What is the atomic weight of hydrogen

The atomic mass and the mole concept


We saw in the last chapter that every atom has a certain mass. Copper atoms, for example, are roughly twice as heavy as sulfur atoms, and sulfur atoms, in turn, are twice as heavy as oxygen atoms, which in turn are 16 times as heavy as hydrogen atoms.

In the last chapter we also kept talking about ominous "mass units". You won't believe it, but you can actually state the mass of an atom, the atomic mass, in units of mass, one then speaks of "units" and abbreviates the unit of measurement with u.

The atomic mass of a copper atom is then approx. 65 u, while a sulfur atom weighs 32 u. An oxygen atom has a mass of 16 u, and a hydrogen atom has a mass of 1 u. There are no lighter atoms. The heaviest atoms known have a mass greater than 260 u.

But why not measure atomic mass in grams?

For the same reason why the weight of a mosquito is not given in tons. Mosquitoes weigh about 2 to 2.5 milligrams; anyone can understand this. At least when you know that 1 milligram is one thousandth part of a gram. But you could also say that mosquitoes weigh 0.0000000025 tons. There is not much that could be done with this statement.

Atoms, too, are far too light to be reasonably weighed in tons or kilograms. Even with grams or milligrams you don't get much further, even with micrograms (a microgram is a thousandth of a milligram). The smallest atoms (hydrogen) are so light that approx. 602,200,000,000,000,000,000,000 of them weigh 1 gram. This incredibly large number is easier to find than 6,022 * 1023 express. This important number even has its own name: LOSCHMIDT's number, after one of its discoverers, and is abbreviated with L.

LOSCHMIDT's number:
L = 6.022 * 1023

Mass of an H-atom:
m (H) = 1 / (6.022 * 1023) g = 1 / L g.

The number L and the mole

So 1 g of hydrogen contains exactly L atoms. A helium atom is now exactly four times as heavy as an H atom. That means, 4 grams of helium also contains L atoms, or 1 g of helium contains L / 4 He atoms.

An oxygen atom is exactly 16 times as heavy as a hydrogen atom. So L O atoms have a mass of 16 grams, and 1 g of oxygen contains L / 16 atoms.

Why does one really want to know how many grams of L atoms of any element weigh?

Because you can count on it excellently. Take iron and sulfur, for example. An iron atom always combines with a sulfur atom to form iron sulfide. A dozen Fe atoms then always combine with a dozen S atoms to form a dozen FeS units. If you don't know the term "dozen": a dozen is always 12 pieces. A dozen eggs is 12 eggs.

Let's come back to the number L again. The LOSCHMIDT number is a similar unit of measure to a dozen, only a little more: 6.022 * 1023 instead of 12. That means: L Fe atoms always combine with L S atoms to form L FeS particles.

Some people wonder why chemists keep looking at the periodic table when they want to know something. This is because in the periodic table all elements are classified according to their atomic mass. And if you know the atomic mass of an element, you also know how many grams of L atoms of this element weigh.

L copper atoms always have a mass of 63.5 g; the atomic mass of copper is namely M (Cu) = 63.5 g / mol.

L sulfur atoms always have a mass of 32.0 g; this is because sulfur has an atomic mass M (S) of 32.0 g / mol.

Right, if you paid attention, you have already noticed: The amount of L atoms is always referred to as 1 mol, but the unit is written with a small "m", ie "mol". It's similar to the meter. The unit of length is called "meter", but the unit that is written after the numbers consists only of the letter "m". So if we read "1 mole" it means something like "1 meter", and if we read "1 mole" it is like "1 m".

If we know that 1 Cu atom always reacts with 1 S atom, we also know that 1 mol of copper always reacts with 1 mol of sulfur, i.e. always 63.5 g of copper with 32 g of sulfur.


With the help of this knowledge you can now make great calculations.

example 1

For example, how much oxygen do I need if I want to burn 100 g of hydrogen?

First you set up the reaction equation, of course you have to be familiar with it, otherwise it won't do much:

$ O_ {2} + 2 H_ {2} \ to 2 H_ {2} O $

From this equation one can now read that 1 mol of oxygen always reacts with 2 mol of hydrogen to form 2 mol of water.

2 mol of hydrogen have a mass of 4 grams, 1 mol of oxygen has a mass of 32 grams, so hydrogen and oxygen always react in a mass ratio of 4:32 or 1: 8. So if I want to burn 1 gram of hydrogen, I need 8 grams Oxygen. According to this, 800 g of oxygen are required for 100 g of hydrogen.

Example 2

I would like to let the elements copper and sulfur react to form copper sulfide, and the reaction should run optimally; In the end, neither unused copper nor residual sulfur should be left in the test tube. If I now weigh 2 grams of copper into the test tube; How much sulfur do I have to add so that the reaction proceeds optimally and therefore as violently as possible?

First of all, let's look at the reaction equation:

$ 2 Cu + S \ to Cu_ {2} S $

2 mol of copper always react with 1 mol of sulfur to form 1 mol of copper sulfide. The atomic mass of copper is 63.5 g / mol, that of sulfur 32.0 g / mol. So 127 g of copper always react with 32 g of sulfur. Now we have to do some math, but that shouldn't be so difficult for us. We just divide everything by 127, and then we find that 1 g of copper always reacts with 32/127 = 0.252 g of sulfur. We had 2 g of copper in the test tube, so we have to weigh 0.504 g of sulfur.

At this point we stop with the bills; Students always find them a bit boring, and these stoichiometric calculations are slowly disappearing from the guidelines or core curricula, although they are really important if you want to understand chemistry properly.

Example 3

A really difficult calculation for very committed schoolchildren: How much carbon is actually left after I burn a bag of household sugar completely?

Table sugar or sucrose has the molecular formula C.12H22O11. The reaction with oxygen to form water and carbon dioxide is similarly complex. Here is the reaction equation:

$ C_ {12} H_ {22} O_ {11} + 12 O_ {2} \ to 12 CO_ {2} + 11 H_ {2} O $

For 1 mole of sucrose, 12 moles of oxygen are required to completely burn the sugar. The molar mass of a compound is the sum of all atomic masses. So we have to add 12 times the atomic mass of carbon, 22 times the atomic mass of hydrogen and 11 times the atomic mass of oxygen to get the molar mass of sucrose.

12 * M (C) + 22 * ​​M (H) + 11 * M (O) =
12 * 12g + 22 * ​​1g + 11 * 16g =
342 g

These 342 grams of sugar react with 384 g of oxygen (12 * 16 * 2), so 1 g of sugar reacts with 384/342 g of oxygen. A bag of sugar usually contains 1 kg or 1000 g of sugar, so we need 384/342 * 1000 = 1122.8 g of oxygen. That's a lot.

If you then consider that the molar volume of gases at room temperature is 22.4 liters, i.e. that 1 mol of a gas has a volume of 22.4 liters, and that 32 g of oxygen are 1 mol, then you get a huge one Volume of oxygen required for combustion.

32 g oxygen = 1 mol = 22.4 liters

1122.8 g oxygen = 35.09 mol = 786 liters

So it takes almost a cubic meter of oxygen to burn a bag of sugar. Normal air contains 21% oxygen, so we need five times as much air, namely 3742.7 liters or around 3.7 cubic meters.