Lesson 7 of 11Quantities in Chemical Reactions

Unit 3 · Lesson 7 of 11

Hydrates

Formulas of hydrates like $CuSO_{4} \cdot 5 H_{2} O$ , and finding the number of water molecules per formula unit from mass-loss-on-heating data.

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Key things to remember

1A hydrate's formula is $salt \cdot x H_{2} O$ ; the dot means "combined with," not multiplication.
2Mass of water lost on heating = mass of hydrate $-$ mass of anhydrous salt remaining.
3 $x = \frac{mol H _{2} O lost}{mol anhydrous salt},$ rounded to the nearest whole number.
4 $% H_{2} O = \frac{mass H _{2} O}{mass hydrate} \times 100%$
5 $M (hydrate) = M (anhydrous salt) + x (18.02 g/mol)$
6Water left behind makes $x$ too low; any other mass lost (spattering, decomposition) makes $x$ too high.

Many ionic compounds crystallize with a fixed number of water molecules trapped inside their crystal lattice — these are hydrates. The water is held loosely enough that heating the crystal drives it off as vapour, leaving behind the pure ionic compound (the anhydrous salt) with a visibly different appearance — for instance, blue $CuSO_{4} \cdot 5 H_{2} O$ crystals turn into white anhydrous $CuSO_{4}$ powder when heated. Because the mass lost is entirely water, you can use it, together with the mass of anhydrous salt left behind, to figure out exactly how many water molecules were attached per formula unit.

Naming and Writing Hydrate Formulas

A hydrate formula is written as $salt \cdot x H_{2} O$ , where the raised dot means "combined with," not multiplication, and $x$ is the fixed whole-number count of water molecules per formula unit — for example, $CuSO_{4} \cdot 5 H_{2} O$ (copper(II) sulfate pentahydrate) has 5 water molecules per formula unit of $CuSO_{4}$ . The number is fixed, not "however damp the crystal happens to be", because the water molecules sit in definite positions in the crystal, many of them clustered around the metal ions. The name is the ordinary ionic name followed by a Greek prefix and the word "hydrate".

Formula	Name	Common name or use
$CuSO_{4} \cdot 5 H_{2} O$	copper(II) sulfate pentahydrate	blue vitriol; turns white when dehydrated
$MgSO_{4} \cdot 7 H_{2} O$	magnesium sulfate heptahydrate	Epsom salt
$CaSO_{4} \cdot 2 H_{2} O$	calcium sulfate dihydrate	gypsum, used in drywall
$Na_{2} CO_{3} \cdot 10 H_{2} O$	sodium carbonate decahydrate	washing soda
$CoCl_{2} \cdot 6 H_{2} O$	cobalt(II) chloride hexahydrate	pink; the anhydrous salt is blue (humidity indicator paper)

Prefixes: mono 1, di 2, tri 3, tetra 4, penta 5, hexa 6, hepta 7, octa 8, nona 9, deca 10.

Because the water is part of the formula, it is part of the molar mass: $M (hydrate) = M (salt) + x (18.02) .$ For copper(II) sulfate pentahydrate that is $159.62 + 5 (18.02) = 249.72 g/mol,$ so more than a third of the blue crystal's mass is water. This matters in the lab: if a recipe needs $0.100 mol$ of copper(II) sulfate and you weigh out the pentahydrate, you need $24.97 g$ , not the $15.96 g$ you would weigh for the anhydrous powder.

Fig 1Finding a hydrate's formula follows the same mass-to-mole-ratio logic as an empirical formula.

Finding $x$ from Heating Data

The standard experiment: weigh an empty crucible, add the hydrate and weigh again, then heat gently. Cool the crucible (ideally in a desiccator so it can't pick up moisture from the air) and reweigh. Repeat heating, cooling and weighing until two readings agree; this is called heating to constant mass, and it is the only evidence that all the water is gone. The mass that disappeared is water, and the mass that remains is anhydrous salt.

Mass of water lost = mass of hydrate before heating $-$ mass of anhydrous salt after heating.
Convert the mass of water lost to moles using $M (H_{2} O) = 18.02 g/mol$ .
Convert the mass of anhydrous salt to moles using its own molar mass.
$x = \frac{mol H _{2} O}{mol anhydrous salt},$ rounded to the nearest whole number.

x = \frac{n ( H _{2} O )}{n ( anhydrous salt )} % H_{2} O = \frac{mass H _{2} O}{mass hydrate} \times 100%

Why a mole ratio? The formula $CuSO_{4} \cdot 5 H_{2} O$ says "5 water molecules per 1 formula unit of salt", which is a count, so it can only be found by comparing moles, exactly as in an empirical formula. Comparing grams would fail because a water molecule ( $18.02 g/mol$ ) is far lighter than a formula unit of salt. Since the salt's mole value is the one you divide by, $x$ comes out as "waters per formula unit". Experimental data give something like $4.98$ or $6.03$ ; round to the nearest whole number.

Use the anhydrous salt's molar mass, not the hydrate's

The mass remaining after heating is the anhydrous salt only — convert it to moles using the anhydrous salt's molar mass (e.g.

