Tag Archives: chemistry

The C-320 Fullerene Polyhedron

Posted on by
2

The duals of the geodesic domes are polyhedra with hexagonal and pentagonal faces. This particular one has 320 vertices, with those vertices representing carbon atoms in the molecular version of this solid. Here is C320 as a polyhedron.

C320 Dual of Geodesic Icosa

The next image shows this molecule as a ball-and-stick model.

C320 ball and stick.gif

Finally, here it is as a space-filling molecular model.

C320 space filling.gif

All three images were created with Stella 4d: Polyhedron Navigator. This is the page to visit if you want to try Stella for yourself: http://www.software3d.com/Stella.php.

An Eighty-Atom Fullerene Molecule

Posted on by
1

c80-fullerene

The fullerene molecule that gets the most attention is C60, so I’m giving C80 a little bit of the spotlight, for balance. I made this using polyhedral modeling software called Stella 4d; you can try it for yourself at this website.

For Science Teachers: A Safer Alternative to Liquid Mercury

Posted on by
Reply

Liquid mercury, in schools, poses three major problems:

  1. It is extremely toxic,
  2. It has a high vapor pressure, so you can be poisoned by invisible mercury vapor leaving any exposed surface of liquid mercury, and
  3. Playing with liquid mercury is a lot of fun.

These are compelling reasons to leave use of mercury to those at the college level, or beyond. In the opinion of this science teacher, use of liquid mercury in science classes, up through high school chemistry, inside or outside thermometers, is a bad idea. If the bulb at the bottom of a thermometer, as well as the colored stripe, looks silvery, as in the picture below (found on Wikipedia), then that silvery liquid is mercury, and that thermometer should not be used in labs for high school, let alone with younger children. Your local poison control center can help you find the proper thing to do with mercury in your area; it should definitely not just be thrown away, for we do not need this serious environmental toxin in landfills, where it will eventually reach, and poison, water. Red-stripe thermometers without any silvery line, on the other hand, are far safer, although broken glass can still cause injury.

Maximum_thermometer_close_up_2

I turned ten years old in 1978, and, by that time, I had already spent many hours playing (unsupervised) with liquid mercury, pouring it hand-to-hand, etc., so I know exactly how irresistible a “plaything” mercury can be, to children. Luck was on my side, and I suffered no ill effects, but I can state from experience that children should not be tempted with highly-toxic “mercury as a toy,” for it’s not a toy at all. Mercury spills require special “hazmat” training to clean up safely; anyone encountering such a spill who does not have such training should simply notify the proper authorities. In the USA, this means evacuating the area immediately, and then calling 911 — from far enough away to keep the caller from breathing invisible mercury vapor.

Fortunately, there is a safe alternative which can give students a chance to experiment with a room-temperature metal: an alloy of three parts gallium to one part indium, by mass. Gallium’s melting point is between normal human body temperature and room temperature, so it can literally melt in your hand (although a hot plate is faster). Indium, on the other hand, has a melting point of 156.6°C. For this reason, I will not buy a hot plate unless it can reach higher that that temperature. (Note: use appropriate caution and safety equipment, such as goggles and insulated gloves, with hot plates, and the things heated with them, to avoid burns.)

Once both elements are massed, in the proportions given above, they can then be melted in the same container. When they melt and mix together, they form an alloy which remains liquid at room temperature.

Some might wonder how mixing two elements can create an alloy with a melting point below the melting points of either of the two ingredients, and the key to that puzzle is related to atomic size. Solids have atoms which vibrate back and forth, but don’t move around each other. In liquids, the atoms are more disordered (and faster), and easily slip around each other. In solid, room-temperature gallium, all the atoms are of one size, helping the solid stay solid. Warm it a little, and it melts. With pure indium, this applies, also, but you have to heat it up a lot more to get it to melt. If the two metals are melted and thoroughly mixed, though, and then frozen (a normal freezer is cold enough), the fact that the atoms are of different sizes (indium atoms are larger than gallium atoms) means the atoms will be in a relatively disordered state, compared to single-element solids. In liquids, atoms are even more disordered (that is, they possess more entropy). Therefore, a frozen gallium/indium alloy, with two sizes of atoms, is already closer to a disordered, liquid state, in terms of entropy, than pure, solid gallium or indium at the same temperature. This is why the gallium-indium mixture has a melting point below either individual element — it requires a lower temperature to get the individual atoms to flow past each other, if they are already different atoms, with different sizes.

