GCSE basic Knowledge and empirical formula
Specification:
-Atomic number, mass number, isotope, Avogadro constant (NA), relative isotopic mass, relative atomic mass (Ar), relative formula mass and relative molecular mass (Mr)
- The concept of amount of substance (Moles) and its use to perform calculations involving: mass of substance, empirical and molecular formulae.
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Atomic number and mass number:
The atomic number is the smallest number found next to the element. It contains the number of protons and the number of electrons. Number of protons= number of electrons. For example, on the diagram on the left, the 3 will be the number of protons and the number of electrons.
The mass number is the largest number found next to the element. It contains the number of protons and the number of neutrons in the nucleus. So from the diagram, we know Lithium has a mass number of 7. We know there are already 3 protons therefore there are 4 neutrons in Lithium. This will be how you need to work out the number of neutrons from the element.
Mass number - atomic number = no. of neutrons
Isotopes and relative isotopic mass:
Isotopes are the element that has the same number of protons and the same number of electrons but they have different numbers of neutrons. Therefore they have different mass numbers but the same atomic number. They also have the same electronic configuration so they will react the same way in chemical reactions.
We compare the isotopic mass to carbon-12 as this is the international standard relative isotopic mass and therefore we compare the mass of the atom of an isotope with 1/12th of the mass of an atom of carbon-12.
To find the relative isotopic mass we use this equation:
For example:
There is Chlorine-35 with 75.77% and Chlorine-37 with 24.23%. Find the relative isotopic mass.
((35*75.77)+(37*24.23))/100=35.49
Relative atomic mass, relative molecular mass and relative formula mass:
The relative atomic mass (Ar) is the weighted mean mass of an atom of an element compared with 1/12th of the mass of an atom of carbon-12.
The relative molecular mass relative formula mass (Mr) is the sum of the relative atomic masses of the atoms in the numbers shown in the formula. The relative formula mass has the symbol, M r. For example Cl2 is 35.45+35.45=70.9 for the Mr of Cl2.
You can do this with other elements as well, for example NaCl. Na is 23 and Cl is 37.45. 23+35.45= 58.45 which is the Mr of NaCl.
Moles and the calculation for the moles:
The mole is the unit for the amount of substance. 1 mol is the amount of substance that contains the same number of particles as there are atoms in 12.0 g of carbon-12.
We calculate the number of moles with the equation:
Mass/Mr=Moles
By rearranging the equation we can find when we have been given the number of moles and the Mr the mass of the substance and the same if we have been given the moles and the mass of the substance, we can find the Mr.
In 1 mole we can find that there are 6.02x1023 particles. This would be called the Avogadro's constant (NA).
Empirical formula:
It is the way to tell you the simplest ratio of various atoms present in a substance like for example ethane which is C2H6. The ratio of the number of carbon to hydrogen will be 1:3 which gives an empirical formula of CH3.
You can be asked to find the empirical formula of a compound with the percentage of each element in the compound or by knowing the number of moles.
For example:
Calculate the empirical formula of each of the following substances.
H=1 , N=14 ,O=16, P=31 , S=32 , Cu=64
What is the empirical formula of the phosphorus oxide that has 43.7% by mass of phosphorus and 28.4% by mass oxygen?
Your first step will be to use the equation mass/mr to find the moles of each compound.
P: 43.7/31 which give = 1.4097 moles of phosphorus to4d.p
O: 28.4/16 which give = 1.7750 moles of oxygen to 4d.p
Your next step will be to divide the smallest number of moles with each number of elements you have.
P: 1.4097/1.4097 would give = 1
O: 1.7750/1.4097 would give =1.26
Therefore the ratio is 1:2 hence the formula for phosphorus oxide is PO2. Therefore we know there is 1.26*2=2.52 oxygen and 1 phosphorus.
Because there is some decimal number we have to find a number that multiplies to give a whole number as we can’t just round up the number.
2.52*2= 5.04 which can be round to 5 oxygen atoms and we have to do the same of the phosphorus so 1*2 gives 2 atoms. Therefore this gives the formula to be P2O5
Molecular formula:
They are multiple ratios of the empirical formula therefore if we take the example from P2O5 we can have some molecular formula to be P6O15. This is just multiplied by 3 from the Empirical formula.