature or leaks make $V$ too small, so $M_r$ comes out too high.
Combine $M_r$ with percentage composition to get the empirical and molecular formulae.
The vapour is not perfectly ideal. Real gases are closest to ideal at high temperature and low pressure, so the result is only an approximation.

Let's practice! 7 exercises in this section

★☆☆ At room temperature and pressure, one mole of any gas occupies approximately…★☆☆ A volatile liquid is vaporised in a graduated gas syringe. State the measurements needed…★☆☆ Convert each temperature to kelvin: a) $20\ ^\circ\text{C}$ b) $100\ ^\circ\text{C}$ c)…