liquid metals

Those who have experience with actual liquid mercury will notice some important differences between it and this gallium-indium alloy, although both do appear to be silver-colored liquids. (This is why mercury is sometimes called “quicksilver.”) For one thing, their densities are different. A quarter, made of copper and nickel, will float on liquid mercury, for the quarter’s density is less than that of mercury. However, a quarter will sink in liquid 3:1 gallium-indium alloy. To float a metal on this alloy, one would need to use a less-dense metal, such as aluminum or magnesium, both of which sink in water, but float in liquid Ga/In alloy.

Other differences include surface tension; mercury’s is very high, causing small amounts of it on a floor to form little liquid balls which are difficult (and dangerous) to recapture. Gallium-indium alloy, by contrast, has much less surface tension. As a result, unlike mercury, this alloy does not “ball up,” and it will wet glass — and doing that turns the other side of the glass into a mirror. Actual mercury will not wet glass.

The most important differences, of course, is that indium and gallium are far less toxic than mercury, and that this alloy of those two elements has a much lower vapor pressure than that of mercury. Gallium and indium are not completely non-toxic, though. Neither indium nor gallium should be consumed, of course, and standard laboratory safety equipment, such as goggles and gloves, should be worn when doing laboratory experiments with these two elements.

Have you noticed what silver’s been doing lately? The price of silver is literally on fire!

Posted on by
Reply

silver is literally on fire

Because of the price of silver being literally on fire, they will not be buying and selling troy ounces of metallic silver when the markets open in New York tomorrow morning. Instead, they will be selling “oxide ounces” of silver oxide, in sealed-plastic capsules of this black powder, with an oxide ounce of silver oxide being defined as that amount of silver oxide which contains one troy ounce of silver.

silver oxide capsule

A troy ounce of silver is 31.1 grams of that element, which has a molar mass of 107.868 g/mole. Therefore, a troy ounce of silver contains (31.1 g)(1 mol/107.868 g) = 0.288 moles of silver. An oxide ounce of silver oxide would also contain oxygen, of course, and the formula on the front side of a silver oxide capsule (shown above; information on the back of the capsule gives the number of oxide ounces, which can vary from one capsule to another) is all that is needed to know that the number of moles of oxygen atoms (not molecules) is half the number of moles of silver, or (0.288 mol)/2 = 0.144 moles of oxygen atoms. Oxygen’s non-molecular molar mass is 15.9994 g, so this is (0.144 mol)(15.9994 g/mol) = 2.30 g of oxygen. Add that to the 31.1 g of silver in an oxide ounce of silver oxide, and you have 31.1 g + 2.30 g = 33.4 grams of silver oxide in an oxide ounce of that compound.

In practice, however, silver oxide (a black powder) is much less human-friendly than metallic silver bars, coins, or rounds. As you can easily verify for yourself using Google, silver oxide powder can, and has, caused health problems in humans, especially when inhaled. This is the reason for encapsulation in plastic, and the plastic, for health reasons, must be far more substantial than a mere plastic bag. For encapsulated silver oxide, the new industry standard will be to use exactly 6.6 g of hard plastic per oxide ounce of silver oxide, and this standard will be maintained when they begin manufacturing bars, rounds, and coins of silver oxide powder enclosed in hard plastic. This has created a new unit of measure — the “encapsulated ounce” — which is the total mass of one oxide ounce of silver oxide, plus the hard plastic surrounding it on all sides, for a total of 33.4 g + 6.6 g = 40.0 grams, which will certainly be a convenient number to use, compared to its predecessor-units.  

# # #

[This is not from The Onion. We promise. It is, rather, a production of the Committee to Give Up on Getting People to Ever Understand the Meaning of the Word “Literally,” or CGUGPEUMWL, which is fun to try to pronounce.]

 

 

A Tour of the Periodic Table of the Elements, Part 1

Posted on by
Reply
Periodic-Table-of-Elements 1st one with alkali metal and such

(click to enlarge)

In this, and the some upcoming posts, I’ll be showing you various collections of elements on the horizontally-extended version of the periodic table — one that includes the f-block elements in their proper place, rather than relegating them to two separate rows below the other elements. (I’m also suggesting the purple letters A – N for the usually-unrecognized groups in the f-block, and keeping the group numbers 1-18, with which many are familiar, for other groups).

