A Very Large, Salty Cube

The Conundrum

The Royal Society of Chemistry posted this chemical conundrum in July:

Imagine that all the sodium chloride dissolved in the earth’s seas and oceans were crystallised into one giant cube-shaped crystal. What length would the side of the cube be? Suggest what difference it would make if all the salt was removed from the seas and oceans.

I decided to undertake this challenge, which was perhaps a foolish decision as a medical student considering that I had only done Chemistry and Mathematics to A-Level and Geography to GCSE, but it seemed like a fun challenge and it would be interesting to see just how much salt there is in the oceans. My solution (pardon the pun) is as follows, with most values rounded to 3 significant figures.

The Solution

To obtain the length of the cube, I have to first calculate the volume of the cube, and take the cube root of the volume. In order to calculate the volume of the cube, I will divide the total mass of sodium chloride in seawater by the density of a giant cube-shaped crystal of sodium chloride.

Unfortunately, I’m not quite able to carry out the necessary experiments to find the total mass of NaCl in seawater (it would probably involve boiling off all the water in the seas and weighing the solutes – in any case, living in the Midlands, I don’t even have good access to the sea!). Thus, I will use values taken from other sources of the concentration of ions in seawater and of the total volume of seawater and use the equation mass = concentration × volume to work out the total mass.

From a handbook made by the US Department of Energy, I found out that:

Moles of chloride per kilogram of seawater = 0.546 mol kg^-1

Moles of sodium per kilogram of seawater = 0.469 mol kg^-1

According to this source, the chloride ions are in excess and so the maximum moles of sodium chloride obtainable per kilogram of seawater is equivalent to that of the sodium ions, 0.469 mol kg^-1, as the excess chloride ions would be bonded to other cations such as magnesium or calcium present in seawater.

I then however have a problem with converting the concentration of NaCl into units of moles per volume, as opposed to moles per mass of seawater. The problem lies with the fact that seawater changes its density depending on the conditions. In general, the greater the temperature, the less dense the seawater is, but according to the book Physics for Scientists and Engineers, the density of seawater is 1030 kg m^-3 under standard conditions.

I am forced to make an assumption that all of seawater is under standard conditions (what these standard conditions are, I am uncertain, although the IUPAC standard is 0°C and 100kPa) but of course this isn’t true. Ocean and sea water temperature can vary from just below 0°C (in the icy Arctic, kept liquid because of the saltwater solution) up to 25°C at the surface of some equatorial regions.  At the depths of the ocean, the pressure exerted by water can rise over 100,000 kPa.

So taking the density of seawater as 1030 kg m^-3, the sodium chloride concentration in seawater in moles per volume is 0.469 mol kg^-1 × 1030 kg m^-3 = 483 mol m^-3 = 0.483 mol dm^-3

Using the value from a 2008 encyclopedia, the volume of total seawater on earth (of all oceans and seas) is 1.5 x 10²¹ dm³; therefore total moles of sodium chloride in oceans and seas can be calculated by 1.5 × 10²¹ dm³ multiplied by 0.483 mol dm^-3

= 7.25 × 10²⁰ mol of NaCl

I’m also assuming that the concentration of sodium chloride ions is the same throughout the whole body of water, although this is unlikely to be true as the solubility of the ions will depend on temperature.

Next, I have to work out the total mass of NaCl in seawater to which 7.25 × 10²⁰ mol of NaCl would correspond. The periodic table gives us the following values.

Molar mass of Na = 22.99 g mol^-1

Molar mass of Cl = 35.45 g mol^-1

Therefore molar mass of NaCl is 58.44 g mol^-1 or 0.0584 kg mol^-1

Therefore the total mass of NaCl in seawater can be obtained by: 7.25 × 10²⁰ mol × 0.0584 kg mol^-1

= 4.24 × 10¹⁹ kg

So the volume of the giant cube would be mass ÷ density, and now, having found the mass, I need to work out the density of the cube.

To find the density of the giant crystal, I could just use the NaCl density value taken from a source, which would have been a simple case of obtaining some salt, weighing the salt, and measuring the volume and using the equation density = mass ÷ volume. But where is the fun in that?!

Instead I’m going to work out the density of a NaCl crystal by envisioning how the ions pack together. I will also use the equation density = mass ÷ volume, but instead of measured values, I will take the mass as the mass of all the ions in a unit cell of a sodium chloride crystal, and the volume as the volume of a unit cell. A unit cell is the simplest repeating unit of a crystal, like a three-dimensional tessellation.

According to Wikipedia (who said it wasn’t a good source of information?!), NaCl has a face-centred cubic structure with a unit cell length of 564.02 picometres, which has been determined by x-ray crystallography. This means that the repeating structure is of a cube (which is very handy as the giant cube is just a larger scale version of the unit cell) with a length of 564.02 × 10^-12 metres.

The mass of all the ions in a unit cell can then be calculated from this information as the mass in each unit cell in kg = number of ions in each unit cell × Mr of NaCl in kg mol^-1 ÷ Avogadro’s number (number of ions per mol of substance).

The number of ions in each unit cell is more difficult to work out.

Because the crystal unit cell structure is face-centred cubic (as shown in the diagram on the right from the Queen’s University in Canada), in each of the eight corners, one ion (shown in purple) is shared equally across eight cubes (the other seven are not shown for each purple ion), which means that in each unit cell, there is one whole purple ion (1/8 × 8 = 1).

