Moles to Grams

Here we explain how to convert moles to grams. The conversion from grams to moles can be found here.   This area of ​​chemistry is called stoichiometry .

The mass in grams of a mole of a substance (that is, the mass in grams per mole) is called the molar mass of that substance. 

The molar mass (in g/mol) of a substance is numerically always equal to the formula weight of the substance (in ame = atom mass unit or also called u = unit). The atomic mass can be found at the top right of every periodic table under atomic weight .

The substance NaCl has e.g. B. a formula weight of 58.5 ame and a molar mass of 58.5 g/mol. The table below contains further examples of calculations using the mole unit.

The entries for N and N 2  in the table make it clear that it is important to precisely name the chemical form of a substance when specifying an amount of substance in moles. Suppose it is stated that 1 mol of nitrogen is produced in a certain reaction. You might conclude that this means 1 mol of nitrogen atoms (14.0 g). However, unless otherwise stated, this probably means 1 mol of nitrogen molecules N2 (  28.0 g), because N2 is  the common chemical form of the element. To avoid such misunderstandings, the chemical form of the substance should be explicitly stated. By specifying the chemical formula N 2 , such misunderstandings are avoided.

Substance name
formula
Formula weight
in name
Molar mass
in (g/mol)
Number & type of particles present in a mole
Atomic nitrogen  N 14.0 14.0
6.022 * 10 23  N atoms
Molecular nitrogen  N2 _ 28.0 28.0
6.022 * 10 23  N 2 molecules
2 * (6.022 * 10 23)  N atoms
Silver  Ag 107.9 107.9
6.022 * 10 23  Ag atoms
Silver ions  Ag + 107.9 107.9 (1 6.022 * 10 23 Ag + ions
Barium chloride  BaCl2 _ 208.2 208.2
6.022 * 10 23  BaCl 2 units
6.022 * 10 23  Ba 2+  ions
2 * (6.022*1023) Cl –  ions

 

1) Remember that the mass of the electron can be neglected and therefore ions and atoms have essentially the same mass.

Example:

What is the mass in grams of 1,000 mol of glucose/sugar, C6 H12 O6  ?

 

Solution:

First: analysis. The chemical formula is given and we are supposed to calculate the molar mass from it.

Procedure: The molar mass of a substance can be calculated by adding the atomic weights of the atomic components together.

Glucose has a formula weight of 180.0 ame. One mole of this substance has a mass of 180.0 g, so the substance C 6 H 12 O 6  has a molar mass of 180.0 g/mol.

Verification: The order of magnitude of our answer seems plausible and g/mol is the correct unit for specifying molar mass.

 6 C – Atoms = 6 (12.0 ame) = 72.0 ame
12 H – Atoms = 12 (1.0 ame) = 12.0 ame
 6 O – atoms = 6 (16.0 ame) = 96.0 ame
--------------------------------------
                             180.0 ame or 180.0 u written
tl;dr: forget about the decimal places in atomic weights - as long as it's about fertilizer.
 
In order to be complete, one more detail needs to be explained. The information shown here assumes an ideal atomic weight, which cannot be found in the periodic table - with a few exceptions. If you look at the atomic weight of hydrogen, it should be exactly 1.0 u (or 1.0 ame). However, 1.0080 is given.
 
This is where reality gets in the way. There is almost no element in nature that occurs without isotopes. The atoms in the periodic table are "sorted" according to the number of protons. But the number of neutrons can vary. For magnesium, for example, only about 78.6% contains 12 neutrons in any sample (i.e. no matter where you find magnesium on Earth). 10.1% have 13 neutrons and 11.3% of them have 14 protons. This is how you get an atomic weight of magnesium of 24.327 u. It works out like this:
 
786 24 Mg isotopes with a mass of 24 u provide a mass of 18864 u.
101  25 Mg isotopes with a mass of 25 u provide a mass of 2525 u.
113  26 Mg isotopes with a mass of 25 u provide a mass of 2938 u.
The total gives the weight of 1000 Mg atoms: 24,327 u. So statistically one Mg atom weighs 24.327 u.
 
If you put together fertilizer according to your own formulas, this inaccuracy should only come into play for very (very) large quantities. This last paragraph only serves to remind you of the chemistry lessons you may have already forgotten at school and to clear up the confusion about the crooked numbers. 
ID: 544
Category: Technology
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