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 CuSO4. 5H2O

 

(a) RMM of CuSO4. 5H2O is: 64 + 32 + 4 X 16 + 5 ( 2 + 16 ) = 250
Hence, % of oxygen = 100% X ( 9 X 16 ) / 250 = 57.6 %

 

(b) % of water = 100% X ( 5 X 18 ) / 250 = 36.0 %

 

 

2. Calculate the mass of nitrogen in 15 g of urea, CO(NH2)2.

 

RMM of urea = 12 + 16 + 2 (14 + 2) = 60

Hence, mass of Nitrogen = 15 g X ( 2 X 14 ) / 60 = 7.0 g

 

 

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 = Pb3O4

 

 

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 = CH2

(b) Let molecular formula be (CH2)n

Then, n X 14 = 70; this implies that n = 70 / 14 = 5.
Hence, molecular formula = (
CH2)5 = C5H10.

 

 

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.
Mols = mass / RMM = 14g / 28 = 0.5 mols

 

Step 2: Convert moles to particles.
Since 1 mol contains 6.02 X 1023
molecules (Avogadro's number)
0.5 mol will contain = 0.5 X 6.02 X 1023
molecules = 3 X 1023 molecules

 

d) volume of 11 grams of carbon dioxide at rtp

Step 1: Convert mass to moles.
Mols = mass / RMM = 11g / 44 = 0.25 mols

 

Step 2: Convert moles to volume.
Since 1 mol of any gas occuoies a volume of 24 dm3 at rtp
0.25 mol will occupy = 0.25 X 24 = 6 dm3

 

 

6. 8.0 g of sodium hydroxide (NaOH) pellets are dissolved in 200 cm3

of water. What is the concentration of the solution in mol/dm3 ?

Mol = mass / RMM = 8 g / 40 = 0.2 mol
Volume =
200 cm3 = 0.2 dm3

Using the relation CONC = MOLE / VOLUME,
Concentration = 0.2 mol /
0.2 dm3 = 1 mol / dm3

 

 

7. Calculate the mass of sodium hydroxide NaOH needed to produce

a concentration of 1 mol/dm3 when dissolved in 250 cm3 of water.

Mol = mass / RMM (eq1) ; Mol = conc X volume (eq2)
From eq1, mass = mol X RMM (eq3)
Substituting (eq2) into (eq3), we have:

mass = ( conc X volume ) X RMM

= ( 1 mol/dm3 X 0.25 dm3 ) 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 = (42-9) 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 = BH3

 

 

 

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 SO2 + O2 => 2 SO3. Determine the (a) Limiting Reagent and

(b) find the mass of the sulfur trioxide produced.

(a) 2 SO2 + O2 => 2 SO3.

From given eqn, mol of SO2 / mol of O2 = 2 / 1
In actual experiment, mol of SO2 / mol of O2 = 3.5/1.1 = 3.18 / 1
Comparing the mole ratios of experimental & calculated,
we observe that SO2 is in excess ( 3.18 > 2 );

Hence, oxygen is our limiting reagent.

(b) From the given eqn, 1 mol O2 gives 2 mols SO3
Since O2 is our limiting reagent,
Hence, 1.1 mols of O2 will give = 1.1 X 2 = 2.2 mols of 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
Mol of methane = 32 g / 16 = 2 mol
Mol of oxygen gas = 96 g / 32 = 3 mols

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
Hence, 3 mol of O2 will react with = 1.5 mol of CH4.
However, 2 mol of CH4 were used in the expt.
Thus, excess reagent = 2 mol - 1.5 mol = 0.5 mol of CH4

 

(c) From eqn, 2 mols of O2 gives 1 mol of CO2
Hence, 3 mols of O2 will give 1.5 mols of CO2

 

 

 

11. The equation for the reaction between iron(III) oxide and carbon

monoxide is given as: Fe2O3 + 3 CO 2 Fe + 3 CO2

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
OR 160 g of Fe2O3 gives 2 X 56 g of 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
OR 160 g of Fe2O3 gives 3 X 24 dm3 of CO2
Hence, 40 g of Fe2O3 will give = 40 X (3 X 24) / 160 = 18 dm3
CO2

 

 

12. An acid-alkali 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


From eqn, mol of H2SO4 / mol NaOH = 1 / 2
Cross multiplying, we have 2 X mol H2SO4 = mol NaOH (eq1)

 

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 cm3

 

 

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 SO2 + O2 ===> 2 SO3

140 g of SO3 was produced in the reaction. Calculate the % yield of SO3.

From the eqn, 2 SO2 + O2 ===> 2 SO3


From eqn, 2 mol SO2 gives 2 mol SO3 [eq1]
Simplifying this, [eq1] becomes: 1 mol SO2 gives 1 mol SO3 [eq2]

[Eq2] can also be written as:
1 X 64 g of SO2 gives 1 X 80 g of SO3
Hence, 128 g of SO2 will give = 128 X 80 / 64 = 160 g of SO3

However, only 140 g of SO3 was actually produced in the reaction.
Therefore, % yield = ( 140 g ) X 100% / ( 160 g ) = 87.5 %

 

 

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

dm3 of carbon dioxide was produced at rtp. Calculate the % purity of the

calcium carbonate sample.

Write the eqn, CaCO3 + 2 HNO3 ===> Ca(NO3)2 + H2O + CO2


From the eqn, 1 mol CaCO3 gives 1 mol 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.
Therefore, % purity = (1.35 dm3) X 100% / (1.44 dm3) = 93.75 %

 

 

15. Complex Calculations involving Volumetric Analysis (Titration)

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