Chemistry – Ben Craven

It depends what you mean by see. Single air molecules scatter light (that’s why the sky glows) so with a dark background and an absurdly intense light source you would presumably be able to visually detect a single atom suspended in a vacuum.

But that doesn’t really feel like seeing to me. The question I’m going to answer is: what is the smallest number of atoms that I can quickly assemble using the stuff in my flat, that I can see with my unaided eye by ordinary reflection in typical room lighting?

I’m sure I could look this up somewhere but there’s no fun in that.

My assemblage of atoms was a tiny pencil dot made on white printer paper. There it is, on the right. The dot was definitely visible but so small that I needed to draw marks nearby so that I didn’t lose it.

I estimate that the number of atoms in that minute mark was about 10¹³, with an uncertainty of at least a factor of 10 in both directions.

In other words, 10 million million, very roughly.

That’s a lot. We talk about atoms very casually, drawing diagrams of chemical structures and so on, and it’s easy to forget how exceedingly tiny they are. It’s useful to do experiments like this one now and again to remind ourselves that atoms really are small beyond our comprehension.

The experiment

I made the dot by rubbing the end of a propelling pencil to a point and then touching the point lightly against a sheet of white paper.

To estimate the thickness of the layer of pencil lead, I held the pencil perpendicular to some paper and scribbled until the lead had a flat end to it. I then adjusted it so that, as far as I could tell, 1 mm of lead protruded from the pencil. Then, using normal pencil pressure, I drew lines 10cm long until the exposed millimetre of lead had all worn away. I took care to hold the pencil perpendicular to the paper so that the lines I drew were the full width of the lead. I could draw 520 such lines with the millimetre of lead, a total of 52 metres of line.

Calculation 1: volume and mass of the dot

I used the thickness (don’t confuse this with the width) of the lines described in the previous paragraph as a proxy for the thickness of the dot (and in doing so introduced probably the biggest uncertainty in the whole procedure). Propelling pencils leads appear to come in sizes of 0.5 mm, 0.7 mm and 0.9 mm and larger. Holding mine against a ruler showed that it was clearly a 0.7 mm lead. The volume of the initial protruding cylinder of lead was therefore

π × (0.35 mm)² = 0.385 mm³ or 3.85 × 10^-10 m³

If the line lines I drew were uniformly 0.7 mm wide (and that’s quite a big if – tilting the pencil will make them narrower) then I can equate the volume of the lines and the volume of the lead cylinder thus:

3.85 × 10^-10 m³ = (52 m) × (0.7 × 10^-3 m) × t

where t is the average thickness of the layer of lead on the paper in metres. This gives us

t = 1.06 × 10^-8 m

It certainly doesn’t deserve 3 significant figures but I’m going to leave more reasonable rounding to the end.

To measure the area of my dot, I took a photograph of it next to the finest scale on my ruler, which is 100ths of an inch (see earlier). Things aren’t made any easier by the non-roundness of the dot, but if I were to say that the dot was 1/300 of an inch in each direction, I don’t think I’d be too far wide of the mark. That makes its area

(1/300 in)² × (25.4 × 10^-3m in^-1)² = 7.17 × 10^-9 m²

(the 25.4 × 10^-3 being the conversion from inches into metres). Using our estimate for the thickness of the pencil layer above, this makes the volume of the dot 7.60 × 10^-17 m³.

Next, we need to know the density of the pencil lead. If it was a clay brick, its density would be 2400 kg m^-3, and if it was pure graphite its density would be in the range 2090-2230 kg m^-3, so it’s a reasonable guess that the density of the graphite/clay mix is about 2300 kg m^-3.

So using the volume calculated earlier, the mass of my pencil dot is about

(2300 kg m^-3) × (7.60 × 10^-17 m³)= 1.75 × 10^-13 kg

Calculation 2: how many atoms in a kilogram of pencil lead?

From the Cumberland Pencil Company, cited here, I infer that an HB pencil lead is roughly 50% clay and 50% graphite. They don’t say whether that’s before or after firing (the clay will lose water on firing). I’m going to assume that it’s after firing, but given all the other uncertainties in this calculation, I don’t think it’ll matter much if I’m wrong.

Clay is variable in composition, but a typical constituent of fired clay appears to be various minerals or combinations of minerals of overall composition Al₂Si₂O₇. The relative “molecular” mass of such a compound/mixture will be 220. The relative atomic mass of carbon (in the graphite) is 12, so a 50:50 (by mass) mix of clay and graphite will need about 18 atoms of carbon for every unit of Al₂Si₂O₇, giving a total of 29 atoms per unit of this mixture/compound, and a relative “molecular” mass of 440.

440 g of pencil lead is therefore one mole of pencil lead, and with 29 atoms per elementary entity of this mole, it will contain about 29 × 6.02 × 10²³ = 1.75 × 10²⁵atoms; this is about 3.97 × 10²⁵ atoms per kilogram. (6.02 × 10²³ is Avogadro’s number: the number of elementary entities in a mole of a substance)

Calculation 3: number of atoms in the dot

From calculation 1, we know that the dot weighs 1.75 × 10^-13 kg, and therefore with 3.97 × 10²⁵ atoms per kilogram, the number of atoms in the dot is

(1.75 × 10^-13 kg) × (3.97 × 10²⁵ kg^-1) = 6.95 × 10¹²

Rounded more reasonably, this is 10¹³ atoms in the pencil dot.

With the uncertainties in the size of the dot and the composition of the lead, I wouldn’t want to quote the answer any more precisely than this.

