# 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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## Pages in this set

### Page 1

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…

### Page 2

= 0.75 mole
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…

### Page 3

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…

### Page 4

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…

### Page 5

that 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…

### Page 6

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…

### Page 7

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…

### Page 8

balance a chemical equation, you can only skip such questions.

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…

### Page 9

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…

### Page 10

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.…