Applications: Water of Crystallisation and Relative Atomic Mass

25 min

From a titre to a formula

Green crystals of iron(II) sulfate are used to treat iron deficiency and to correct alkaline soils. Their formula is written $\mathrm{FeSO_4{\cdot}7H_2O}$ : seven water molecules are built into the crystal for every formula unit of iron(II) sulfate. How was the "7" found? Not by looking at a crystal, but by weighing a sample, titrating it, and letting the arithmetic reveal the molar mass. The same chain of reasoning finds the relative atomic mass of an unknown metal or the molar mass of an unknown acid.

Definition

Water of crystallisation: water molecules held in fixed proportion within the crystal of a salt, written after a dot: $\mathrm{Na_2CO_3{\cdot}10H_2O}$ . A salt containing it is a hydrate; without it, anhydrous.

Basicity of an acid: the number of $\mathrm{H^{+}}$ ions one molecule can donate. Ethanedioic acid, $\mathrm{H_2C_2O_4}$ , is dibasic.

Relative atomic mass, $A_{r}$ : the average mass of an atom of an element relative to one-twelfth of the mass of a carbon-12 atom.

Figure 11.6: Every application in this lesson follows the same chain.

For a hydrate $\mathrm{X{\cdot}}x\mathrm{H_2O}$ :

$M(\text{hydrate}) = M(\mathrm{X}) + 18.02x \qquad x = \frac{M(\text{hydrate}) - M(\mathrm{X})}{18.02}$

The answer must come out close to a whole number; if it does not, check the working before blaming the experiment.

Worked example

Problem. 3.575 g of washing soda, $\mathrm{Na_2CO_3{\cdot}}x\mathrm{H_2O}$ , is dissolved and made up to 250.0 cm³. 25.00 cm³ portions need 25.00 cm³ of 0.1000 mol dm⁻³ HCl with methyl orange. Find $x$ . ( $M(\mathrm{Na_2CO_3}) = 105.99$ )

Solution.

$n(\mathrm{HCl}) = 0.1000 \times 0.02500 = 2.500 \times 10^{-3}$ ; $\mathrm{Na_2CO_3 + 2HCl}$ , so $n(\mathrm{Na_2CO_3}) = 1.250 \times 10^{-3}$ in 25.00 cm³, and $0.01250\,\mathrm{mol}$ in 250.0 cm³.

$M = \frac{3.575}{0.01250} = 286.0\,\mathrm{g\,mol^{-1}} \qquad x = \frac{286.0 - 105.99}{18.02} = 9.99 \approx 10$

Answer: $\mathrm{Na_2CO_3{\cdot}10H_2O}$ .

Worked example

Problem. 6.950 g of hydrated iron(II) sulfate, $\mathrm{FeSO_4{\cdot}}x\mathrm{H_2O}$ , is dissolved in dilute sulfuric acid and made up to 250.0 cm³. 25.00 cm³ portions need 25.00 cm³ of 0.02000 mol dm⁻³ $\mathrm{KMnO_4}$ . Find $x$ . ( $M(\mathrm{FeSO_4}) = 151.90$ )

Solution.

$n(\mathrm{MnO_4^{-}}) = 0.02000 \times 0.02500 = 5.000 \times 10^{-4}$ ; ratio 1 : 5, so $n(\mathrm{Fe^{2+}}) = 2.500 \times 10^{-3}$ per aliquot and $0.02500\,\mathrm{mol}$ in total.

$M = \frac{6.950}{0.02500} = 278.0\,\mathrm{g\,mol^{-1}} \qquad x = \frac{278.0 - 151.90}{18.02} = 7.00$

Answer: $\mathrm{FeSO_4{\cdot}7H_2O}$ .

Worked example

Problem. 1.381 g of the carbonate of a Group 1 metal, $\mathrm{M_2CO_3}$ , is made up to 250.0 cm³. 25.00 cm³ portions need 20.00 cm³ of 0.1000 mol dm⁻³ HCl. Find $A_{r}$ of M and identify it.

Solution. $n(\mathrm{HCl}) = 2.000 \times 10^{-3}$ ; $n(\mathrm{M_2CO_3}) = 1.000 \times 10^{-3}$ per aliquot, $0.01000\,\mathrm{mol}$ in total.

$M(\mathrm{M_2CO_3}) = \frac{1.381}{0.01000} = 138.1 \qquad 2A_r = 138.1 - 60.01 = 78.09 \qquad A_r = 39.0$

Answer: $A_r \approx 39.0$ : the metal is potassium (39.10).

