Notes on Earth Science activities that incorporate Chemistry into Earth Science, from the Worlds of Change Workshop 2003 “Doing Chemistry While Exploring Earth Science, by Russ Colson, MSUM.

Chemical partitioning activity notes:  Why do pollutants sometimes end up in the water and sometimes in the soil?

Condensation from the air notes:  How does rain come from thin air?



Chemistry of course!  But the partitioning of elements between different Earth reservoirs is a powerful tool for understanding other aspects of our Earth as well, ranging from the formation of Earth’s Core, Mantle, and Crust to why ore deposits occur where they do.

Some background to put the chemistry in the context of the overall Earth Science puzzle

Some Philosophical opening remarks:
    Who we are as a society reflects in part the fact that natural resources are not evenly distributed.  This heterogeneous distribution encouraged ancient societies to learn to communicate and get along.  Trade in resources further encouraged the trade of ideas and cultures as well.  Uneven distribution of resources encouraged the occupation of territories that might have otherwise been avoided.  Many societies became influential in the world of ideas in part because they were at crossroads of trade where they absorbed ideas from many places, digested them, and spread them to the world in a new form because of their position at a center of trade.  Ancient Palestine was at such a crossroads of trade for many centuries.  Much of our culture, writing, and religion can be traced to this region of the world.  Thus, to some extent, we can think of the distribution of resources on Earth as a nursemaid beckoning us to explore, think, share ideas, and learn to get along with other people.

Chemical Differentiation: Process for global change

Chemical differentiation is the process by which a single chemical substance is divided into two or more new substances each with a different chemical composition than the original (thus the words "chemical" and "different" in "chemical differentiation").  The number of processes by which earth (and other planets) can be differentiated is almost infinite.  However, every process can be described as having two steps:


1)  Chemical separation:  a process whereby a single substance becomes two different substances.  For example, if we partly evaporate salt water, the salt stays with the liquid part (making it saltier) and the vapor part is mostly salt free.  Thus we have two new compositions.

2)  Physical separation:  Once we have two new substances, we need some process for physically separating them.  For example, when water is partly evaporated, the vapor part is less dense and rises up out of the liquid into the air, thereby separating the two.

Chemical Separation: 

The key concepts in understanding chemical separation are phase change and partitioning.


1)  A particular collection of atoms can exist in a number of different phases, depending on the temperature, pressure, concentrations of various elements, pH, oxygen activity, and other properties of the system (e.g. water exists as solid at low temperature, liquid at higher temperature, and gas at higher temperature still).  As seen in the example of water, the stable phase can change as conditions change.  Phase changes occur in many situations.  Mollusks crystallize solid CaCO3 out of the sea to make shells, olivine and other minerals crytallizes from molten rock (magma) as it cools within the earth or at the earth's surface, ore minerals can crystallize from hot water within cracks in the rock of the earth, grease coagulates out of your greasy soup as it cools, raindrops or snowflakes condense out of air, and so on.

2)  Partitioning is a measure of how a particular element partitions, or distributes itself, between two phases.  Every element of the periodic table will have a natural, or equilibrium, distribution between any two phases.  The distribution will depend on temperature, pressure, composition and other properties of the system.  Partitioning can be defined as the concentration of an element in one phase divided by the concentration in another phase.  For example, the concentration of salt in liquid water (in the thought experiment above) is much higher than the salt concentration in the water vapor.  Thus the partition coefficient (D) for salt in liquid water/salt in vapor is very large.  The partition coefficient is therefore a measure of the tendency of an element to go into one phase rather than another.  If D>>1, most of the element will go into the phase put on top of the ratio.  If D<<1, most of the element will go into the phase put on bottom of the ratio.  If D is nearly equal to 1, then the element "likes" each phase about equally.

Example calculation of a partition coefficient:  Partition coefficients are measured experimentally in a laboratory by measuring the concentrations of an element in phases that are held in chemical equilibrium.  Suppose that you do an experiment to measure the partitioning of a pollutant (we'll call it Death-X) between liquid water and clay.  You add 1 gram of Death-X to 10 grams of water and 5 grams of clay.  Concentration is defined as grams of Death-X for each gram of water or clay.  You stir the clay into the water, allowing the Death-X to reach equilibrium between clay and water.  You measure the amount of Death-X in the water (defined as grams Death-X) and find it to be 0.1 gram.