M (CuSO_{4}) = 159.62 g/mol

), never the molar mass of the whole hydrate.

Sources of Error and Their Effect on x

What went wrong	Effect on the measured water lost	Effect on calculated x
Not heated to constant mass (some water left)	too small	too low
Anhydrous salt reabsorbed moisture before weighing	too small	too low
Solid spattered out of the crucible	too large	too high
Heated so strongly the salt decomposed (e.g. $CuSO_{4}$ losing $SO_{3}$ )	too large	too high

Any extra mass loss is blamed on water, so it raises x; any water left behind lowers x.

Exam questions love this reasoning. Trace the error through the calculation: every gram lost counts as water, so anything else that escapes (spattered solid, decomposition gases) inflates $n (H_{2} O)$ , and it also shrinks the anhydrous mass, which lowers $n (salt)$ ; both push $x$ up. Water that stays behind does the opposite. The colour change is a useful check too: blue $CuSO_{4} \cdot 5 H_{2} O$ becoming uniformly white means dehydration is complete, and a black residue means the salt has started to decompose.

Recognising the question type

Masses before and after heating, or a percent of water, with an unknown x: find x by the mole ratio. A known hydrate formula with "percent water" or "mass of water driven off": use the hydrate's molar mass, including the

x (18.02)

term, just like a percent-composition question.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

2.50 g

sample of hydrated copper(II) sulfate,

CuSO_{4} \cdot x H_{2} O

, is heated to constant mass, leaving

1.60 g

of anhydrous

CuSO_{4}

(

M = 159.62 g/mol

). Find

x

Example 2

2.03 g

sample of hydrated magnesium chloride,

MgCl_{2} \cdot x H_{2} O

, is heated until all water is driven off, leaving

0.952 g

of anhydrous

MgCl_{2}

(

M = 95.21 g/mol

). Find

x

Example 3

Calculate the percent water by mass in

CuSO_{4} \cdot 5 H_{2} O

Example 4

A hydrate of barium chloride,

BaCl_{2} \cdot x H_{2} O

, is found to contain

14.75%

water by mass (

M (BaCl_{2}) = 208.23 g/mol

). Find

x

, then say how the answer would change if the student had not heated to constant mass.

Where marks get lost

Forgetting to subtract the two masses first, and treating the total hydrate mass as if it were the mass of water lost.
Using the molar mass of the hydrate (salt + water) instead of the molar mass of the anhydrous salt alone when finding moles of salt.
Getting a value like $x = 4.98$ and reporting it as-is instead of recognizing it should round to the nearest whole number, 5.
Writing the hydrate formula without the centred dot, or placing the coefficient $x$ incorrectly, e.g. writing $x CuSO_{4} H_{2} O$ .
Leaving out the water when calculating a hydrate's molar mass (using $159.62$ instead of $249.72 g/mol$ for $CuSO_{4} \cdot 5 H_{2} O$ ), which throws off percent water and any mass-to-mole conversion.

Flashcards 9

What does the dot mean in

CuSO_{4} \cdot 5 H_{2} O

How do you find the mass of water lost on heating a hydrate?

Formula to find

x

in a hydrate

What is an anhydrous salt?

+ 5 more

Practice

2.86 g

sample of

Na_{2} CO_{3} \cdot x H_{2} O

is heated to constant mass, leaving

1.06 g

of anhydrous

Na_{2} CO_{3}

(

M = 105.99 g/mol

). Find

x

Calculate the percent water by mass in

MgCl_{2} \cdot 6 H_{2} O

(

M = 203.33 g/mol

). Give 2 decimal places.

After heating a hydrate sample, the mass stops decreasing further. What does this tell you?

Why is the mass remaining after heating a hydrate used with the anhydrous salt's molar mass, not the hydrate's molar mass?

5.00 g

sample of

CuSO_{4} \cdot 5 H_{2} O

(

M = 249.72 g/mol

) is heated completely. What mass of water is driven off?

Order the correct steps for finding

x

in a hydrate formula from heating data.

1Convert mass of water to moles
2Divide mol water by mol anhydrous salt
3Convert mass of anhydrous salt to moles
4Subtract to find mass of water lost

Match each hydrate to its correct name.

CuSO_{4} \cdot 5 H_{2} O

MgCl_{2} \cdot 6 H_{2} O

BaCl_{2} \cdot 2 H_{2} O

Na_{2} CO_{3} \cdot 10 H_{2} O

A hydrate of magnesium sulfate,

MgSO_{4} \cdot x H_{2} O

(

M (MgSO_{4}) = 120.38 g/mol

), contains 51.17% water by mass. Find

x

While heating a hydrate, some of the solid spatters out of the crucible. How does this affect the calculated value of

x

Calculate the molar mass of cobalt(II) chloride hexahydrate,

CoCl_{2} \cdot 6 H_{2} O

(Co = 58.93, Cl = 35.45). Give 2 decimal places.

g/mol

10.0 g

sample of

BaCl_{2} \cdot 2 H_{2} O

(

M = 244.27 g/mol

) is heated to constant mass. What mass of anhydrous

BaCl_{2}

(

M = 208.23 g/mol

) remains? Give 3 sig figs.

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