Mole
concept calculations A Quick Review
prepared by Alex Teoh / 8 Nov 2004  0630 hours
1. Calculate the percentage % of:
(a) oxygen (b) water
in CuSO_{4}. 5H_{2}O
(a) RMM of CuSO_{4}. 5H_{2}O is: 64 + 32 +
4 X 16 + 5 ( 2 + 16 ) = 250 (b) % of water = 100% X ( 5 X 18 ) / 250 = 36.0 % 
2. Calculate the mass of nitrogen in 15 g of urea, CO(NH_{2})_{2}.
RMM of urea = 12 + 16 + 2 (14 + 2) = 60 
3. An oxide of lead contains 9.34% by mass of oxygen.
Find its empirical
formula.
Element 
% 
RAM 
% / RAM 
Divide by smallest no. 
Ratio 

Pb 
90.66 
207 
0.438 
1 
X 3 = 
3 
O 
9.34 
16 
0.584 
1.33 
X 3 = 
3.99 
Empirical formula = Pb_{3}O_{4}
4. A hydrocarbon of mass 5 g was found to contain 4.286 g of
Carbon
by analysis. Find
(a) the empircal formula of the hydrocarbon and
(b) the molecular formula if its relative molecular mass is
given as 70.
Element 
mass 
RAM 
mass / RAM 
Divide by smallest no. 
Ratio 
C 
4.286 
12 
0.357 
1 
1 
H 
0.714 
1 
0.714 
2 
2 
(a) Empirical formula = CH_{2}
(b) Let molecular
formula be (CH_{2})_{n}
Then, n X 14 =
70; this implies that n = 70 / 14 = 5.
Hence, molecular formula = (CH_{2})_{5
}= C_{5}H_{10}.
5. Calculate the:
a) number of moles of atoms in 24 g of oxygen.
Mols = mass / RAM
= 24 g / 16 = 1.5 mols
b) number of moles of molecules in 24 g of
oxygen.
Mols = mass / RMM
= 24 g / 32 = 0.75 mols
c) number of molecules in 14 g of nitrogen gas.
Step 1: Convert mass to moles. Step 2: Convert moles to particles. 
d) volume of 11 grams of carbon dioxide at rtp
Step 1: Convert mass to moles. Step 2: Convert moles to volume. 
6.
8.0 g of sodium hydroxide (NaOH) pellets are dissolved in 200 cm^{3}
of water. What is the concentration of the
solution in mol/dm^{3}^{
}?
Mol = mass
/ RMM = 8 g / 40 = 0.2 mol Using the relation
CONC = MOLE / VOLUME, 
7. Calculate the mass of sodium hydroxide
NaOH needed to produce
a concentration
of 1 mol/dm^{3 }when dissolved
in 250 cm^{3} of water.
Mol = mass
/ RMM (eq1) ; Mol = conc X volume (eq2) mass = (
conc X volume ) X RMM = ( 1 mol/dm^{3}
X 0.25 dm^{3} ) X 40 = 10 g ^{ } 
8. A compound of
mass 42 g contains only the elements  Boron and
Hydrogen. On
analysis, it was found that the compound contained
9 g of Hydrogen.
(a) Find the number of moles of Boron atoms B, present in
the
compound.
(b) Find the number of moles of Hydrogen atoms H, present in
the
compound.
(c) Find the empirical (simplest) formula of this compound.
(a) Mols of Boron atoms = mass / RAM = (429) g / 11 = 3
mols B (b) Mols of Hydrogen atoms = mass / RAM = 9 g / 1 = 9 mols
H (c) 3 mols of B combine with 9 mols of H or 1 mol of B
combines with 3 mols of H or 1 atom of B
chemically combines with 3 atoms of H Hence,
empirical formula = BH_{3} 
9.
