# Tutorial on Molar Calculation

Our teacher gave us this tutorial to help us w/ our molar calculations when we started learning chem. He wrote it himself. There are a lot of examples for different levels and requirements. It basically includes everything except for gas molar calculations because we hadn't learned that at the time.

This was prepared according to the HKDSE curriculum in Hong Kong, which is between the GCSE level and the A Levels, but I think that molar calculations are the same in all syllabuses. Hope you find this useful!

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• Created by: saraht
• Created on: 23-05-11 14:50

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Tutorial on Molar Calculation (Chemistry II)
This tutorial is a step-to-step guide on how to handle questions on molar
calculation. Please refer to the summary on p.106 to p.107 for the details on what
Hint: Molar calculation has a very simple principle yet many possible applications in
real life. When you see an unfamiliar situation, no need to feel frustrated or scared,
Equation required: (no. of particles / 6.02 x 1023) = no. of mole = (mass / molar
mass)
The Fundamental Level
Conversion between mass and number of mole, and comparing the amount of
different substances
Mass of substances cannot be compared directly as the particles inside would have
different masses. For example, glucose molecule (C6H12O6) has a molecular mass of
180 whereas that of carbon dioxide is 44. So 10 g of glucose would have less number
of molecules compare to 10 g of carbon dioxide. Hence for fair comparison between
different substances, it needs to be done according to their number of mole instead.
Example 1
Calculate the number of mole of molecules in 4 g of hydrogen gas and 4 g of
helium. Calculate the number of atoms in these gases. Which one of them
contains more atoms?
No. of mole of hydrogen gas (H2) = = 2 moles
No. of mole of helium gas (He) = = 1 mole
No. of hydrogen atoms
= 2.4 x 1024
No. of helium atoms = 1 mole 6.02 1023 = 6.02 x 1023
Hence in the same mass of hydrogen gas and helium gas, hydrogen gas contains 4
times the number of atoms to that of helium.
Example 2
Calculate the mass of ozone which contains the same number of molecules as
24.0 g of oxygen gas.
No. of mole of oxygen molecules in 24.0 g oxygen gas =

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Mass of ozone having 0.75 mole of molecules =
= 36.0 g
Example 3
What is the mass of sodium carbonate that has the same number of ions as 10.0
g of magnesium chloride?
Formulae of sodium carbonate and magnesium chloride are Na2CO3 and MgCl2
respectively. Each mole of them contains 5 moles and 3 moles of ions respectively.
Number of mole of MgCl2 in 10.0 g = =
0.105 mole
Number of mole of ions in the MgCl2 = =

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Percentage by mass of S = = 24.3 %
Percentage by mass of O = = 48.4 %
Alternative, Percentage by mass of O = (100 ­ 21.2 ­ 6.06 ­ 24.3) % = 48.4 %
Example 5
Calculate the mass of carbon in 50.0 of glucose (C6H12O6).
Percentage by mass of C
=
=
Mass of C in 50.0 g glucose =
Alternatively,
Mass of C in 50.0 glucose =
= 20.0 g
Example 6
Nitrogen-containing compounds are important fertilizers for agriculture.…read more

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Percentage of N in ammonium chloride =
= 26.2 %
Percentage of N in potassium nitrate =
= 13.8 %
Hence ammonium nitrate is the best fertilizer with 35 % by mass of nitrogen inside.
The Intermediate Level
Determining empirical formula and molecular formula
Empirical formula shows the simplest ratio between the numbers of particles of
different elements inside a compound. It is also the only formula for ionic

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O exists as O2 molecules in nature.
Example 8
Upon heating 25.0 g of a sample of mercury oxide, 1.847 g of oxygen gas was
liberated. Deduce the empirical formula of mercury oxide.
Mass analysis: mass of oxygen in the sample = 1.847 g
mass of mercury in the sample = 25.0 ­ 1.847 =
23.153 g
Hg O
Mass (g) 23.153 1.847
No. of mole
= 0.115 = 0.115
Simplest integral ratio
=1 =1
The empirical formula of the mercury oxide is HgO. (i.e.…read more

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Mg O
Mass (g) 12.6 20.9 ­ 12.6 = 8.3
No. of mole
Simplest integral ratio
The empirical formula of the metal oxide formed is MgO.
From empirical formula to atomic mass
Sometimes we need to deduce the relative atomic mass of an unknown element in a
compound. If the empirical formula of the compound is known, the relative atomic
mass required can be calculated from the ratio between the elements involved and
the relative atomic mass of some known elements.…read more

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Sometime we are looking for the ratio between different substances instead of the
ratio between atoms, such as finding the number of water of crystallization.
Example 13
Upon being heated strongly, the mass of a sample of hydrated nickel(II)
sulphate, NiSO4.nH2O, changed from 16.20 g to 8.93 g. Calculate the number of
water of crystallization in hydrated nickel(II) sulphate.
Mass of NiSO4 = 8.93 g
Mass of H2O = 16.20 ­ 8.93 =7.27 g
NiSO4 H 2O
Mass (g) 8.93 7.27

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For example, 2H2(g) + O2(g) 2H2O(l)
no. of particles 2 1
2
no. of particles 1.2 x 1024 6.02 x 1023
1.2 x 1024
no. of mole 2 1
2
molar mass (g mol-1) 1.0 x 2 16.0 x 2 1.0 x 2 +
16.0
reacting masses (g) 1.0 x 2 x 2 16.0 x 2 (1.0
x 2 + 16.0) x 2
The equation shows that 2 hydrogen molecules react with 1 oxygen molecule and

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Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s)
No. of mole of AgNO3 present =
= 0.0218 mole
No. of mole of Cu required = = 0.0109 mole
Mass of Cu required = 0.0109 63.5 = 0.692 g
Example 17
Sodium oxide dissolves in water to produce sodium hydroxide. Calculate the
mass of sodium hydroxide produced when 10.4 g of sodium oxide is dissolved
in water.
Na2O(s) + H2O(l) 2NaOH(aq)
No. of mole of Na2O used = = 0.168
mole

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Compounds A, B and C have molar masses 105 g mol-1, 234 g mol-1 and 148 g
mol-1 respectively. They react according to the equation 2A + B 3C. Calculate
the mass of B required to completely react with 4.0 g of A and the mass of C
produced.
No. of mole of A reacted = = 0.038 mole
No. of mole of B required = = 0.019 mole
Mass of B required = 0.019 234 = 4.45 g