  Question 1:  Is most of the Death-X in the water or the clay?
  Question 2:  What is the partition coefficient for Death-X in Clay/Death-X in water?
  Question 3:  If you spilled the Death-X on a clay-rich layer of sediment above your neighbors well (oops), do you think it would be likely to reach the water table and get into your neighbors wellwater?


Question 1:  Since 0.1 grams Death-X is in the water, the rest, 0.9 grams, must be in the clay.  Thus, most of the Death-X

is in the clay.


Question 2:  The concentration of Death-X in clay is 0.15 (0.9 grams Death-X in 5.9 grams total Death-X + clay).  The

concentration Death-X in water is 0.0099 (0.1 grams Death-X in 10.1 grams total Death-X + clay).  The partition

coefficient for Death-X in clay/water is thus 15.


Question 3:  Since the Death-X tends to stay with the clay, rain water seeping down toward the water table and the well

intake will have the Death-X removed from it by the clay, and little will reach you neighbors well (hee, hee, maybe you

won't have to tell least until everything dies at the surface of the ground).




1)  Oil deposits:  Oil forms when microscopic bits of organic matter (dead planktonic creatures mainly) are buried in rock.  Pressure and temperature generate an oily material which is less dense than the water.  The water and oil both exists in the pores of the rock.  Because the oily material is less dense than water, it rises buoyantly until it either escapes to the Earth's surface (and is lost to human use) or it is trapped in the Earth and accumulates.

2)  Core and Crust Differentiation:  The most significant event in the chemical history of Earth was the separation of a metallic core from the bulk material of the planet.  This occurred as a metal phase formed from the silicate, rock-like material of a hot earth (possibly partly molten).  The metal phase, being more dense than the silicate, sank to the center of the Earth.  A smaller differentiation, but one with large consequences for the surface of the Earth where we live, was the formation of Earth's crust.  This occurred substantially later than the formation of the core.  Areas of partial melting occurred within the otherwise solid mantle.  The liquid had a different composition and was less dense than the mantle, rising toward the surface.  This less dense material, through many intermediate differentiation steps, became the crust.

Question 1:  Au (gold) and Ni (nickel) have partition coefficients metal/silicate that are very high.  Where do we expect most of Earth's gold and nickel to be found?

 3)  Chromium Ore Formation:  Chromite, an ore of chromium, crystallizes from basaltic magma.  Because it is more dense than the magma, it sinks.  This can produce layers of concentrated chromite at the bottom of former magma chambers (former magma chambers because we can not go into magma chambers when they are still molten, and because magma chambers occur deep in the Earth where we can't easily go, we need to allow erosion to excavate these chambers for us after they have cooled). 



Some simple experiments students can do to measure the partitioning of a ‘pollutant’ (food coloring) between water and two different types of sediment or rock:

In this experimental activity, students learn about 1)  the different chemical properties of sediments,  2) the concept of partitioning which is used widely in geochemical studies in Earth Science, and  3) the most basic concept in analytical chemistry, the concept of the standard.

Partitioning Lab

Materials:  paper cups, beakers, sand, clay (or cornstarch), graduated cylinders, food coloring, brain.

DO THIS FIRST:  Put 5 drops of red food coloring in 50 ml water and mix.  Add 20 ml of clay (or cornstarch) and stir

until mixed.  Let this sit for 30 minutes without disturbance while you continue the lab.

Goal:  Measure the partitioning between sand and water and between clay and water (we may use cornstarch as a substitute for clay).  

We will use red food coloring to represent some element or pollutant partitioned between each of these pairs.

This experiment should be quantitative, meaning that you need to get a real number.  In addition, you need to consider units.  We will use drops of red food coloring per ml as units.  Note that this does not mean you have to use this concentration or amount in your experiment!

