Chem1 General Chemistry Virtual Textbook
Understanding density and buoyancy
The density of an object is one of its most important and easily-measured physical properties. Densities are widely used to identify pure substances and to characterize and estimate the composition of many kinds of mixtures. The purpose of this lesson is to show how densities are defined, measured, and utilized, and to make sure you understand the closely-related concepts of buoyancy and specific gravity
You didn't have to be in the world very long to learn that the mass and volume of a given substance are directly proportional, although you certaintly did not first learn it in these words which are now the words of choice now that you have become a scholar.
These plots show how the masses of three liquids vary with their volumes. Notice that
The only difference between these plots is their slopes. Denoting mass and volume by m and V respectively, we can write the equation of each line as m = ρV, where the slope ρ (rho) is the proportionality constant that relates mass to volume. This quantity ρ is known as the density, which is usually defined as the mass per unit volume: ρ = m/V.
The volume units millilitre (mL) and cubic centimetre (cm3) are almost identical and are used interchangably in this course.
The general meaning of density is the amount of anything per unit volume. What we conventionally call the "density" is more precisely known as the "mass density".
Density can be expressed in any combination of mass and volume units; the most commonly seen units are grams per mL (g mL–1, g cm–3), or kilograms per litre.
Ordinary commercial nitric acid is a liquid having a density of 1.42 g mL–1, and contains 69.8% HNO3 by weight. a) Calculate the mass of HNO3 in 800 ml of nitric acid. b) What volume of acid will contain 100 g of HNO3?
Solution: The mass of 800 mL of the acid is (1.42 g mL–1) × (800 mL) = 1140 g. The weight of acid that contains 100 g of HNO3 is (100 g) / (0.698) = 143 g and will have a volume of (143 g) / (1.42 g mL–1) = 101 mL.
It is sometimes more convenient to express the volume occupied by a unit mass of a substance. This is just the inverse of the density and is known as the specific volume.
A glass bulb weighs 66.3915 g when evacuated, and 66.6539 g when filled with xenon gas at 25°C. The bulb can hold 50.0 mL of water. Find the density and specific volume of xenon under these conditions.
Solution: The mass of xenon is found by difference: (66.6539 – 66.3915)g = 0.2624 g. The density ρ = m/V = (0.2624 g)/(0.050 L) = 5.248 g L–1. The specific volume is 1/(5.248 g L–1 = 0.190 L g–1.
A quantity that is very closely related to density, and which is frequently used in its place, is specific gravity.
Specific gravity is the ratio of the mass of a material to that of an equal volume of water. Because the density of water is about 1.00 g mL–1, the specific gravity is numerically very close to that of the density, but being a ratio, it is dimensionless.
The presence of "volume" in this definition introduces a slight complication, since volumes are temperature-dependent owing to thermal expansion. At 4°C, water has its maximum density of almost exactly 1.000 g mL–1, so if the equivalent volume of water is assumed to be at this temperature, then the density and specific gravity can be considered numerically identical. In making actual comparisons, however, the temperatures of both the material being measured and of the equivalent volume of water are frequently different, so in order to specify a specific gravity value unambiguously, it is necessary to state the temperatures of both the substance in question and of the water.
Thus if we find that a given volume of a substance at 20°C weighs 1.11 times as much as the same volume of water measured at 4°C, we would express its specific gravity as
Although most chemists find density to be more convenient to work with and consider specific gravity to be rather old-fashioned, the latter quantity is widely used in many industrial and technical fields ranging from winemaking to urinalysis.
In general, gases have the lowest densities, but these densities are highly dependent on the pressure and temperature which must always be specified. To the extent that a gas exhibits ideal behavior (low pressure, high temperature), the density of a gas is directly proportional to the masses of its component atoms, and thus to its molecular weight. Measurement of the density of a gas is a simple experimental way of estimating its molecular weight (more here).
Liquids encompass an intermediate range of densities. Mercury, being a liquid metal, is something of an outlier. Liquid densities are largely independent of pressure, but they are somewhat temperature-sensitive.
The density range of solids is quite wide. Metals, whose atoms pack together quite compactly, have the highest densities, although that of lithium, the lighest metallic element, is quite low. Composite materials such as wood and high-density polyurethane foam contain void spaces which reduce the average density.