For the yellow ions which are located at the six faces of the cube, one ion is shared across two cubes (the other half of the cube is not shown for each yellow ion). The diagram seems to show only three yellow half-ions as the other three half-ions are hidden from view; in reality, there would be six half-ions per unit cell. This means that in each unit cell, there are three whole yellow ions (1/2 × 6 = 3).

This means that in total, there will be four ions inside a unit cell, which would be an average of two Na⁺ ions and two Cl^– ions to make the charge of a unit cell neutral.

So the mass of a unit cell = 4 ions × 0.0584 kg mol^-1 ÷ (6.02 × 10²³ ions mol⁻¹)

= 3.88 × 10^-25 kg

The volume of a face-centred cubic cell is obtained by simply cubing the unit cell length (L) as the unit cell is simply a cube.

= (564.02 × 10^-12 m)³

= 1.79 × 10^-28 m³

The density of a giant cube-shaped crystal of NaCl (assuming that the density of a giant cube-shaped crystal of NaCl is identical to that of a unit cell of NaCl)

= mass ÷ volume

= 3.88 × 10^-25 kg ÷ 1.79 × 10^-28 m³

= 2163 kg m^-3 (close to Wikipedia’s 2165 kg m^-3 value!)

Therefore the volume of the giant cube-shaped crystal

= mass ÷ density

= 4.24 × 10¹⁹ kg ÷ 2163 kg m^-3

= 1.96 × 10¹⁶ m³

And the length of the side of the cube is just the cube root of the volume = 269,511.3845 m

= 270,000 m to 3 significant figures

So there you have it! If all the sodium chloride dissolved in the earth’s seas and oceans were crystallised into one giant cube-shaped crystal, the length of the side of the cube would be around 270km, which is around the same distance as the straight line from Oxford to Guernsey, or Oxford to Dunkirk. Comparatively, the altitude of the official boundary between the atmosphere and outer space is 100km above sea level.

A Deadly ‘A Salt’ on Sea Levels

Tackling the next part of the conundrum was even harder, but I gave it a shot anyway. If all salts were suddenly removed from the oceans and the sea, aside from the great (and probably fatal) osmotic damages to many marine organisms as well as the sudden lack of supply of NaCl for human consumption, there would be a significant change in the volume of oceans and seas.

The density of pure water is 1000 kg m^-3 under standard conditions. Going back to the textbook values, assuming seawater is under standard conditions, the total mass of seawater on earth can be calculated by  1.5 x 10¹⁸ m³ x 1030 kg m^-3 = 1.545 x 10²¹ kg.

According to Wikipedia, on average, seawater in the world’s oceans has a salinity of about 3.5%. Therefore the mass of water in seawater = 1.545 x 10²¹ kg x (100-3.5)% = 1.490925 x 10²¹ kg. Again, assuming standard conditions, the volume of seawater with salts removed (i.e. pure water)= 1.490925 x 10²¹ kg ÷ 1000 kg m^-3 = 1.490925 x 10¹⁸ m³.

This is a difference in volume of 9.08 x 10¹⁵ m³ or 0.605% percentage reduction. Although this seems like a very small percentage difference, this volume is 2.42 times the volume of the Mediterranean Sea. The change in sea levels due to the loss of volume is less clear as the sea levels are affected not just by the amount of water but also due to the temperature of the water.

However, assuming that the earth were a complete sphere (with a radius of 6,371,000 m) and all the water of the oceans lay on top of land, then the decrease in sea levels could be expected to be around 17.8m to 3 significant figures.

Radius of the earth = 6,371,000 m

Volume of a sphere = (4/3) π r³

The change in volume can be given by the original radius, minus the new radius, as shown below, where h is the change in sea levels expected:

(4/3)  π (6371000)³ –  (4/3) π (63710000 ­+ h)³ = 9.075 x 10¹⁵ m³

When solving for h, the value of h = –17.79191817m, or a decrease of 17.8m to 3 significant figures. However, of course this value is not a valid estimation as oceanic topology is not as is assumed for the calculation.

I would like to hear from other science students (or recreational mathematicians!) to see what figures they obtain, especially for the second part of the challenge. Was there a factor I missed out? A calculation I got wrong? Let me know in the comments section below!

Sources

Beichner, R. J. (2000). Physics for Scientists & Engineers. Orlando: Saunders College Publishing.

Current sea level rise. (n.d.). Retrieved from Wikipedia: http://en.wikipedia.org/wiki/Current_sea_level_rise

Dickson, A. G., & Goyet, C. (1994). Handbook of Methods for the Analysis of the Various Parameters of the Carbon Dioxide System in Sea Water. U. S. Department of Energy.

Earth radius. (n.d.). Retrieved from Wikipedia: https://en.wikipedia.org/wiki/Earth_radius

Face centered cubic. (n.d.). Retrieved from Department of Chemistry at Queen’s University, Ontario, Canada: http://www.chem.queensu.ca/people/faculty/mombourquette/firstyrchem/Solids/FCC.htm

Ionic Radius. (n.d.). Retrieved from Wikipedia: http://en.wikipedia.org/wiki/Ionic_radius

Periodic Table of the Elements by WebElements. (2012). Retrieved from WebElements: http://www.webelements.com/

Properties of water. (n.d.). Retrieved from Wikipedia: https://en.wikipedia.org/wiki/Properties_of_water

Seawater. (n.d.). Retrieved from Wikipedia: http://en.wikipedia.org/wiki/Seawater

Sodium chloride. (n.d.). Retrieved from Wikipedia: http://en.wikipedia.org/wiki/Sodium_chloride

Van Nostrand’s Scientific Encyclopedia, 10th Edition. (2008). Wiley-Interscience.