Because this experiment is quantitative, you need to have some way of determining how much red food coloring is in the water and in the sand or clay.  HINT:  Most quantitative chemical work involves comparing what you want to analyze against a standard solution, or a series of standard solutions whose concentrations you know (because, for example, you made the standard mixtures).  Comparisons can be visual and still be considered quantitative. For example, you could make a series of standards solutions with different numbers of drops of food coloring, and compare these to your unknown solution.

HINT:  get organized first, figure out what you want to do.

HINT: your standard solutions need to be in the same range of compositions as your experiments.  The concentration of food coloring can't be so low as to be undetectable with your eye, nor so high as to make variations in concentration undetectable.  (e.g. you may want to have standard solutions whose concentrations range from 1 to 5 drops per 50ml water).  Also, you probably want only about 20 ml or thereabouts of standards solution, so that you don't need so much of the unknown solution to compare against.

HINT:  you can figure out the amount of food coloring in the clay or sand by difference between what you started with and what is left in the water.

HINT:  you can calculate the amount of food coloring by determining the concentration, then scaling to an amount of water.  This lab involves understanding how to determine concentration and what it means mathematically.  For example, you don't have to have 50ml of water to have a concentration of 4 drops per 50 ml.  I want you to try to figure this out on your own, but if it doesn't make sense or you aren't sure, be sure to talk to me about it because if you don't understand what concentration is, you can't understand these experiments.

In your report: be sure to explain your experimental procedures carefully, your experimental results, and how you interpret those results, about 1.5 pages.  In addition, your report should include your measurement and calculation of the following

partition coefficients (again make sure you understand what concentration is!).



A simple activity with beans to help students understand the concept of partitioning (also an interesting exercise in proportions, amounts, and concentrations):

Let’s consider a simplified puzzle for the separation of the Earth’s core from the original Earth.  If we remove Iron (Fe) from the original Earth, how will the composition of the remaining Earth change (the part that became the mantle and crust of the Earth)?

Let’s only consider 3 components for the Earth:  Si (plus oxygen), Fe (plus oxygen), and Mg (plus oxygen).  These are three of the most important components of the Earth.  We can symbolize them with Red Beans (for Si), Black Beans (for Fe), and Lima Beans (for Mg).

Let’s start with 200 total beans to represent the original Earth, 50 Red Beans (Si), 50 Lima Beans (Mg), and 100 Black Beans (Fe).

What percent of the original Earth (in this imaginary problem) is Si?  What percent is Mg?  Fe?  (25%, 25%, 50%)

Let’s now consider the separation of the Core (for our thought problem it will be entirely Fe - in the real case it is more complex than this).

First, make a prediction:  How do you expect the concentration of Fe, Si, and Mg in the remaining material (which becomes the mantle) to change?

Expectation for Fe:  increase, decrease, or stay the same

Expectation for Si:  increase, decrease, or stay the same

Expectation for Mg:  increase, decrease, or stay the same

Now, let’s do the problem with the beans.  Suppose that the partition coefficient for Fe (metal/silicate) = 3.  Apportion your beans into two piles, one representing the core (all Fe), and one representing the Mantle (Si, Mg, and Fe), such that the partition coefficient for Fe (concentration of Fe in metal/concentration of Fe in silicate) = 3.

Now, calculate the concentrations in the silicate material:

Fe concentration =

Si concentration=

Mg concentration=

(note:  you should end up with 25 black beans (Fe) in the core, and in the mantle you end up with 25 each of black beans (Fe), Lima beans (Mg), and Red beans (Si), resulting in concentrations of 33.33% for each, for an increase in the concentration of Mg and Si and a decrease in Fe.)

  • How does rain come from thin air?

     Why, it comes from clouds, of course!  But where do the clouds come from?  How do you make a cloud?


Understanding how rain or clouds form, a study within Earth Science, depends on understanding the chemistry of air, and how water can dissolve in the air, and how its ability to dissolve in the air is a function of temperature.


