The atom is a tiny particle and cannot be weighed directly, so its mass is calculated concerning another atom taken as a reference. So in this article, we will give a complete guide about the molar mass of oxygen. In the past, the smallest atom, that of hydrogen, was taken as a reference; we then moved on to the sixteenth part of the oxygen atom, while from 1961. Today the unit of measurement is the dalton and denoted by u, and not the gram.
In chemistry, the terms weight and mass are often used indistinctly, with the physical meaning of mass; although this is not strictly correct, we have preferred to use weight instead of mass here. So the atomic weight (PA) – or relative atomic mass – of an element is the mass of its atoms. Expressed in daltons, of that element’s atoms. The mass of the element and the atomic mass unit, therefore relative. For example, the PA of hydrogen is 1.008 u, which derives from the atomic weights.
The molecular weight of H 2 O is:
- atomic weights
- therefore
- molecular weight
The molar mass of oxygen: Avogadro’s number
In the laboratory, it is not possible to work with such small measures:
- 1 u = 1.66 10 -24 g
- 1 g = 6.023 10 23 u
for this, a particular number has been introduced, the Avogadro number ( N ). Which allows you to pass from dalton to grams, thus connecting the microscopic world with the macroscopic one.
If, for example, a carbon atom weighs 12 u, 6,023 · 10 23 carbon atoms weigh 12 g.
Amount
The mole is a quantity of substance (atoms or molecules), equal to the atomic (or molecular) weight, made up of an Avogadro number of atoms or molecules.
m = mass of the sample expressed in grams
M = molar mass: PA or PM expressed in g / mol.
The mole is indicated by n, and the unit of measurement is mol.
One mole of carbon corresponds to 12 g and contains 6,023 · 10 23 atoms.
One mole of hydrogen corresponds to 1 g and contains 6,023 · 10 23 atoms.
the molar mass of oxygen
The molar mass ( M ), i.e., the mass of one mole, is equal to the relative atomic (or molecular) mass expressed in grams.
So, the unit of measurement of the molar mass is g / mol.
The atomic mass and molar mass of oxygen have the same numerical value but different measurement units; for example, a carbon atom weighs 12 u, so the molar mass is 12 g / mol.
Determination of the percentage composition
According to the available data, one proceeds in the following two ways to determine the percentage composition of a compound.
1.We have the mass of the individual elements available.
Suppose that the decomposition of a compound gave the following result:
- 14 g silicon Si
- 0 g sodium Na
- 86 g oxygen O
The total mass of the compound, being respected Lavoisier’s law is:
- 14 g + 30.0 g + 20.86 g = 60 g
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The percentages are, therefore:
2.We know the formula of the compound.
We want to determine the percentage of CaSO 4 elements.
In the periodic table, we find the PA of the elements:
- PA Ca = 40 g / mol
- And PA S = 32 g / mol
- PA O = 16 g / mol
PM CaSO4= 40 + 32 + 16 • 4 = 136 g / mol
We can now determine the percentages (the units of measurement are simplified).
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Determination of the minimum formula
Here, too, we have two possibilities to determine the formula based on the data. We find the minimum formula because to get to the molecular formula. You also need to know the molecular weight calculated experimentally.
1.We have the mass of the individual elements available.
Suppose that the decomposition of a compound gave the following result:
- 3 g mercury Hg
- 7 g oxygen O
The total mass of the compound, being respected Lavoisier’s law is:
- PM = 46.3 g + 3.7 g = 50 g
In the periodic table, we find the PA of the elements:
- PA Hg = 200.5 g / mol
- PA O = 16 g / mol
We now find the piers.
Dividing everything by 0.23, we get a ratio of 1: 1, so the minimum formula is HgO.
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2.We know the percentage of the elements.
- C = 40.02%
- H = 6.66%
In the periodic table, we find the PA of the elements:
- PA C = 12 g / mol
- PA H = 1 g / mol
The percentage tells me how many grams of the element are present in 100 g of the compound. I can now find the docks. Dividing by 3.33, we obtain a ratio of 1: 2: 1, i.e., However, the minimum formula is CH 2 O.
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Determination of the molecular formula
If we know the molecular weight experimentally, we can arrive at the molecular formula.
Starting from the last example, it was experimentally determined that the MW = 180 g / mol.
We calculate the molecular weight starting from the minimum formula CH 2 O:
Molecular weight
By dividing the experimental PM with the theoretical one, we obtain the multiplication factor:
The molecular formula is C 6 H 12 O 6.
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The molar mass of oxygen and the molar volume
volumeThe relative atomic mass of the carbon-12 atom is equal to 12 u. If we consider a quantity in grams (g) of carbon equal to its relative atomic mass, 12 g. It contains an enormous number of atoms: 6,022 · 10 23. Similarly, we consider a quantity expressed in grams of any atom or molecule equal to the relative atomic mass or the relative molecular mass.