In a certain reaction, 3.5 mols of sulfur dioxide was reacted with
1.10 mols of oxygen to form sulfur
trioxide. The equation is:
2 SO_{2} + O_{2} => 2
SO_{3}. Determine the (a) Limiting Reagent and
(b) find the mass of the sulfur trioxide
produced.
(a) 2 SO_{2} + O_{2}
=> 2 SO_{3}. From given eqn, mol of SO2 / mol
of O2 = 2 / 1 Hence, oxygen
is our limiting reagent. (b) From the given eqn, 1 mol O2
gives 2 mols SO3 
10.
Methane reacts with excess oxygen to form carbon dioxide and
water vapor. In a certain reaction, 32 g
of methane was mixed
with 96 g of oxygen and ignited.
a) Determine the Limiting reactant.
b)
Find the amount of excess reactant left over.
c) Find the amounts of carbon dioxide
produced.
a) Equation: CH4 + 2O2 ==> CO2
+ 2H2O Mole ratio(calc) = mol of O2 / mol
of CH4 = 2 / 1 * Convert all reactant
masses into mols for easier operation Mole ratio (expt) = mol O2 / mol
CH4 = 3 / 2 = 1.5 / 1 If we compare mole ratio(expt)
with mole ratio(calc), we find that (a) Oxygen is our limiting
reactant because ( 1.5
< 2 ) (b) According to eqn, 2 mols of O2
reacts with 1 mol of CH4 (c) From eqn, 2 mols of O2 gives 1
mol of CO2 
11. The equation for the reaction between iron(III) oxide
and carbon
monoxide is
given as: Fe_{2}O_{3} +
3 CO è 2 Fe + 3
CO_{2}
In a certain
experiment, 40 g of iron(III) oxide was reacted with
excess carbon
monoxide. Calculate the:
(a) mass of Iron, Fe produced.
(b) volume of carbon dioxide gas produced.
a) From eqn, 1 mole Fe2O3 gives 2 mol Fe Hence, 40 g of Fe2O3 will give = 40 X ( 2 X 56 ) / 160 = 28 g of Fe b) From eqn, 1 mol of Fe2O3 produces 3 mol of CO2 
12. An acidalkali titration using methyl orange was carried
out. It was
found that a
certain volume of 0.4 mol/dm3 sulfuric acid
neutralised 25
cm3 of 0.5 mol/dm3 of sodium hydroxide.
Calculate the
volume of sulfuric acid used for neutralisation.
Write the eqn, 2
NaOH + H2SO4 ==> Na2SO4 + 2 H2O
Since Mol = Conc X Volume, Eq1 becomes: 2 X Conc (acid) X Vol (acid) = Conc (alkali) X Vol
(alkali) Hence, Volume (acid) = (
0.5 X 25 ) / ( 2 X 0.4 ) = 15.6 cm^{3} 
Question
13 onwards 
for Pure Chem 5068 students
13. In an experiment, 128 g of sulfur dioxide was reacted
with oxygen to produce
sulfur
trioxide. The equation for the reaction is:
2 SO_{2} + O_{2} ===> 2 SO_{3}
140 g of SO_{3 }was produced in the reaction. Calculate the % yield of SO_{3.}
From the eqn, 2 SO_{2} + O_{2} ===> 2 SO_{3}

_{ }
_{ }
14. An impure sample of calcium carbonate contains calcium
sulfate as an
impurity. When
excess dilute nitric acid was added to 6 g of the sample, 1.35
dm^{3} of
carbon dioxide was produced at rtp. Calculate the % purity of the
calcium carbonate sample.
Write the eqn,
CaCO3 + 2 HNO_{3} ===> Ca(NO3)2 + H2O + CO2
or we say that, 1 X 100 g CaCO3 gives 1 X 24 dm3 of CO2 @
rtp Hence, 6 g of CaCO3 will give = 24 dm3 X 6 g / 100 g =
1.44 dm3 However, only 1.35 dm3 of CO2 was actually produced in the
reaction. 
15. Complex Calculations involving Volumetric Analysis
(Titration)
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