Some background to put the chemistry in the context of the overall Earth Science puzzle:
     Clouds are formed from water that was dissolved in the air and condenses out in small droplets.  We can watch water dissolve into the air when we boil water.  The water becomes part of the air.  When we set our cold Coke can out on a hot, muggy day, we can see the water undissolve, it condenses back out of the air.  From these two observations, we can infer how the solubility of water in air (that is, how much water dissolves in the air) depends on temperature, which is the key to understanding how we get clouds.  If you drew a graph showing the relationship between air temperature and it’s capacity to hold water, what would the graph look like?  
     Dewpoint is the temperature at which water will start to condense from air as it is cooled.  Dewpoint is therefore a measure of how much water is in the air.
     Relative Humidity is the amount of water in the air/amount of water the air can hold at that temperature x 100%.  Relative humidity is therefore a measure of how much water is in the air compared to how much water the air could possibly hold at that temperature.
    Since cold air can hold less water than warm air, we can squeeze water out of the air simply by changing the temperature of the air (to make dew, or fog, or snow or clouds)!  If we cool the air, it will hold less dissolved water, and some of that water will condense out as liquid or solid water.
     So how can we cool air?
     One way is simply to radiate energy away from the Earth at night (black body radiation).  This cools the ground, which cools the air next to the ground.  Under these conditions we may get dew, or frost, or fog as water condenses out of the air near the ground.
     But the big way to cool air for making clouds and rain is by adiabatic decompression (isn't that the coolest thing to say?).  When air is compressed, its temperature rises.  When air is decompressed, its temperature falls.  You can observe this with a manual tire pump.  The pump gets hot as you compress the air into your tire.  Cans of compressed air can be purchased (for cleaning items you don't want to get wet).  The nozzle of these get very cold as the compressed air decompresses on its way out.  But maybe the easiest way to observe this effect is with a balloon.  Blow up a balloon, then hold it against your cheek and let the air out quickly.  You can feel the balloon cool as the air in it decompresses.
     Because pressure is lower higher up in the atmosphere, rising air will decompress and cool adiabatically.  Air may rise because it is warm and buoyant (warm air is less dense and rises), or air may be forced up over mountains by winds, or air may be forced up by a cold or warm front.  Regardless of how the air is made to rise, the rising results in decompression and cooling.  The cooling can squeeze out the water dissolved in the air, and we get clouds and precipitation.
      The evaporation and condensation of water is one of the main ways that Earth moves energy (remember that weather is the effort to even-out Earth's energy distribution).  To evaporate water, you have to add energy to it (think about adding heat to boiling water, or how water evaporating from your skin takes energy from you and makes you cooler).  That energy doesn't simply vanish from existence.  When the water is re-condensed as rain or snow, we get that energy back.  This is a major source of energy driving storms such as hurricanes, thunderstorms, and even winter storms.  When water evaporates from the Gulf of Mexico, travels on winds up to Fargo-Moorhead and falls as snow in a big winter storm, the effect is to move a huge amount of heat from down closer to the equator to up here closer to the north pole!



Some simple experiments students can do to measure the solubility of water in air as a function of Temperature:


Experiment statement:  In any experiment, you only want two variables to change, the variable that you are changing, and the response to that change.  e.g. You can change Temperature, and measure the  response of the air’s capacity to hold water, keeping pressure, air volume, and other variables constant.


The following are experiments that students in my classes have come up with to measure the relationship between the air’s capacity to hold water and temperature of the air.


Method 1:  Observe the temperature at which a known amount of water can dissolve in a known volume of air at various temperatures:


Get a large, clear, empty pop bottle (the bigger the better, but say 2 liters).  You need the cap.  Dry the bottle as much as possible by subjecting it to low-humidity conditions (warm, dry air).  Put a small pin-hole in the bottle (this is so that pressure does not build inside the bottle when you close it up and change temperature----the air exchange through this pin-hole should be small compared to the overall experiment and not significantly compromise your results). 

Place a drop of water in the bottle (use an eyedropper so that the amount of water added is reproducible), along with a thermometer and cap it.  Place the bottle in cold water, and gradually increase the T of the water until you see the drop of water completely evaporate in the bottle.  Record the air temperature in the bottle.  You can determine how much water is in one drop by dropping 100 drops in a graduated cylinder and dividing the amount by 100.