All substances tend to expand as they are heated, causing the same mass to occupy a greater volume, and thus lowering the density. For most solids, this expansion is relatively small, but it is far from negligible; for liquids, it is greater. The volumes of gases, as you may already know (see here for details), are highly temperature-sensitive, and so, of course, are their densities.
What is the cause of thermal expansion? As molecules aquire thermal energy, they move about more vigorously. In condensed phases (liquids and solids), this motion has the character of an irregular kind of bumping or jostling that causes the average distances between the molecules to increase, thus leading to increased volume and smaller density.
One might expect the densities of the chemical elements to increase uniformly with atomic weight, but this is not what happens; density depends on the volume as well as the mass, and the volume occupied by a given mass of an element, and these volumes can vary in a non-uniform way for two reasons:
The sizes (atomic radii) follow the zig-zag progression that characterizes the other periodic properties of the elements, with atomic volumes diminishing with increasing nuclear charge across each period (more here).
The atoms comprising the different solid elements do not pack together in the same way. The non-metallic solids are often composed of molecules that are more spread out in space, and which have shapes that cannot be arranged as compactly. so they tend to form more open crystal lattices than do the metals, and therefore have lower densities.
The plot below is taken from the popular WebElements site.
Nature has conveniently made the density of water at ordinary temperatures almost exactly 1.000 g/mL ( 1 kg/L). Water is subject to thermal expansion just as are all other liquids, and throughout most of its temperature range, the density of water diminishes with temperature. But water is famously exceptional over the temperature range 0-4° C, where raising the temperature causes the density to increase, reaching its greatest value at about 4°C.
This 4°C density maximum is one of many "anomalous" behaviors of water. As you may know, the H2O molecules in liquid and solid water are loosely joined together through a phenomenon known as hydrogen bonding. Any single water molecule can link up to four other H2O molecules, but this occurs only when the molecules are locked into place within an ice crystal. This is what leads to a relatively open lattice arrangement, and thus to the relatively low density of ice.
Below are three-dimensional views of a typical local structure of liquid water (right) and of ice (left). Notice the greater openness of the ice structure which is necessary to ensure the strongest degree of hydrogen bonding in a uniform, extended crystal lattice. The more crowded and jumbled arrangement in liquid water can be sustained only by the greater amount thermal energy available above the freezing point.
When ice melts, thermal energy begins to overcome the hydrogen-bonding forces so that each H2O molecule, instead of being permanently connected to four neighbors, is now only linked to an average of three other molecules through hydrogen bonds that continually break and re-form. With fewer hydrogen bonds, the geometrical requirements that formerly mandated a more open structural arrangement now diminish, so the entire network tends to collapse, rendering the water more dense. As the temperature rises, the fraction of H2O molecules that occupy ice-like clusters diminishes, contributing to the rise in density that is seen between 0° and 4°.
Whenever a continuously varying quantity such as density passes through a maximum or a minimum value as the temperature or some other variable is changing, you know that two opposing effects are at work.
The 4° density maximum of water corresponds to the temperature at which the breakup of ice-like clusters (leading to higher density) and thermal expansion (leading to lower density) achieve a balance.
Suppose that you place 1000 mL of pure water at 25°C in the refrigerator and that it freezes, producing ice at 0C. What will be the volume of the ice?
Solution: From the graph above, the density of water at 25°C is 0.9997 kg L–1, and that of ice at 0°C 0.917 g L–1.
The density maximum at 4°C has some interesting consequences in the aquatic ecology of lakes. In all but the most shallow lakes, the water tends to be stratified, so that for most of the year, the denser water remains near the bottom and mixes very little with the less-dense waters above. Because water has its density maximum at 4°C, the waters of deep lakes (and of the oceans) usually stay around 4°C at all times of the year. In the summer this will be the coldest water, but in the winter, the surface waters lose heat to the atmosphere and if they cool below 4°, they will be colder than the more dense waters below.
When the weather turns cold in the fall, the surface waters lose heat and cool to 4°C. This more dense layer of water sinks to the bottom, displacing the water below, which rises to the surface and restores nutrients that were removed when dead algae sank to the bottom. This “fall turnover” renews the lake for the next season.
