SAT Subject Test Chemistry




Liquids, Solids, and Phase Changes


Water is so often involved in chemistry that it is important to have a rather complete understanding of this compound and its properties. Pure water has become a matter of national concern. Although commercial methods of purification will not be discussed here, the usual laboratory method of obtaining pure water, distillation, will be covered.

Purification of Water

The process of distillation involves the evaporation and condensation of the water molecules. The usual apparatus for the distillation of any liquid is shown in Figure 26.

This method of purification will remove any substance that has a boiling point higher than that of water. It cannot remove dissolved gases or liquids that boil off before water. These substances will be carried over into the condenser and subsequently into the distillate.


In distillation, first boil and then condense.

Figure 26. Distillation of water.

Composition of Water

Water can be analyzed, that is, broken into its components, by electrolysis. This process shows that the composition of water by volume is 2 parts of hydrogen to 1 part of oxygen. Water composition can also be arrived at by synthesis. Synthesis is the formation of a compound by uniting its components. Water can be made by mixing hydrogen and oxygen in a eudiometer over mercury and passing an igniting spark through the mixture. Again the ratio of combination is found to be 2 parts hydrogen: 1 part oxygen. In a steam-jacketed eudiometer, which keeps the water formed in the gas phase, 2 volumes of hydrogen combine with 1 volume of oxygen to form 2 volumes of steam.

Another interesting method is the Dumas experiment pictured in Figure 27. Data obtained show that hydrogen and oxygen combine to form water in a ratio of 1 : 8 by mass. This means that 1 gram of hydrogen combines with 8 grams of oxygen to form 9 grams of water.


Water ratios of combination:
H2 to O2
2 vol : 1 vol in gaseous state
1 g : 8 g by mass

Figure 27. Synthesis of Water

Some sample problems involving the composition of water are shown below.



An electric spark is passed through a mixture of 12 grams of hydrogen and 24 grams of oxygen in the eudiometer setup shown. Find the number of grams of water formed and the number of grams of gas left uncombined.

Since water forms in a ratio of 1 : 8 by mass, to use up the oxygen (which by inspection will be the limiting factor since it has enough hydrogen present to react completely) we need only 3.0 grams of hydrogen.

3.0 g of hydrogen + 24 g of oxygen = 27 g of water

This leaves 12 g - 3.0 g = 9.0 g of hydrogen uncombined.



A mixture of 8 milliliters of hydrogen and 200 milliliters of air is placed in a steam-jacketed eudiometer, and a spark is passed through the mixture. What will be the total volume of gases in the eudiometer?

Because this is a combination by volume, 8 mL of hydrogen require 4 mL of oxygen. (Ratio H2 : O2 by volume = 2 : 1.)

The 200 mL of air is approximately 21% oxygen. This will more than supply the needed oxygen and leave 196 mL of the air uncombined.

The 8 mL of hydrogen and 4 mL of oxygen will form 8 mL of steam since the eudiometer is steam-jacketed and keeps the water formed in the gaseous state.

(Ratio by volume of hydrogen : oxygen : steam = 2 : 1: 2)

TOTAL VOLUME = 196 mL of air + 8 mL of steam = 204 mL

HEAVY WATER. A small portion of water is called “heavy” water because it contains an isotope of hydrogen, deuterium (symbol D), rather than ordinary hydrogen nuclei. Deuterium has a nucleus of one proton and one neutron rather than just one proton. Another isotope of hydrogen is tritium. Its nucleus is composed of two neutrons and one proton. Both of these isotopes have had use in the nuclear energy field.

HYDROGEN PEROXIDE. The prefix per- indicates that this compound contains more than the usual oxide. Its formula is H 22. It is a well-known bleaching and oxidizing agent. Its electron-dot formula is shown in Figure 28.

Figure 28. Hydrogen Peroxide

Water Calorimetry Problems

A calorimeter is a container well insulated from outside sources of heat or cold so that most of its heat is contained in the vessel. If a very hot object is placed in a calorimeter containing some ice crystals, we can find the final temperature of the mixture mathematically and check it experimentally. To do this, however, certain behaviors must be understood. Ice changing to water and then to steam does not represent a continuous and constant change of temperature as time progresses. In fact, the chart would look as shown in Figure 29.


Water’s heat of vaporization = 40.79 kJ/mol

Figure 29. Changing Ice to Steam


Water’s heat of fusion = 6.01 kJ/mol

From this graph, you see that heat is being used at 0°C and 100°C to change the state of water, but not its temperature. One gram of ice at 0°C needs 80 calories or 3.34 × 102 joules to change to water at 0°C. This is called its heat of fusion. Likewise, energy is being used at 100°C to change water to steam, not to change the temperature. One gram of water at 100°C absorbs 540 calories or 2.26 × 103 joules of heat to change to 1 gram of steam at 100°C. This is called its heat of vaporization. This energy absorbed at the plateaus in the curve is being used to break up the bonding forces between molecules by increasing the potential energy content of the molecules so that a specific change of state can occur.

The amount of heat energy required to melt one mole of solid at its melting point is called its molar heat of fusion. If the quantity of ice melted is one mole (18 grams), then, for ice, it is 6.01 kJ/mole. Likewise, the amount of heat energy required to vaporize one mole of liquid at its boiling point is called its molar heat of vaporization. If the quantity of water vaporized is one mole, then its molar heat of vaporization is 40.79 kJ/mole.

The following two problems are samples of calorimetry problems. The first is solved by using the SI units of joules. The second is done by using calories but is finally converted to joules.


What quantity of ice at 273 K can be melted by 100. joules of heat?

Heat to fuse (melt) a substance = heat of fusion of the substance × mass of the substance.

This quantity can be expressed by the following formula, where q denotes the heat measurement made in a calorimeter:

q=m(mass)×C(heat of fusion)

Solving for m, we get

m = 29.9 × 10−2 g or 0.299 g of ice melted.

Because heat is absorbed in melting, this is an endothermic action.


How much heat is needed to change 100.0 grams of ice at 273 K to steam at 373 K?

To melt 100. grams of ice at 273 K:

Use: m(mass) × C(heat of fusion) = q(quantity of heat)

To heat 100. grams of water from 273 K to 373 K:

Use: m × ΔT × specific heat = q

To vaporize 100. grams of water at 100.°C to steam at 100.°C:

Use: m × heat of vaporization = q

Total heat = 33.44 kJ + 41.8 kJ + 225.72 kJ = 300.96 kJ

Water’s Reactions with Anhydrides

Anhydrides are certain oxides that react with water to form two classes of compounds— acids and bases.

Many metal oxides react with water to form bases such as sodium hydroxide, potassium hydroxide, and calcium hydroxide. For this reason, they are called basic anhydrides or basic oxides. The common bases are water solutions which contain the hydroxyl (OH) ion. Some common examples are:

Na2O + H2O → 2NaOH, sodium hydroxide

CaO + H2O → Ca(OH)2, calcium hydroxide

In general then: Metal oxide + H2O → Metal hydroxide

In a similar manner, nonmetallic oxides react with water to form an acid such as sulfuric acid, carbonic acid, or phosphoric acid. For this reason, they are referred to as acidic oxides or acidic anhydrides. (The term acid anhydride is now used to refer to specific organic compounds.) The common acids are water solutions containing hydrogen ions (H+). Some common examples are:

CO2 + H2O → H2CO3, carbonic acid

SO3 + H2O → H2SO4, sulfuric acid

P2O5 + 3H2O → 2H3PO4, phosphoric acid

In general, then: Nonmetallic oxide + H2O → Acid