Repeat this process for 2, 3, and 4 drops of water.  Plot your results on a graph showing milliliters H2O soluble in 2 liters of air (or whatever your bottle volume is), versus temperature of the air.



Method 2:  Measure the amount of water that dissolves in air at a variety of different temperatures.


Get several bottles (should be at least 2 liters)  and a means to connect them to a smaller bottle or another 2-liter bottle (bottle connectors are available at scientific outlets, used for ‘tornado in a bottle’ demonstrations, or you can use some good tape if you seal it thoroughly).  Poke a pin-hole in the two liter bottle (to allow pressure equalization when air in the bottles is heated or cooled).  Add water to the other bottle (20-30 ml is plenty).  Place a thermometer in the bottles to measure the air temperature.  Seal the two bottles together, taking care that no water spills into the water-free 2-liter bottle.   Bring the entire system to the desired temperature (placing in water bath, and controlling the temperature of the water bath works well, so long as there is no leak -- the pin-hole can be at one end and left extending out of the water bath slightly). 

Give this system time to equilibrate (so that water vapor in both bottles equilibrates with the water present in the one).  Then, detach and seal the bottle NOT containing the water (with bottle cap and tape over pin hole).

Cool the entire bottle in an ice + saltwater bath (salt decreases the temperature of an ice-water bath which will be at 0C without salt).  This will cause most of the water vapor present to condense.  Cut open the 2-liter bottle and swab out the water inside with an absorbent cloth whose mass you have previously determined carefully.  Redetermine the mass of the cloth quickly after swabbing out all the water (quickly, because water may be evaporating as you do it – watch how quickly the mass changes with time to get some idea of how much water may have evaporated before you got it weighed). 

Repeat this for other air temperatures, using additional 2-liter bottles.  Plot your results on a graph showing milligrams H2O soluble in 2 liters of air (or whatever your bottle volume is), versus Temperature of the air.


Note:  For both of these experiments, larger variations in temperature is more likely to yield results that are not obscured by errors in the measurements.


 What are errors that arise in each method?  Which is more precise?  Which is more accurate?  How do the results of the two methods compare?


An activity to illustrate conceptually how solubility of water changes with air temperature and how this affects relative humidity and condensation in the air.


Illustrative activity:  What are condensation/humidity/dewpoint?

The following activity is an illustrative activity, not an experiment.  An illustrative activity helps one understand a concept that has already been figured out through experiments by other people.

This activity was developed for Project Atmosphere and published by the American Meteorological Society (1993), who grant permission for its reproduction so long as it is used for non-commercial educational purposes.

Activity: Water Vapor Investigation

Four 12-ounce clear or translucent drinking cups for every group, enough beans or popped popcorn to fill one cup, scissors, permanent marking pen.

Procedure to make a set of three cups that each hold ½ the volume of the previous cup (12 oz, 6 oz, 3 oz)  (this step is already done for this activity today)
Prepare a set of cups for use in this activity. Fill a 12-ounce cup to the brim with water. Pour from that cup into another until the water levels in both are the same. Trace the water line on the outside of both cups with a permanent marking pen. Empty the water from one of these and cut along the traced line to make a 6-ounce cup. Now pour water remaining in the 12-ounce cup into the third until their water levels are the same. Trace their water levels. Cut the third cup down to the water level to make a cup that holds 3 ounces. You should end up with one 3-ounce cup, one 6-ounce cup that has a line marking the 3-ounce level, and one 12-ounce cup with lines marking 6- and 3-ounce levels. Write a large 0 on the side of the smallest cup, 10 on the middle size, and 20 on the largest cup to indicate 0, 10, and 20 degrees Celsius. Use these Cups as guides to preparing other sets. The fourth and unmarked cup will hold the Styrofoam beans or popped corn.

After completing this investigation, you should be able to:

• Explain with cups of different sizes how the "capacity" of air to hold water vapor varies with temperature.