What do an ice cube and a block of wood have in common? Throw either material into water, and it will float. Well, mostly; each object will have its bottom part immersed, but the upper part will ride high and dry. People often say that wood and ice float because they are "lighter than water", but this of course is nonsense unless we compare the masses of equal volumes of the substances. In other words, we need to compare the masses-per-unit-volume, meaning the densities, of each material with that of water. So we would more properly say that objects capable of floating in water must have densities smaller than that of water.
The apparent weight of an object immersed in a fluid will be smaller than its “true” weight (Archimedes' principle). The latter is the downward force exerted by gravity on the object. Within a fluid, however, this downward force is partially opposed by a net upward force that results from the displacement of this fluid by the object. The difference between these two weights is known as the buoyancy.
The displaced fluid is of course not really confined to the "phantom volume" shown at the bottom of the diagram; it spreads throughout the container and exerts forces on all surfaces of the object and increase with depth, combining to produce the net buoyancy force as shown. See here for another diagram that shows this more clearly.
Dynamics of buoyancy - an interesting physics-mechanics treatment
An object weighs 36 g in air and has a volume of 8.0 cm3. What will be its apparent weight when immersed in water?
Solution: When immersed in water, the object is buoyed up by the mass of the water it displaces, which of course is the mass of 8 cm3 of water. Taking the density of water as unity, the upward (buoyancy) force is just 8 g.
The apparent weight will be (36 g) – (8 g) = 28 g.
Air is of course a fluid, and buoyancy can be a problem when weighing a large object such as an empty flask. The following problem illustrates a more extreme case:
A balloon having a volume of 5.000 L is placed on a sensitive balance which registers a weight of 2.833 g. What is the "true weight" of the balloon if the density of the air is 1.294 g L–1?
Solution: The mass of air displaced by the balloon exerts a buoyancy force of
(5.000 L) × (1.294 g L –1) = 6.470 g. Thus the true weight of the balloon is this much greater than the apparent weight: (2.833 + 6.470) g = 9.303 g.
A piece of metal weighs 9.25 g in air, 8.20 g in water, and 8.36 g when immersed in gasoline. a) What is the density of the metal? b) What is the density of the gasoline?
Solution: When immersed in water, the metal object displaces (9.25 – 8.20) g = 1.05 g of water whose volume is (1.05 g) / (1.00 g cm–3) = 1.05 cm3. The density of the metal is thus (9.25 g) / (1.05 cm3) = 8.81 g cm–3.
The metal object displaces (9.25 - 8.36) g = 0.89 g of gasoline, whose density must therefore be (0.89 g) / (1.05 cm3) = 0.85 g cm–3.
When an object floats in a liquid, the portion of it that is immersed has a volume that depends on the mass of this same volume of displaced liquid.
A cube of ice that is 10 cm on each side floats in water. How many cm does the top of the cube extend above the water level? (Density of ice = 0.917 g cm–3.)
Solution: The volume of the ice is (10 cm)3 = 1000 cm3 and its mass is
(1000 cm3) x (0.917 g cm–3) = 917 g. The ice is supported by an upward force equivalent to this mass of displaced water whose volume is (917 g) / (1.00 g cm–3) = 917 cm3 . Since the cross section of the ice cube is 100-cm2, it must sink by 9.17 cm in order to displace 917 cm3 of water. Thus the height of cube above the water is (10 cm – 9.17 cm) = 0.83 cm.
... hence the expression, “the tip of the iceberg”, implying that 90% of its volume is hidden under the surface of the water.
The most obvious way of finding the density of a material is to measure its mass and its volume. This is the only option we have for gases, but observing the mass of a fixed volume of a liquid is time-consuming and awkward, and measuring the volumes of solids whose shapes are irregular or which are finely divided is usually impractical.
The traditional hydrometer is a glass tube having a weighted bulb near the bottom. The hydrometer is lowered into a container of the liquid to be measured, and comes to a rest with the upper part protruding above the liquid surface at a height (read from a calibrated scale) that depends on the density of the liquid. This will only work, of course, if the overall density of the hydrometer itself is smaller than the density of the liquid to be measured. For this reason, hydrometers intended for general use come in sets. Because liquid densities are temperature dependent, hydrometers intended for precise measurements also contain an internal thermometer so that this information can be collected in the event that temperature corrections will be made.
