Volumes of Gases and Concentrations of Solutions

AS · 10 min

Most reactions in the laboratory involve gases or solutions, which are measured by volume rather than mass. This note extends the mole method to gas volumes (using the molar volume) and to solutions (using concentration), and then applies both to the calculations Cambridge sets most often: gas volumes in combustion, titrations, dilutions and back titrations. These are core Paper 2 and Paper 3 skills, and titration calculations are almost guaranteed in Paper 3.

Volumes of gases

At the same temperature and pressure, equal volumes of all gases contain equal numbers of molecules (Avogadro's law). It does not matter whether the gas is hydrogen or carbon dioxide: the molecules are so far apart that their own size is irrelevant. So one mole of any gas occupies the same volume under the same conditions. This is the molar volume, $V_m$ .

Key result

n = \frac{V}{V_m}

$V_m = 24.0\ \text{dm}^3\ \text{mol}^{-1}$ at room conditions (about 298 K and 101 kPa).
$V_m = 22.4\ \text{dm}^3\ \text{mol}^{-1}$ at s.t.p. (273 K and 101 kPa).

Both values are in the data booklet. $1\ \text{dm}^3 = 1000\ \text{cm}^3$ .

Volume of gas from a mass of solid

Calculate the volume of carbon dioxide, measured at room conditions, produced when $5.00\ \text{g}$ of calcium carbonate reacts with excess hydrochloric acid.

\ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g)}

Solution

$n(\ce{CaCO3}) = \dfrac{5.00}{100.1} = 0.04995\ \text{mol}$

Ratio $1 : 1$ , so $n(\ce{CO2}) = 0.04995\ \text{mol}$ .

$V = 0.04995 \times 24.0 = 1.20\ \text{dm}^3$ (or $1200\ \text{cm}^3$ ).

Reacting volumes of gases

Because volume is proportional to moles for gases under the same conditions, the ratio of gas volumes equals the ratio of coefficients in the equation. You do not need the molar volume at all.

\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}

$50\ \text{cm}^3$ of methane needs $100\ \text{cm}^3$ of oxygen and makes $50\ \text{cm}^3$ of carbon dioxide. If the volumes are measured at room temperature, the water is a liquid and its volume is negligible.

Finding the formula of a hydrocarbon from gas volumes

A classic question: a hydrocarbon $\ce{C_xH_y}$ is burnt in excess oxygen, the products are cooled to room temperature (so water condenses), and the remaining gas is passed through aqueous sodium hydroxide or potassium hydroxide, which absorbs carbon dioxide.

Key result

\ce{C_xH_y + (x + $\frac{y}{4}$)O2 -> xCO2 + $\frac{y}{2}$H2O}

Contraction on passing through alkali $=$ volume of $\ce{CO2}$ $= x \times$ volume of hydrocarbon.
Volume of oxygen used $= \left(x + \tfrac{y}{4}\right) \times$ volume of hydrocarbon.

Exam-style: identifying a hydrocarbon

$20\ \text{cm}^3$ of a gaseous hydrocarbon was exploded with $150\ \text{cm}^3$ of oxygen (an excess). After cooling to room temperature, the gas volume was $110\ \text{cm}^3$ . After passing through aqueous sodium hydroxide, $50\ \text{cm}^3$ of gas remained. All volumes were measured at the same temperature and pressure. Find the molecular formula of the hydrocarbon.

Solution

The sodium hydroxide absorbs $\ce{CO2}$ : volume of $\ce{CO2} = 110 - 50 = 60\ \text{cm}^3$ .

The $50\ \text{cm}^3$ left is unreacted oxygen, so oxygen used $= 150 - 50 = 100\ \text{cm}^3$ .

Divide by the volume of hydrocarbon (20 cm³) to get the mole ratio:

$\ce{CO2}$ : $\tfrac{60}{20} = 3$ , so $x = 3$ .
$\ce{O2}$ : $\tfrac{100}{20} = 5 = x + \tfrac{y}{4} = 3 + \tfrac{y}{4}$ , so $y = 8$ .

The hydrocarbon is $\ce{C3H8}$ , propane: $\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)}$ .

Concentration of solutions

Definition

The concentration of a solution is the amount of solute dissolved in $1\ \text{dm}^3$ of solution. It is usually given in $\text{mol dm}^{-3}$ ; mass concentration is given in $\text{g dm}^{-3}$ .