Checking

We’ve done quite a few steps here. Can we check that this answer looks about right?

Suppose the dot were pure graphite. Its mass would be (2150 kg m^-3) × (7.17 × 10^-17 m³) kg, which is 1.54 × 10^-11 moles and hence about 9.2 × 10¹² atoms in the dot. As carbon atoms are smaller than aluminium or silicon atoms, it’s not surprising that this number is a little bit bigger than the unrounded number of atoms calculated in the dot.

Now suppose that it was pure aluminium, with density 2700 kg m^-3. Its mass would be (2700 kg m^-3)× (7.2 × 10^-17m^-3) = 1.94 × 10^-13 kg which is 7.46 × 10^-12 moles and hence 4.5 × 10¹² atoms in the dot. As aluminium atoms are larger than carbon atoms, it’s not surprising that this number is a little bit smaller than the unrounded number of atoms calculated in the dot.

So our clay mineral calculations look at least plausible. The bit I’m really worried about is the thickness of the dot. Making a mark with a pointed lead and drawing a line with a flat end of the lead are likely to involve different pressures and hence different mark thicknesses. My feeling is that I’m most likely to have underestimated the thickness of the dot, and hence the number of atoms in it.

Recently I spent a few days walking and wild camping among the mountains of the English Lake District. I was moving on every day and carrying everything I needed for the trip on my back. My rucksack got lighter and lighter as I worked my way through my food supply, and as there was no evidence that my body was getting any heavier, I started wondering where all that mass goes.

The food was largely very dry, so I’m not going to concern myself with water (though some water will be generated as the food is metabolised). Of the dry mass, I will excrete some as faeces, and some in my urine. I’ll secrete some skin oils and will also shed some skin. Then there’s hair, toenails and fingernails, and we mustn’t forget the odd bit of snot and earwax. Quite a trail of debris, really.

But what I started to wonder as I walked was: how much of this dry mass do I breathe out? I must lose some mass with each breath, because outbreaths are poorer in oxygen and richer in carbon dioxide than inbreaths, and carbon dioxide is heavier than oxygen. Can we put some numbers to this loss of mass?

It turns out that we can, and the result surprised me: by a big margin, the most important exit route for carbon appears not to be my bottom, but my lungs.

The sums

The volume of gas exchanged on each breath in ordinary breathing is about 0.5 litres. On the way in, the concentration of carbon dioxide is practically nothing (about 0.04%), but on the way out, it’s about 5%. So every 20 breaths results in me breathing out the equivalent of half a litre of pure CO₂, which means 40 breaths to breathe out a litre.

Assuming that for every molecule of carbon dioxide breathed out, there is one molecule of oxygen breathed in, then the extra (non-water) mass breathed out is just the mass of the carbon in the carbon dioxide; we can ignore the oxygen.

How much carbon is there in one litre of CO₂? One mole of carbon is 12 grams, and one mole of a perfect gas occupies roughly 24 litres in everyday conditions. So one litre of carbon dioxide contains one-24th of a mole of carbon, or about 0.5 grams.

So every 40 breaths, I breathe out 0.5 grams of carbon that originally entered my body as food. At rest, I breathe about once every 4 seconds, so it takes 2 × 40 × 4 = 320 seconds to breathe out a gram of carbon. There are 86400 seconds in a day, so over the course of a day, I’ll breathe out 86400/320 = 270 grams of carbon.

I’ve made all sorts of approximations, so let’s say that I breathe out 200-300 grams of carbon daily.

What about the other exit routes for carbon?

This is a much bigger number than I expected. How does it compare to the other output routes for carbon, and do the numbers stack up when we consider how much carbon I ingest?

From various internet sources, fat is something like 75% carbon, carbohydrate is about 40% carbon, and protein is about 50% carbon. On this basis, I’m going to estimate that the dry matter of faeces is in the region of 50% carbon. A typical person produces about 130 grams of faeces a day, of which 75% is water. So the amount of carbon excreted daily in faeces is about 130 × 0.25 × 0.5 = 16 grams.

Wikipedia tells me that we typically produce about 1.4 litres of urine a day, with 6.9 grams per litre of carbon, which gives about 10 grams of carbon excreted by this route per day.

Several internet sources (which may be equally wrong) suggest that we shed about 10 grams of dead skin every day. I don’t know what the moisture content of dead skin is, but it looks like we’re not going to lose more than 5g of carbon by this route every day.

Neglecting the other minor carbon loss processes, this gives a total of about 30 grams of carbon leaving the body daily by routes other than the lungs, compared to 200-300 grams via the lungs.

If these sums are right, I must ingest an equal amount of carbon in my food. This isn’t easy to estimate at all precisely. My diet is largely a mixture of fat, protein, and carbohydrate, but I don’t know what the balance is, and it’ll vary a lot between people. So let’s go to an extreme and suppose that I meet a daily energy requirement of 2500 kilocalories by eating fat only. At 9 kilocalories per gram I’d need to eat 280 grams of grease a day. At 75% carbon, this would be 210 grams of carbon. Doing a similar calculation assuming a diet of undiluted protein or carbohydrate gives a daily carbon intake of 310 grams or 250 grams respectively. So maybe I eat about 250 grams of carbon per day, which tallies reasonably well with the total figure for carbon output that I calculated earlier.

Not the Lake District

The picture at the top of the post wasn’t actually taken in the Lake District. It was taken in Glen Feshie, in the Cairngorms. To keep the weight down, I didn’t take my camera on the trip to the Lakes.

Tag: Chemistry

What’s the smallest number of atoms that I can see?