For this first post, I’ll start with some sets of elements which are familiar to most who have studied the subject, plus some others which are much less well-known.

  • Light blue — the alkali metals.
  • Black background with red symbol and atomic number — hydrogen, which is definitely not an alkali metal, despite it sharing group 1 with them.
  • Dark blue — the alkaline-earth metals.
  • Dark yellow — the lanthanides.
  • Orange — these two elements are included with the lanthanides in some sources, and with the transition metals in others.
  • Bright pink — the actinides.
  • Light pink — these two elements are included with the actinides is some sources, and with the transition metals in others.
  • Red — the transition metals, also known as the transition elements, and d-block elements.
  • Light purple — group 13 is often called the “boron group,” but it also goes by other names, such as the “icosagens” and the “triels.”
  • Dark purple — group 14 is often called the “carbon group,” but it also goes by other names, such as “tetragens” and “crystallogens.” In semiconductor physics, these elements are referred to as group IV elements. 
  • Dark green — group 15 elements are referred to as the pnictogens, or nitrogen-group elements.
  • Bright yellow — bright yellow is used here for the chalcogens, also known as the group 16 elements, or oxygen-group elements.
  • Light green — the halogens.
  • Gray — the noble gases.

One Dozen Precious Metal Cubes: A Problem Involving Geometry, Chemistry, and Finance (Solution Provided, with Pictures)

Posted on by
1

The troy ounce is a unit of mass, not weight, and is used exclusively for four precious metals. At this time, the prices per troy ounce, according to this source for current precious metal prices, for these four elements, are:

  • Gold, $1,094
  • Palladium, $600
  • Platinum, $965
  • Silver, $14.82

(As a side note, it is rare for platinum to have a lower price per troy ounce than gold, as is now the case. I would explain the reasons this is happening, except for one problem: I don’t understand the reasons, myself, well enough to do so. Yet.)

A troy ounce equals 31.1034768 grams, but, for most purposes, 31.103 g, or even 31.1 g, works just fine.

Also, as you can see here, these “troy elements” are all in one part of the periodic table. This is related to the numerous similarities in these elements’ physical and chemical properties, which is itself related, of course, to the suitability of these four elements for such things as jewelry, coinage, and bullion.

1371846764_periodic_table_of_elements

To determine the volume of a given mass of one of these metals, it is also necessary to know their densities, so I looked them up, using Google (they are not listed on the periodic table above):

  • Gold, 19.3 g/cm³
  • Palladium, 11.9 g/cm³
  • Platinum, 21.46 g/cm³
  • Silver, 10.49 g/cm³

In chemistry, of course, one must often deal with elements (as well as other chemicals) in terms of the numbers of units (such as atoms or molecules), except for one problem: this is absurdly impractical, due to the outrageously small size of atoms. Despite this, though, it is necessary to count such things as atoms in order to do much chemistry at all, so chemists have devised a “workaround” for this problem: when counting units of pure chemicals, they don’t count such things as atoms or molecules directly, but count them a mole at a time. A mole is defined as a number of things equal to the number of atoms in exactly 12 grams of pure carbon-12. To three significant figures, this number is 6.02 x 10²³. To deal with moles, since atoms have differing masses, we need to know the molar mass (mass of one mole) of whatever we are dealing with to convert, both directions, between moles and grams. Here are the molar masses of the four troy-measured elements, as seen on the periodic table above, below each element’s symbol.

  • Gold, 196.97 g
  • Palladium, 106.42 g
  • Platinum, 195.08 g
  • Silver, 107.87 g

I’ve given these numbers  as the information needed to solve the following problem: rank one dozen precious metal cubes (descriptions follow) by ascending order of volume. There are three cubes each of gold, palladium, platinum, and silver. Four of the twelve (one of each element) have a mass of one troy ounce each. Another four each have a value, at the time of this writing, of $1,000. The last set of four each contain one mole of the element which composes the cube, and, again, there is one of each of these same four elements in the set.

If you would like to do this problem for yourself, the time to stop reading is now. Otherwise (or to check your answers against mine), just scroll down.

.

.

.

.

.

.

.

.

.

.

.