Owing to the ease with which they can be observed, densities are widely employed to estimate the composition or quality of liquid mixtures or solutions, and in some cases determine their commercial value. This has given rise to many kinds of hydrometers that are specialized for specific uses:
Battery hydrometer - theory
Aquarium salinity hydrometer
A boat with depth markings on its body can be thought of as a gigantic hydrometer!
Sugar and syrup hydrometer
Don't confuse them!
A hydrometer measures the density or specific gravity of a liquid
a hygrometer measures the relative humidity of the air
Hydrometers for general purpose use are normally calibrated in units of specific gravity, but often defined at temperatures other than 25°C. A very common type of calibration is in "degrees" on various arbitrary scales, of which the best known are the Baumé scales. Special-purpose hydrometer scales can get quite esoteric; thus alcohol hydrometers may directly mesure percentage alcohol by weight on a 0–100% scale, or "proof" (twice the volume-percent of alcohol) on a 0-200 scale.
Measuring the density of a solid that is large enough to weigh accurately is largely a matter of determining its volume. For an irregular solid such as a rock, this is most easily done by observing the amount of water it displaces.
A small vessel having a precisely determined volume can be used to determine the density of powdered or granular samples. The vessel (known as a pycnometer) is weighed while empty, and again when filled; the density is found from the weight difference and the calibrated volume of the pycnometer. This method is also applicable to liquids and gases.
In forensic work it is often necessary to determine the density of very small particles such as fibres, flakes of paint or metal, or grains of sand. Neither the weight nor volumes of such samples can be determined directly, so the simplest solution is to place the sample in a series of liquids of different densities, and see if it floats, sinks, or remains suspended within the liquid. A more sophisticated method is to layer two liquids in a vertical glass tube and allow them to slowly mix, creating a density gradient. When a particle is dropped into the tube, it sinks to a depth that matches its density.
This reference provides a brief summary of some of the modern methods of determining density.
The most famous application of buoyancy is due to Archimedes of Syracuse around 250 BC. He was asked to determine whether the new crown that King Hiero II had commissioned contained all the gold that he had provided to the goldsmith for that purpose; apparently he suspected that the smith might have set aside some of the gold for himself and substituted less-valuable silver instead. According to legend, Archimedes devised the principle of the “hydrostatic balance” after he noticed his own apparent loss in weight while sitting in his bath. The story goes that he was so enthused with his discovery that he jumped out of his bath and ran through the town, shouting "eureka" to the bemused people.
If the weight of the crown when measured in air was 4.876 kg and its weight in water was 4.575 kg, what was the density of the crown?
Solution: The volume of the crown can be found from the mass of water it displaced, and thus from its buoyancy: (4876 – 4575) g / (1.00 g cm–3) = 301 cm3. The density is then
(4876 g) / (301 cm3) = 16.2 g cm–3
The densities of the pure metals: silver = 10.5, gold = 19.3 g cm–3,
One of the delights of chemical science is to find way of using the macroscopic properties of bulk matter to uncover information about the microscopic world at the atomic level. The following problem example is a good illustration of this.
Estimate the diameter of the neon atom from the following information:
Density of liquid neon: 1.204 g cm–3; molar mass of neon: 20.18 g.
Solution: This problem can be divided into two steps.
1 - Estimate the volume occupied by each atom. One mole (6.02E23 atoms) of neon occupy a volume of (20.18 g) / (1.204 g cm–3) = 16.76 cm3. If this space is divided up equally into tiny boxes, each just large enough to contain one atom, then the volume allocated to each atom is given by: (16.76 cm3 mol–1) / (6.02E23 atom mol–1) = 2.78E–23 cm3 atom–1.
2 - Find the length of each box, and thus the atomic diameter. Each atom of neon has a volume of about 2.8E–23 cm3. If we re-express this volume as 28E–24 cm3 and fudge the “28” a bit, we can come up with a reasonably good approximation
of the diameter of the neon atom without even using a calculator. Taking the volume as 27E–24 cm3 allows us to find the cube root, 3.0E–8 cm = 3.0E–10 m = 300 pm, which corresponds to the length of the box and thus to the diameter of the atom it encloses.
The accepted [van der Waals] atomic radius of neon is 154 pm, corresponding to a diameter of about 310 pm. This estimate is suprisingly good, since the atoms of a liquid are not really confined to orderly little boxes in the liquid.
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the highlighted terms in the context of this topic.
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