Key result

c = \frac{n}{V} \qquad n = c \times V

$c$ in $\text{mol dm}^{-3}$ , $n$ in mol, $V$ in $\text{dm}^3$ . When the volume is in $\text{cm}^3$ , use $n = \dfrac{c \times V}{1000}$ .

Mass concentration ( $\text{g dm}^{-3}$ ) $=$ concentration ( $\text{mol dm}^{-3}$ ) $\times$ molar mass.

Making up a solution

$4.00\ \text{g}$ of sodium hydroxide is dissolved in water and made up to $250\ \text{cm}^3$ . Calculate the concentration in $\text{mol dm}^{-3}$ and in $\text{g dm}^{-3}$ .

Solution

$n(\ce{NaOH}) = \dfrac{4.00}{40.0} = 0.100\ \text{mol}$

$c = \dfrac{0.100}{0.250} = 0.400\ \text{mol dm}^{-3}$

Mass concentration $= 0.400 \times 40.0 = 16.0\ \text{g dm}^{-3}$ (equivalently $4.00\ \text{g} / 0.250\ \text{dm}^3$ ).

Dilution

Adding water changes the volume but not the amount of solute.

Key result

c_1 V_1 = c_2 V_2

For example, $25.0\ \text{cm}^3$ of $2.00\ \text{mol dm}^{-3}$ acid made up to $250\ \text{cm}^3$ has concentration $\dfrac{2.00 \times 25.0}{250} = 0.200\ \text{mol dm}^{-3}$ : a tenfold dilution.

Titration calculations

In a titration, a solution of known concentration (the standard solution) is added from a burette to a measured volume of another solution (from a pipette) until the reaction is just complete, shown by an indicator. The volume added from the burette is the titre.

Method

Titration calculation

Write the balanced equation and note the mole ratio.
Calculate the moles of the substance whose concentration and volume you know: $n = cV / 1000$ .
Use the ratio to find the moles of the other substance in the volume pipetted.
If the pipetted volume was taken from a larger volumetric flask, scale up (for example $\times 10$ for 25.0 cm³ from 250 cm³).
Calculate the required quantity: concentration ( $n / V$ ), molar mass ( $m / n$ ) or percentage purity.

Acid–alkali titration

$25.0\ \text{cm}^3$ of sodium hydroxide solution is neutralised by $21.30\ \text{cm}^3$ of $0.100\ \text{mol dm}^{-3}$ sulfuric acid. Calculate the concentration of the sodium hydroxide.

\ce{2NaOH(aq) + H2SO4(aq) -> Na2SO4(aq) + 2H2O(l)}

Solution

$n(\ce{H2SO4}) = \dfrac{0.100 \times 21.30}{1000} = 2.130 \times 10^{-3}\ \text{mol}$

Ratio $\ce{NaOH} : \ce{H2SO4} = 2 : 1$ , so $n(\ce{NaOH}) = 4.260 \times 10^{-3}\ \text{mol}$ in $25.0\ \text{cm}^3$ .

c(\ce{NaOH}) = \frac{4.260 \times 10^{-3}}{0.0250} = 0.170\ \text{mol dm}^{-3}

Exam-style: identifying a metal from a titration

$1.38\ \text{g}$ of a Group 1 carbonate, $\ce{M2CO3}$ , is dissolved in water and made up to $250\ \text{cm}^3$ in a volumetric flask. $25.0\ \text{cm}^3$ portions are titrated with $0.100\ \text{mol dm}^{-3}$ hydrochloric acid; the mean titre is $20.00\ \text{cm}^3$ . Identify M.

\ce{M2CO3(aq) + 2HCl(aq) -> 2MCl(aq) + H2O(l) + CO2(g)}

Solution

$n(\ce{HCl}) = \dfrac{0.100 \times 20.00}{1000} = 2.000 \times 10^{-3}\ \text{mol}$

$n(\ce{M2CO3})$ in $25.0\ \text{cm}^3 = 1.000 \times 10^{-3}\ \text{mol}$ .

In $250\ \text{cm}^3$ : $1.000 \times 10^{-2}\ \text{mol}$ .

$M(\ce{M2CO3}) = \dfrac{1.38}{0.01000} = 138\ \text{g mol}^{-1}$

$2A_r(\ce{M}) = 138 - 60.0 = 78$ , so $A_r(\ce{M}) = 39$ . M is potassium.

Back titrations

A back titration is used when the substance being analysed is insoluble or reacts too slowly to titrate directly (for example, calcium carbonate in limestone). A known excess of a reagent is added; after the reaction, the amount of excess left is found by titration. The difference is the amount that reacted.