In the solutions which follow, a rearrangement of the formula for density (d=m/v) is used; solved for v, this equation becomes v = m/d. In order, then, by both volume and edge length, from smallest to largest, here are the twelve cubes:

Smallest cube: one troy ounce of platinum

One tr oz, or 31.103 g, of platinum would have a volume of v = m/d = 31.103 g / (21.46 g/cm³) = 1.449 cm³. A cube with this volume would have an edge length equal to the its volume’s cube root, or 1.132 cm. (This explanation for the calculation of the edge length, given the cube’s volume, is omitted in the items below, since the mathematical procedure is the same each time.)

Second-smallest cube: $1000 worth of gold

Gold worth $1000, at the time of this posting, would have a troy mass, and then a mass in grams, of $1000.00/($1,094.00/tr oz) = (0.914077 tr oz)(31.103 g/tr oz) = 28.431 g. This mass of gold would have a volume of v = m/d = 28.431 g / (19.3 g/cm³) = 1.47 cm³. A cube with this volume would have an edge length of  1.14 cm.

Third-smallest cube: $1000 worth of platinum

Platinum worth $1000, at the time of this posting, would have a troy mass, and then a mass in grams, of $1000.00/($965.00/tr oz) = (1.0363 tr oz)(31.103 g/tr oz) = 32.231 g. This mass of platinum would have a volume of v = m/d = 32.231 g / (21.46 g/cm³) = 1.502 cm³. A cube with this volume would have an edge length of  1.145 cm.

Fourth-smallest cube: one troy ounce of gold

One tr oz, or 31.1 g, of gold would have a volume of v = m/d = 31.1 g / (19.3 g/cm³) = 1.61 cm³. A cube with this volume would have an edge length of 1.17 cm.

Fifth-smallest cube: one troy ounce of palladium

One tr oz, or 31.1 g, of palladium would have a volume of v = m/d = 31.1 g / (11.9 g/cm³) = 2.61 cm³. A cube with this volume would have an edge length of 1.38 cm.

Sixth-smallest cube: one troy ounce of silver 

One tr oz, or 31.103 g, of silver would have a volume of v = m/d = 31.103 g / (10.49 g/cm³) = 2.965 cm³. A cube with this volume would have an edge length of 1.437 cm.

Sixth-largest cube: $1000 worth of palladium

Palladium worth $1000, at the time of this posting, would have a troy mass, and then a mass in grams, of $1000.00/($600.00/tr oz) = (1.6667 tr oz)(31.103 g/tr oz) = 51.838 g. This mass of palladium would have a  volume of v = m/d = 51.838 g / (11.9 g/cm³) = 4.36 cm³. A cube with this volume would have an edge length of  1.63 cm.

Fifth-largest cube: one mole of palladium

A mole of palladium, or 106.42 g of it, would have a volume of v = m/d = 106.42 g / (11.9 g/cm³) = 8.94 cm³. A cube with this volume would have an edge length of 2.07 cm.

Fourth-largest cube: one mole of platinum

A mole of platinum, or 195.08 g of it, would have a volume of v = m/d = 195.08 g / (21.46 g/cm³) = 9.090 cm³. A cube with this volume would have an edge length of 2.087 cm.

Third-largest cube: one mole of gold

A mole of gold,  or 196.97 g of it, would have a volume of v = m/d = 196.97 g / (19.3 g/cm³) = 10.2 cm³. A cube with this volume would have an edge length of  2.17 cm.

Second-largest cube: one mole of silver

A mole of silver, or 107.87 g of it, would have a volume of v = m/d = 107.87 g / (10.49 g/cm³) = 10.28 cm³. A cube with this volume would have an edge length of 2.175 cm.

Largest cube: $1000 worth of silver

Silver worth $1000, at the time of this posting, would have a troy mass, and then a mass in grams, of $1000.00/($14.82/tr oz) = (67.48 tr oz)(31.103 g/tr oz) = 2099 g. This mass of gold would have a volume of v = m/d = 2099 g / (10.49 g/cm³) = 200.1 cm³. A cube with this volume would have an edge length of  5.849 cm.

Finally, here are pictures of all 12 cubes, with 1 cm³ reference cubes for comparison, all shown to scale, relative to one another.

dozen cubes

A third of these cubes change size from day-to-day, and sometimes even moment-to-moment during the trading day, if their value is held constant at $1000 — which reveals, of course, which four cubes they are. The other eight cubes, by contrast, do not change size — no precious metal prices were used in the calculation of those cubes’ volumes and edge lengths, precisely because the size of those cubes is independent of such prices, due to the way those cubes were defined in the wording of the original problem.

