Download Using mass to calculate molecular formula

Using mass to calculate molecular formula
From chemical analysis we know that toluenesulphonamide has the following
percentage composition by mass: C:49.13%, H:5.25%, N:8.18%, O:18.69% and
S:18.75%. What is the chemical formula of the compound?
Solution: Express the percentages as mass in grams, and then as moles, i.e. 100 g of
the substance contains:
49.13 g
49.13 g C: 12.01 g mol-1
= 4.09 moles of C atoms
5.25 g
5.25 g H = 1.01 g mol-1
= 5.25 moles of H atoms
8.18 g
8.18 g N: 14.00 g mol-1
= 0.58 moles of N atoms
18.69 g
18.69 g O: 15.99 g mol-1.
= 1.17 moles of O atoms
18.75 g
18.75 g S: 32.06 g mol-1
= 0.58 moles of S atoms
The ratio of moles of the constituent elements = ratio of atoms in the molecule.
Fractions of atoms cannot exist; convert the calculated ratio to the least whole
number ratio by dividing by the smallest mol number (0.58).
C: (4.09/0.58) = 7.05,
H: (5.25/0.58) = 9.05, N(0.58/0.58)= 1.00,
O: (1.17.0.58) = 2.01, S:().58/0.58) =1.00
i.e. C7H9NO2S
Note that the resulting ratios show small deviations from a whole-number ratio because
of experimental error in the original data and can be neglected.
Empirical formula and Molecular formula
Benzene consists of 7.69% H and 92.31%C. Converting this to a formula gives CH. This
is the simplest integer ratio. In fact a molecule of benzene has the formula C6H6.
Empirical formula CH – simplest whole number ratio.
Molecular formula C6H6 – actual number of atoms in the molecule.
Percentages of elements in a compound can only give the empirical formula. Need extra
information to determine the molecular formula.
Non-molecular compounds
There are three kinds of non-molecular substances:
1. Mono-atomic substances: e.g. He, Ne, Kr, Ar and Xe (known as ‘Noble Gas’
elements).
C
2. Giant molecules: e.g. silicon carbide i.e. SiC or (SiC)n. In this
compound there are equal numbers of silicon and carbon
Si
Si
Si
C
C
C
Si
atoms. Each silicon atom is surrounded by and bonded to four
carbon atoms, while each carbon is similarly attached to four
silicon atoms. The size of the particle of solid silicon carbide
is the only limitation on the number of atoms in the network –
which is so large as, for all intents and purposes, to be effectively infinite.
(Can include metals in this group of non-molecular compounds.)
3. Ionic Compounds and Ionic Bonding
An ion is an atom or a molecule with an electrical charge.
e.g. Na+ - the sodium ion - is the sodium atom less one electron. (11 protons in the
nucleus, 10 electrons around the nucleus.)
Cl- - the chloride ion – is the chlorine atom with an extra electron. ( 17 protons in the
nucleus, 18 electrons around the nucleus.)
O
_
nuclei. (How many electrons??)
N
O
NO3- - the nitrate ion - one electron more than the total charge on the
O
Ionic bonding is the electrostatic attraction between the ions. Na+ Cl- ;
(Na+)2 SO42-; K+ NO3- ; Ca2+ (F-)2 ; (NH4+)(ClO4-) ; etc (overall neutral)The ions are held
together by electrostatic attraction between the opposite charges.(Ionic Bonding).
Since an ionic compound must be neutral overall there must be equal numbers of charges
on the positive ions (cations) as there are on the negative ions (anions) and this
determines the formula for that compound.
Example: Sodium chloride, NaCl, 'table salt'
Both Giant molecules and Ionic compounds do not contain molecules. They have an
empirical formula but not a molecular formula.
Chemical Reactions and Chemical Equations
Compounds react to form other compounds – chemical reactions. Reactions are described
by a chemical equation.
e.g.
2H2 + O2  2H2O
Hydrogen and oxygen react to form water
Hydrogen and oxygen are the REACTANTS, water is the PRODUCT
As a chemical reaction does not change the types of atoms in the reaction, only
rearranges them, there must be as many atoms of each type on the left hand side (LHS) as
on the right hand side (RHS). (A balanced chemical equation.)
The equation shows the amounts of reactants and products in proportion:
2 molecules of hydrogen react with 1 molecule of oxygen to form 2 molecules of water.
2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water.
4 g of H2 react with 32 g of O2 to form 36 g of H2O
A chemical equation shows the reactant and product molecules and the amounts in which
they react or are formed.
e.g.
HCl + NaOH  H2O + NaCl
Hydrochloric acid reacts with sodium hydroxide to form water and sodium chloride
(common salt).
1 mole of HCl (36.5 g) reacts with 1 mole of NaOH (23+16+1=40 g) to form 1 mole of
water (18 g) and 1 mole of NaCl (23+35.5=58.5 g)
CH4 + 2O2  CO2 + 2H2O
methane + oxygen  carbon dioxide + water
Check that the number of atoms of each type is correct and that the masses are conserved.
If 3.4 g of methane are burnt in excess oxygen, what mass of carbon dioxide is formed.
13.4 g of CH4 is 13.4/16 = 0.837 mol of CH4
Equation shows that 1 mol CH4 produces 1 mol CO2, therefore 0.837 mol CO2, mass
0.837(12+2x16) = 36.8 g are formed.
If 0.7 mol of CH4 is reacted with 0.5 mol O2, what is the composition of the mixture after
the reaction?
0.5 mol O2 reacts with 0.25 mol CH4 to form 0.25 mol of CO2 and 0.5 mol H2O.
Therefore mixture contains 0.25 mol of CO2 , 0.50 mol H2O, 0.45 mol CH4. (Limiting
Reagents)
Balance the equation:
NH4NO3  N2 + O2 + H2O
(ammonium nitrate decomposes to give nitrogen gas, oxygen gas and water)
balance H: NH4NO3  N2 + O2 + 2H2O
balance O: NH4NO3  N2 + 1/2O2 + 2H2O
N is balanced.
Whole numbers: 2NH4NO3  2N2 + O2 + 4H2O
If 7.4 g of ammonium nitrate decomposes, how many moles of the products are formed?
Molecular mass NH4NO3 = 2x14+4x1+3x16 = 80 g mol-1
7.4 g is 7.4/80 = 0.0925 mol NH4NO3, therefore 0.0925 mol N2, 0.0462 mol O2 and 0.185
mol H2O are formed.
More information in Chemical Equations.
Chemical equations also show the state of the compounds.
NH4NO3 (s)  N2 (g) + O2 (g) + H2O (l)
HCl (aq) + NaOH (aq)  H2O + NaCl (aq)
solid, liquid, gas, aqueous solution …
Types of Chemical Reactions
Some types of reactions have particular names:
Decomposition:
CaCO3 (s)  CaO (s) + CO2 (g)
Combustion (reaction with oxygen):
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g)
Hydrogenation (reaction with hydrogen):
N2 (g) + 3H2 (g)  2NH3 (g)
Combination
Formation
Hydrolysis
Precipitation
Disproportionation