It will contain the same number of units (atoms or molecules) equal to Avogadro’s number (i.e., 10 23 ). That contains several particles (atoms, molecules, or ions) equal to that contained in 12 g of carbon – 12. So the mole is a fundamental quantity of the SI that defines the unit of amount of substance. So, 1 mole of atoms contains 6,022 10 23 atoms1 moles of molecules containing 6,022 · 10 23 molecules1 mole of ions containing 6.022 · 10 23 ions1 mole of atoms element = quantity in grams. That element corresponding to its relative atomic mass (mar). Example:
sodium sea (Na) = 22.99 u; the mass of 1 mol of Na = 22.99 mass of 1 mole of molecules of an element or compound = quantity in grams. That element or compound corresponding to its relative molecular mass (MMR). Example:
MMR of water (H 2 O) = 18.016 u. The molar mass of oxygen the various substances (atoms, molecules, ions) are in the same numerical ratio to each other as the respective relative atomic or molecular masses, with the advantage of being evaluated with the balance. From Avogadro’s law and the definition of the mole. Under the same conditions of temperature and pressure, it occupies the same volume.
Molecule
According to Dalton’s atomic model, an element is a pure substance consisting of only one type of atom. However, let us consider water, which is a compound. So, the water molecule is a grouping of two hydrogens and one oxygen atom.
The molecule is the smallest particle of a compound that possesses the chemical properties of that compound. Moreover, each water molecule has all the chemical properties of water and the physical ones. It boils at 100 ° C and freezes at 0 ° C. So the molecule is the stable form of matter.
O 2: The oxygen formula says that the oxygen molecule is made up of two O atoms.
H 2 O: The water formula says that two atoms of hydrogen and oxygen from the water formula. These are crude formulas since they merely indicate the quality and number of atoms that form them.
On the other hand, structural formulas represent the structure, that is, the atoms’ position in the molecule.
- the structural formula of O 2
- the structural formula of H 2 O
Any matter is made up of molecules.
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The molar mass of oxygen: Molecular weight
The molecular mass is given by the sum of the masses of the atoms that compose it. However, we will use the expression molecular weight PM. Even if it is improper because the quantities under consideration are masses (not forces), however, in many versions of the periodic table and the element’s initials and the atomic number. So, the atomic weight (atomic mass) value of the same, expressed in atomic mass units [u], appears.
In this way, it becomes effortless to calculate a chemical compound’s molecular weight, considering that the molecular weight is expressed in [g / mol].
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Example
Calculates the molecular weight of sulfuric acid H 2 SO 4.
- 2H = 2 × 1.008 = 2.016
- 1S = 1 × 32.064 = 32.064
- 4O = 4 × 15.9994 = 63.998
MW = 2.016 + 32.064 + 63.998 = 98.078 g / mol
Calculates the molecular weight of potassium chloroiridiate K 2 IrCl 6.
- 2K = 2 × 39.1 = 78.2
- 1Ir = 1 × 192.2 = 192.2
- 6Cl = 6 × 35.45 = 212.7
- MW = 78.2 + 192.2 + 212.7 = 483.1 g / mol
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The molar mass of oxygen: Amount
The mass of an atom or a molecule is always tiny while during chemical reactions. A mole is the amount of substance that contains 6.02 · 10 23 elementary particles. That is atoms, molecules, or ions.
Formally, the mole is for the International System (SI), the unit of measurement of matter’s quantity. Even if in this system, the definition of the mole is the following.
The relationship that links the mass in grams of a substance to the number of moles is as follows:
- n = number of moles
- m = mass in grams [g]
- PM = molecular weight [g / mol]
The number 6.02 10 23 is also called Avogadro’s constant and has the symbol N.
- N = 6.02 10 23 particles / mol
from the dimensional point of view, the unit of measurement of the constant is mol -1.
Moreover, the Avogadro’s importance lies in allowing the passage from the microscopic level (atom or molecule) to the macroscopic level (molecular weight). If necessary, we can refer to the following formula:
- p = number of particles (atoms, molecules, ions, etc …)
- n = number of moles
- N = Avogadro’s constant
How many grams of H 2 S are there in 0.4 moles of H 2 S?
- 016 + 32.064 = 34.08 g / mol
- 4mol × 34.08 g / mol = 13.632g
How many grams of H and S are contained in 0.4 moles of H 2 S?
There are
- 2 x 0.4 = 0.8 u of H
- 1 x 0.4 = 0.4 u of S
- H = 0.8 × 1.008 = 0.806 g
- S = 0.4 × 32.064 = 12.82 g
The molar mass of oxygen: Chemical reactions
The equation consists of two members, the left one, representative of the reactants. And the right one, representative of the reaction products. So, Therefore, it is possible to start from elements to obtain a compound (synthesis). And start from a compound to obtain elements (analysis).
So the coefficients of the substances participating in a reaction indicate, according to what number of moles the various reactants participate in the reaction. Therefore, notes the moles of a reactant. So it is possible to deduce the moles of any other reagent which intervenes in the same reaction. So go back to the weight of this using the formula.
600g of HCl must be neutralized with Ca (OH) 2, according to the reaction:
Ca (OH) 2 + 2HCl → CaCl 2 + 2H 2 O
Determines how many grams of calcium hydroxide are needed.
PM (HCl) = 36.5
600 g of HCl correspond to
from the reaction coefficients, it can be seen that to neutralize 2 moles of HCl. It takes one mole of Ca (OH) 2. Therefore to neutralize 16.5 moles of HCl, it takes 8.25 moles of Ca (OH) 2.
PM [Ca (OH) 2 ] = 74 whereby 8.25 × 74 = 612 g of Ca (OH) 2 .