Testing a Climate Change Claim: Does Burning a Gallon of Gasoline Really Put Twenty Pounds of Carbon Dioxide Into the Atmosphere?

Posted on by
3
OLYMPUS DIGITAL CAMERA

OLYMPUS DIGITAL CAMERA

When I read, recently, that “every gallon of gas you save not only helps your budget, it also keeps 20 pounds of carbon dioxide out of the atmosphere,” (source) I reacted with skepticism. Twenty pounds? That seemed a bit high to me.

Just because it seems high, to me, though, doesn’t mean it’s wrong, any more than “about.com” putting it on the Internet makes the statement correct. No problem, I thought: I’ll just do the math, and check this for myself.

So, in this problem, we start with a gallon of gasoline. In units I can more easily work with, that’s 3.785 liters (yes, they rounded it incorrectly on the gas can shown above). Google tells me that the density of gasoline varies from 0.71 to 0.77 kg/L (reasonable, since it floats on water, but is still heavy to lug around), so I’ll use the average of that range, 0.74 kg/L, to find the mass of a gallon of gasoline: (3.785 L)(0.74 kg/L) = 2.8 kg.

Next, I need to find out how much of that 2.8 kg of gasoline is made of carbon. That would be an easy chemistry problem if gasoline were a pure chemical, but it isn’t, so I’ll estimate. First, I’ll ignore the elements which only make up a minor part of gasoline’s mass, leaving only hydrogen and carbon to worry about. Next, I consider these things:

  1. Alkanes larger than methane (major gasoline components), whether branched-chain or not, have slightly more than two hydrogen atoms for every carbon atom. This ratio doesn’t exceed three, though, and is equal to three only for ethane, which has too high a boiling point to remain in liquid form at typical temperatures and pressures, anyway.
  2. Cycloalkanes, another major component of gasoline, have exactly two hydrogen atoms for every carbon atom.
  3. Aromatic hydrocarbons can have fewer than two hydrogen atoms for every carbon atom, and these chemicals are also a major component of gasoline. For the simplest aromatic hydrocarbon, benzene, the H:C ratio drops to its lowest value: 1:1.

For the reasons above, I’m choosing two to one as a reasonable estimate for the number of atoms of hydrogen for every atom of carbon in gasoline. Carbon atoms, however, have twelve times the mass of hydrogen atoms. Gasoline is therefore ~12/14ths carbon, which reduces to ~6/7. I can now estimate the mass of carbon in a gallon of gasoline: (6/7)(2.8 kg) = 2.4 kg.

So how much carbon dioxide does that make? Well, first, does a car actually burn gasoline completely, so that every carbon atom in gasoline goes out the tailpipe as part of a carbon dioxide molecule? The answer to this question is simple: no.

However, this “no” doesn’t really matter, and here’s why. In addition to carbon dioxide, automobile exhaust also contains carbon present as carbon monoxide, unburned carbon, and unburned or partially-burned hydrocarbons. To have an environmental impact, though, it isn’t necessary for a given carbon dioxide molecule to come flying straight out of the tailpipe of one’s car. In our oxygen-rich atmosphere, those carbon atoms in car exhaust which are not yet fully combusted (to each be part of a carbon dioxide molecule) are quite likely to end up reacting with oxygen later on — certainly within the next year, in most cases — and the endpoint of carbon reacting with oxygen, once combustion is complete, is always carbon dioxide. What’s carbon monoxide, then? Well, one way to look at it is this: molecules of carbon monoxide are simply half-burned carbon atoms. When the other half of the burning (combustion) happens, later, carbon dioxide is the product. So, for purposes of this estimate, I am assuming that all 2.4 kg of carbon in a gallon of gasoline ends up as carbon dioxide — either directly produced by the car, or produced in other combustion reactions, later, outside the car.

The molar mass of elemental carbon is 12 grams. For oxygen, the corresponding figure is 16 g, but that becomes 32 g of oxygen in carbon dioxide, with its two oxygen atoms per molecule. Add 12 g and 32 g, and you have carbon dioxide’s molar mass, 44 g. Therefore, 12 g of carbon is all it takes to make 44 g of carbon dioxide; the rest of the mass comes from oxygen in the air. By use of ratios, then, I can now find the mass of carbon dioxide formed from burning a gallon of gasoline, with its 2.4 kg of carbon.