Exam-hard: purity by back titration

A $1.00\ \text{g}$ sample of impure calcium carbonate is added to $50.0\ \text{cm}^3$ of $0.500\ \text{mol dm}^{-3}$ hydrochloric acid (an excess). When the reaction has finished, the excess acid needs $40.10\ \text{cm}^3$ of $0.200\ \text{mol dm}^{-3}$ sodium hydroxide for neutralisation. Calculate the percentage purity of the calcium carbonate. Assume the impurities do not react with acid.

Solution

Acid added: $n(\ce{HCl}) = \dfrac{0.500 \times 50.0}{1000} = 0.02500\ \text{mol}$

Excess acid: $n(\ce{NaOH}) = \dfrac{0.200 \times 40.10}{1000} = 0.008020\ \text{mol}$ , and $\ce{HCl + NaOH -> NaCl + H2O}$ is $1 : 1$ , so excess $n(\ce{HCl}) = 0.008020\ \text{mol}$ .

Acid that reacted with the carbonate $= 0.02500 - 0.008020 = 0.01698\ \text{mol}$ .

$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$ , so $n(\ce{CaCO3}) = \dfrac{0.01698}{2} = 0.008490\ \text{mol}$ .

Mass $= 0.008490 \times 100.1 = 0.850\ \text{g}$ .

\text{purity} = \frac{0.850}{1.00} \times 100 = 85.0\%

Watch out

Convert $\text{cm}^3$ to $\text{dm}^3$ by dividing by 1000. Forgetting this gives answers 1000 times too big.
Apply the mole ratio in the right direction. With $\ce{H2SO4}$ and $\ce{NaOH}$ , the alkali has twice the moles of the acid, not half.
Remember the scale-up factor when only a portion of a made-up solution is titrated.

Exam tip

Write $n = cV/1000$ with the numbers substituted; each line of working earns credit.
Give titration-derived concentrations to 3 or 4 significant figures, matching the data (a titre of 21.30 cm³ is 4 s.f.; a concentration of 0.100 is 3 s.f.).
For gas-volume questions, state the conditions you assumed (room temperature and pressure) and the molar volume used.

Summary

One mole of any gas occupies $24.0\ \text{dm}^3$ at room conditions and $22.4\ \text{dm}^3$ at s.t.p.: $n = V / V_m$ .
For gases at the same temperature and pressure, volume ratio = mole ratio.
In combustion analysis by volume: contraction with alkali gives $\ce{CO2}$ ; oxygen used gives $x + y/4$ .
$c = n / V$ ( $\text{mol dm}^{-3}$ ); $n = cV/1000$ with $V$ in $\text{cm}^3$ . Dilution: $c_1V_1 = c_2V_2$ .
Titrations: moles of the known, ratio, moles of the unknown, scale up if necessary.
Back titration: amount reacted = amount added minus excess found by titration.

Practice

Question

Calculate the volume at room conditions of $0.0500\ \text{mol}$ of hydrogen, and the amount in moles of $600\ \text{cm}^3$ of oxygen at room conditions.
Calculate the mass of magnesium that reacts with excess acid to give $240\ \text{cm}^3$ of hydrogen at room conditions: $\ce{Mg + 2HCl -> MgCl2 + H2}$ .
Calculate the concentration of a solution made by dissolving $2.65\ \text{g}$ of anhydrous sodium carbonate in water and making it up to $250\ \text{cm}^3$ .
What volume of $0.500\ \text{mol dm}^{-3}$ hydrochloric acid reacts exactly with $1.00\ \text{g}$ of calcium carbonate?
$10\ \text{cm}^3$ of a gaseous hydrocarbon was exploded with $100\ \text{cm}^3$ of oxygen. After cooling to room temperature the volume was $75\ \text{cm}^3$ , and after passing through aqueous potassium hydroxide it was $35\ \text{cm}^3$ . Find the molecular formula.
$25.0\ \text{cm}^3$ of ethanoic acid is neutralised by $23.50\ \text{cm}^3$ of $0.120\ \text{mol dm}^{-3}$ sodium hydroxide. Calculate the concentration of the ethanoic acid.
What volume of $0.800\ \text{mol dm}^{-3}$ sulfuric acid is needed to make $500\ \text{cm}^3$ of $0.0500\ \text{mol dm}^{-3}$ sulfuric acid?
$7.15\ \text{g}$ of hydrated sodium carbonate, $\ce{Na2CO3.xH2O}$ , is made up to $250\ \text{cm}^3$ . $25.0\ \text{cm}^3$ of this solution needs $25.00\ \text{cm}^3$ of $0.200\ \text{mol dm}^{-3}$ $\ce{HCl}$ to react completely: $\ce{Na2CO3 + 2HCl -> 2NaCl + H2O + CO2}$ . Find $x$ .
Calculate the total volume of gas, at room conditions, produced when $4.10\ \text{g}$ of calcium nitrate decomposes completely: $\ce{2Ca(NO3)2(s) -> 2CaO(s) + 4NO2(g) + O2(g)}$ .
A $0.500\ \text{g}$ sample of fertiliser containing ammonium sulfate is warmed with $50.0\ \text{cm}^3$ of $0.200\ \text{mol dm}^{-3}$ $\ce{NaOH}$ (an excess) until no more ammonia is given off: $\ce{NH4+ + OH- -> NH3 + H2O}$ . The remaining $\ce{NaOH}$ needs $26.00\ \text{cm}^3$ of $0.100\ \text{mol dm}^{-3}$ $\ce{HCl}$ . Calculate the percentage by mass of $\ce{(NH4)2SO4}$ in the fertiliser.