Finding the molar mass and basicity of an acid

If the molar mass of an acid is known, titrating a weighed sample with NaOH gives its basicity: $\text{basicity} = \frac{n(\mathrm{NaOH})}{n(\text{acid})}$ . If the basicity is known, the same data give the molar mass. For a metal that dissolves in acid, a back titration of the unused acid gives the amount of metal, and so its $A_{r}$ .

Warning

Typical errors: using the anhydrous molar mass when the hydrate was weighed; forgetting the scale-up factor; and leaving hydrated crystals in warm air, where some lose water (efflorescence) and $x$ comes out too small. Store hydrates in stoppered bottles.

Safety: iron(II) sulfate is harmful if swallowed in quantity; iron tablets are a leading cause of accidental poisoning in young children, so keep them out of reach. Wear goggles with dilute acids. First aid (CDC/NIOSH, sulfuric acid): flush eyes or skin with water immediately; if swallowed, get medical attention immediately.

Tip

Before calculating, predict roughly. Washing soda is "10-water"; Epsom salt $\mathrm{MgSO_4{\cdot}7H_2O}$ and green vitriol $\mathrm{FeSO_4{\cdot}7H_2O}$ are "7-water"; blue vitriol $\mathrm{CuSO_4{\cdot}5H_2O}$ is "5-water". If your $x$ is 3.4 for washing soda, look for a missing factor of 10 or 2, not a new compound.

MaChemGuy: Quick Revision, Further Titration Calculations. Watch how a solid is dissolved and made up in a volumetric flask, then follow the scale-up from one aliquot to the whole sample.

✦

Practice

Calculate the percentage by mass of water in $\mathrm{CuSO_4{\cdot}5H_2O}$ (M = 249.68).
1.26 g of hydrated ethanedioic acid, $\mathrm{H_2C_2O_4{\cdot}}x\mathrm{H_2O}$ , is made up to 250.0 cm³. 25.00 cm³ portions need 20.00 cm³ of 0.1000 mol dm⁻³ NaOH. Find $x$ . ( $M(\mathrm{H_2C_2O_4}) = 90.03$ )
0.1000 g of an acid of molar mass 90.0 g mol⁻¹ needs 22.2 cm³ of 0.100 mol dm⁻³ NaOH. Find its basicity.
0.120 g of a metal M, which forms $\mathrm{M^{2+}}$ ions, dissolves in 50.00 cm³ of 0.500 mol dm⁻³ HCl (excess). The solution is made up to 250.0 cm³, and 25.00 cm³ portions need 15.13 cm³ of 0.1000 mol dm⁻³ NaOH. Calculate $A_{r}$ of M and identify it.
9.84 g of $\mathrm{MgSO_4{\cdot}}x\mathrm{H_2O}$ contains 0.0400 mol of $\mathrm{MgSO_4}$ (M = 120.37). Find $x$ .
A student's washing-soda crystals had been left open on a warm shelf for a week. Predict how her value of $x$ compares with 10, and explain.

Show answersHide answers

$\frac{5 \times 18.02}{249.68} \times 100 = 36.1\,\%$ .
$n(\mathrm{NaOH}) = 2.000 \times 10^{-3}$ ; dibasic, so $n(\text{acid}) = 1.000 \times 10^{-3}$ per aliquot, 0.01000 mol total; $M = 126.0$ ; $x = \frac{126.0 - 90.03}{18.02} = 2.0$ .
$n(\text{acid}) = \frac{0.1000}{90.0} = 1.11 \times 10^{-3}$ ; $n(\mathrm{NaOH}) = 2.22 \times 10^{-3}$ ; ratio 2: dibasic.
Excess HCl $= 10 \times 0.1000 \times 0.01513 = 0.01513\,\mathrm{mol}$ ; HCl used $= 0.02500 - 0.01513 = 0.00987\,\mathrm{mol}$ ; $n(\mathrm{M}) = 0.004935\,\mathrm{mol}$ ( $\mathrm{M + 2HCl}$ ); $A_r = \frac{0.120}{0.004935} = 24.3$ : magnesium.
$M = \frac{9.84}{0.0400} = 246.0$ ; $x = \frac{246.0 - 120.37}{18.02} = 6.97 \approx 7$ .
Less than 10. The crystals lose some water of crystallisation to the air (efflorescence), so each gram contains more $\mathrm{Na_2CO_3}$ , the calculated molar mass is too small, and so is $x$ .

1From a titre to a formula

2Finding the molar mass and basicity of an acid

From a titre to a formula

Finding the molar mass and basicity of an acid