Here is the ratio needed: 2.4 kg of carbon / unknown mass of carbon dioxide produced = 12 g of carbon / 44 g of carbon dioxide. By cross-multiplication, and using “x” for the unknown, this equation becomes (12 g)(x) = (44 g)(2.4 kg), which simplifies to 12x = 105.6 kg, so, by division, x = 8.8 kg of carbon dioxide produced from one gallon of gasoline.

Each kilogram (of anything), near sea level on this planet, weighs about 2.2 pounds. This 8.8 kg of carbon dioxide, then, translated into “American,” becomes (8.8 kg)(2.2 pounds/kg) = 19.36 pounds. Given the amount of estimation I had to do to obtain this answer, this is close enough to twenty pounds for me to conclude that the statement I was examining has survived my testing. In other words, yes: burning a gallon of gasoline goes, indeed, put about twenty pounds of carbon dioxide into the atmosphere.

However, I’m not quite finished. Carbon dioxide is an invisible gas, making it quite difficult to picture what twenty pounds of it “looks” like. To really understand how much carbon dioxide this is, it would be helpful to know its volume. So I have a new problem: 8.8 kg of pure carbon dioxide is trapped, in a balloon, at standard temperature and pressure. What is the balloon’s volume?

To solve this problem, one needs the density of carbon dioxide under these conditions, which chemists refer to as “STP” (standard temperature and pressure). According to Google, the density of carbon dioxide at STP is about 1.98 kg per cubic meter. Since 8.8 kg / (1.98 kg/m³) = 4.44 m³, this means that a gallon of gasoline can produce enough carbon dioxide, held at STP, to inflate this balloon to a volume of 4.44 cubic meters. For the benefit of those who aren’t used to thinking in metric units, that volume equals the volume of a perfect cube with an edge length of ~5.4 feet. You could fit a bunch of people into a cube that large, especially if they were all on friendly terms.

Now, please consider this: all of that was from one gallon of gasoline. How many gallons of gas do you typically buy, when you fill up your car’s gas tank? Well, multiply by that number. How many times do you fill up your car, on average, in a year? Multiply again. Next, estimate the total number of years you will drive during your lifetime — and multiply again. You now have your own personal, lifetime carbon dioxide impact-estimate from just one activity: driving.

This may sound like a change of topic, but it isn’t: what’s the hottest planet in the solar system? Even though Mercury is much closer to the Sun, the answer is Venus, and there is exactly one reason for that: Venus has a thick atmosphere which is chock-full of carbon dioxide. In other words, yes, the planet nearest the Earth, and the brightest object in the sky (behind the Sun and the Moon), Venus, is one of the best warnings about global warming known to exist, and we’ve known this for many decades. One wonders if any theologian has ever speculated that the creator of the universe designed Venus this way, and then put it right there “next door,” on purpose, specifically as a warning, to us, about the consequences of burning too much carbon.

The science, and the math which underlies it, are both rock-solid: climate change is real. Lots of politicians deny this, but that’s only because of the combined impact of two things: their own stupidity, plus lots of campaign contributions from oil companies and their political allies. Greed and stupidity are a dangerous combination, especially when further combined with a third ingredient: political power. Voting against such politicians helps, but it isn’t enough. One additional thing I will do, immediately, is start looking for ways to do the obvious, in my own life: reduce my own consumption of gasoline. Since I’m putting this on the Internet, perhaps there will be others persuaded to do the same.

The truth may hurt, but it’s still the truth: the United States is a nation of petroleum junkies, and we aren’t just harming ourselves with this addiction, either. It’s time, as a people, for us to invent, and enter, fossil-fuel rehab.

[Image credit: “GasCan” by MJCdetroit – Own work. Licensed under CC BY 3.0 via Wikimedia Commons – https://commons.wikimedia.org/wiki/File:GasCan.jpg#/media/File:GasCan.jpg]

Riddle: How did the chemist accidentally kill his dog?

Posted on by
Reply

Answer: He fed him a whole can of aluminum phosphate.

Aluminum phosphate

Disclaimers: (1) no actual dogs were harmed in the making of this awful pun, and (2) yes, I actually did the math regarding the toxicity of aluminum phosphate. Don’t feed it to your dog!