Answers

$V = 0.0500 \times 24.0 = 1.20\ \text{dm}^3$ . $n = 0.600 / 24.0 = 0.0250\ \text{mol}$ .
$n(\ce{H2}) = 0.240 / 24.0 = 0.0100\ \text{mol} = n(\ce{Mg})$ ; mass $= 0.0100 \times 24.3 = 0.243\ \text{g}$ .
$n = 2.65 / 106.0 = 0.0250\ \text{mol}$ ; $c = 0.0250 / 0.250 = 0.100\ \text{mol dm}^{-3}$ .
$n(\ce{CaCO3}) = 1.00 / 100.1 = 9.99 \times 10^{-3}\ \text{mol}$ ; $n(\ce{HCl}) = 2 \times 9.99 \times 10^{-3} = 0.01998\ \text{mol}$ ; $V = 0.01998 / 0.500 = 0.0400\ \text{dm}^3 = 40.0\ \text{cm}^3$ .
$\ce{CO2} = 75 - 35 = 40\ \text{cm}^3$ , so $x = 40 / 10 = 4$ . Oxygen used $= 100 - 35 = 65\ \text{cm}^3$ , so $x + y/4 = 6.5$ and $y = 10$ . $\ce{C4H10}$ .
$n(\ce{NaOH}) = 0.120 \times 23.50 / 1000 = 2.820 \times 10^{-3}\ \text{mol}$ (1 : 1 with $\ce{CH3COOH}$ ); $c = 2.820 \times 10^{-3} / 0.0250 = 0.113\ \text{mol dm}^{-3}$ .
$V_1 = c_2V_2 / c_1 = 0.0500 \times 500 / 0.800 = 31.3\ \text{cm}^3$ .
$n(\ce{HCl}) = 0.200 \times 25.00 / 1000 = 5.00 \times 10^{-3}$ ; $n(\ce{Na2CO3})$ in 25.0 cm³ $= 2.50 \times 10^{-3}$ ; in 250 cm³ $= 0.0250\ \text{mol}$ . $M = 7.15 / 0.0250 = 286$ . $106.0 + 18.0x = 286$ , so $x = 10$ .
$M(\ce{Ca(NO3)2}) = 164.1$ ; $n = 4.10 / 164.1 = 0.02498\ \text{mol}$ . Gas: 2 mol of salt gives $4 + 1 = 5$ mol of gas, so $n(\text{gas}) = 0.02498 \times 2.5 = 0.06246\ \text{mol}$ ; $V = 0.06246 \times 24.0 = 1.50\ \text{dm}^3$ .
$\ce{NaOH}$ added $= 0.200 \times 50.0 / 1000 = 0.0100\ \text{mol}$ . Remaining $= 0.100 \times 26.00 / 1000 = 0.00260\ \text{mol}$ . Reacted with $\ce{NH4+}$ : $0.00740\ \text{mol}$ , so $n(\ce{NH4+}) = 0.00740$ and $n(\ce{(NH4)2SO4}) = 0.00370\ \text{mol}$ . Mass $= 0.00370 \times 132.1 = 0.489\ \text{g}$ . Percentage $= 0.489 / 0.500 \times 100 = 97.8\%$ .

Volumes of gases

Reacting volumes of gases

Finding the formula of a hydrocarbon from gas volumes

Concentration of solutions

Dilution

Titration calculations

Back titrations

Practice

Console