Download Formula masses ( Ar and Mr) The Ar is the relative atomic ... and more Schemes and Mind Maps Chemistry in PDF only on Docsity!
Formula masses ( Formula masses (Formula masses (
Formula masses (A
AA
Ar
rr
r and andand
and M
MM
Mr
rr
r) ))
The A
r
is the relative atomic massrelative atomic massrelative atomic massrelative atomic mass in the periodic table. The A
r
values of the atoms in a formula are added to get the formula formulaformula
formula
massmassmassmass or M
r
(sometimes called molecular mass)
H
2
SO
4
= 1x2 + 32 + 16x4 = 98
Mg(NO
3
2
= 24 + (14 + 3x16)x2 = 148
Percentage CalculationsPercentage CalculationsPercentage CalculationsPercentage Calculations
If you need to find the percentage of an element in a compound,
you use the formula:
Percentage = Ar x No of atoms x 100
Mr of compound
e.g.
Find the percentage of nitrogen in ammonium nitrate (NH
4
NO
3
Mr of N = 15
No of N atoms = 2
Mr of ammonium nitrate = 14 + 4 x 1+ 14 + 3 x 16 = 80
Percentage of nitrogen = 2x14x 100 = 35%
80
Empirical FormulaeEmpirical FormulaeEmpirical FormulaeEmpirical Formulae
- This is the formula which shows the lowest whole number ratio of the
atoms
- e.g. Molecular formula = C
2
H
4
, Empirical Formula = CH
2
- To calculate an empirical formula
- Find the mass (or %) of each element present
- Divide each of these masses by the relative mass of that element
- Divide each number obtained in stage 2 by the smallest of those
numbers.
- This should give whole numbers which can be used in the empirical
formula.
20g of a compound of Silicon with hydrogen contains
17.5g of silicon. Find the empirical formula.
Equation CalculationsEquation CalculationsEquation CalculationsEquation Calculations
In these, you will always be given the mass of one substance and
be asked to find the mass of another substance
MOLES; MOLES; MASS !!MOLES; MOLES; MASS !!MOLES; MOLES; MASS !!MOLES; MOLES; MASS !!
Moles
MolesMoles
Moles of the one you know
Moles
MolesMoles
Moles of the one you don’t
Now work out the unknown mass
massmass
mass
Element Si H
Mass of element
present
17.5 20-17.5 = 2.
Relative Mass (Ar) 28 1
Mass ÷ Ar 17.5÷28 = 0.625 2.5÷1 = 2.
Divide through by
smallest.
0.625 ÷ 0,625 = 1 2.5 ÷ 0.625 = 4
Whole Number Ratio 1 4
Empirical Formula SiH
4
Mass
M
r
Moles
Paper chromatography is a form of
separation that allows analysis of food
additives or dyes in pens.
Instrumental methods:
- Great for analysis of small amounts
- Rapid
- Sensitive
- Accurate
G as C hromatography linked to M ass
S pectroscopy (GC-MS) is an example
of an instrumental method
Gas chromatography allows the separation of a mixture of compounds
The time taken for a substance to travel through the column helps to
identify the substance
The mass spectrometer attached to the gas chromatography column
allows the mass of the substance to be observed as it leaves the column,
which also helps to identify the substance
The mass spectrometer can also give the relative molecular mass (Mr) of
each substance separated – this is seen at the molecular ion peakmolecular ion peakmolecular ion peakmolecular ion peak.