• Use the cups and Styrofoam beans as a model to explain relationships between the "capacity" of air to hold water vapor and the actual amount of water vapor in the air.

Cups of different sizes are used in this exercise to represent the "capacity" of air to hold water vapor at 0, 10, and 20 degrees C.  Beans are poured into the cups to represent the water vapor actually in the air.

Start the activity by filling the large unmarked cup approximately level with beans. Tap the cup gently on your table or desk top to help the material settle as you fill the cup. This is the supply of beans you will use in this exercise.

This activity involves the use of four cups. The unmarked cup is used to store the beans. The largest marked cup is twice the capacity of the mid-size cup. The mid-size cup is twice as big as the small cup. The marked cups represent the "capacity" of air to hold water vapor at 0, 10, and 20 degrees Celsius. Each cup is labeled by the temperature related to its capacity.

1. Fill the small 0-degree cup with beans until the contents are level with the top of the container. Pour the contents into the mid-size 10-degree cup. Repeat this until the 10-degree cup is level full. Now pour filled 10-degree cups into the 20-degree cup until it is full. Assuming the cups represent the capacities of air to hold water vapor at 0, 10, and 20 degrees, complete the following statement:

The capacity of the air to hold water vapor approximately ______________ when the temperature increases 10 Celsius degrees.

2. Starting with a filled 20-degree cup, pour its contents into the 10-degree cup until it is level full. Now pour the contents of the 10-degree cup into the 0 -degree cup until filled to the brim. Now complete the following statement:

The capacity of the air to hold water vapor approximately ________________when the temperature lowers 10 Celsius degrees.

3. Now empty the 20-degree cup and pour filled 0-degree cups into the large cup until it is full. Based on this, complete the following statement:

The capacity of the air to hold water vapor increases approximately ____________times when the temperature rises 20 Celsius degrees.

4. According to the same observations, what happens to the capacity of air to hold water vapor as the temperature falls 20 Celsius degrees?

5. When air cools, its capacity to hold water vapor decreases, and any excess water vapor must condense. This can be demonstrated by attempting to pour all the beans from a filled 20-degree cup into the 10-degree cup. Level the top on the 10-degree cup. The overflow represents the water vapor that condensed out. In this example of the 10-degree cooling, how much of the water vapor condensed to liquid as the temperature dropped 10 degrees?

6. Air filled to its capacity with water vapor is called saturated air. If saturated air at 20 degrees is cooled 20 degrees, how much of its water vapor must condense?
  7. Saturated air has a relative humidity of 100%. Relative humidity is a measure of the amount of water vapor actually in the air compared to the amount it would hold if saturated at the same temperature. Pour a full 0-degree cup into a 10-degree cup to determine what the relative humidity would be if air saturated at 0 degrees is warmed 10 degrees with no addition of water vapor. What is it? What would the relative humidity be if that same air were warmed another 10 degrees to 20 degrees Celsius?
  8. Explain in your own words why in cold weather the relative humidities in heated buildings (without humidifiers) are quite low.

9. Dew point is another common humidity measure. It is the temperature to which air has to be cooled (without changing the amount of water vapor in the air) to become saturated. Whenever air is saturated, its temperature and dew point will be the same. What is the approximate dew point of air at 20 degrees with a relative humidity of 50%? To find out, fill the 20-degree cup half full. Then, pour it into the 10-degree cup.
  10. What is the dew point of air saturated at 0 degrees when the air temperature is raised to 10 degrees without the addition of water vapor? To find out, pour a filled 0-degree cup into 10-degree cup and ask yourself whether or not the dew point changed.

11. If saturated air at 20 degrees is cooled to 10 degrees, what is its final dew point? To help find your answer, attempt to pour a filled 20-degree cup into 10-degree cup while asking yourself how much water vapor the 10-degree cup is holding compared to its capacity.

12. In general, when saturated air is cooled, what happens to its capacity to hold water vapor, its dew point, and its relative humidity? Refer to your observations made above.

13. Describe, in your own words, the water vapor and temperature relationships which must exist for cloud, dew, and frost formation.