topic 6 chemical calculations

7/22/2019 Topic 6 Chemical Calculations

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INTRODUCTION

Hi there and welcome to the topic on chemical solutions. Let us start this topic bylooking at our surroundings. Do you realise that chemical reactions are a dailyoccurrence? Take the kitchen as an example. Do you notice that chemicalreactions happen while you are cooking your favourite dishes? How about in thegarden? In the garden, chemical reactions happen during photosynthesis andwhen plants go through the process of decay.

We will explore more on chemical reactions as we learn how to write and balancethe chemical equations, calculate the relative mass and relative molecular mass,and the mole and Avogadro constant. Then, we will study the concept of molesof gases and solutions. Last but not least, we will learn two chemical formulaeof compound empirical and molecular, and how to calculate the empiricalformula.

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66

Chemical

Calculations

LEARNING OUTCOMES

By the end of this topic, you should be able to:1. Balance chemical equations;

2. Calculate the relative mass and relative molecular mass;

3. Calculate the mole and Avogadro constant;

4. Describe the concept of moles of gases, and moles and solutions;

5. Differentiate between the empirical formula and molecular formulaof compounds; and

6. Calculate the empirical formula of a compound.



TOPIC 6 CHEMICAL CALCULATIONS 142

WRITING AND BALANCING CHEMICALEQUATIONS

Before we learn how to write and balance chemical equations, firstly, let us learnthe meaning of chemical reaction.

How about a chemical equation?

A chemical equation can be written in words or using a chemical formula. Thus,do you know that in a chemical reaction, there are starting substances and newsubstances? Starting substances are known as reactants. They react with eachother to form new substances called products.

In a chemical equation, the reactants are written on the left side of the equationwhile the products are written on the right side. There is another importantsymbol to be added; an arrow () is placed between the reactants and productsin order to show the direction of the reaction.

6.1

A chemical reaction is a process in which one or more substancesreact with each other to form one or more new substances.

A chemical equation is a precise description of a chemical reaction.




According to the Law of Conservation of Mass, matter can neither be created nordestroyed. Therefore the number of atoms before and after a chemical reaction isthe same. From this, we can conclude that a chemical equation must be balanced.

How do we write a chemical equation? Writing a chemical equation involvesthree important steps:

(a) Write Down the Unbalanced EquationFor chemical equations, the reactants are written on the left side of theequation while the products are written on the right side.

(b) Balance the EquationThe Law of Conservation of Mass is applied to balance the chemical

equation. Remember, you only need to adjust the coefficients in front of theformulae. The subscript numbers in the formulae must not be changed asthis will change the formula of the substances. Each atom must be balancedand the equation must be checked again after balancing it.

(c) Indicate the State of the Reactants and ProductsThere are several indicators to identify the state of the reactants andproducts:

(i) Use (g) for gaseous substances;

(ii) Use (s) for solids;

(iii) Use (l) for liquids; and

(iv) Use (aq) for solution in water.

Let us apply these steps by constructing a chemical equation for the reaction ofhydrogen (H2) and oxygen (O2) to form water (H2O). Hydrogen and oxygen reacttogether to form water, and energy is released. Now, let us write the chemicalformulae and chemical equation for this reaction:

Reactants (Starting substances) = Products (New substances)




Therefore, we can write the chemical equation in words for this reaction as:

Hydrogen + Oxygen = Water

Chemical formula for hydrogen gas is H2

Chemical formula for oxygen gas is O2

Chemical formula for water molecule is H2O

The chemical equation for this reaction can be written as:

H2 + O2 H2O

As can you see from the above chemical equation, the reactants are hydrogen andoxygen as they are placed on the left side of the equation. The product is waterand it is written on the right side of the equation. However, the number ofoxygen atoms on the left side and the right side of the equation are unbalanced.

So how do we balance it? In order to balance the chemical equation, 2 moles ofhydrogen gas need to react with 1 mole of oxygen gas to form 2 moles of water.Therefore, we can balance the equation by adding a coefficient of 2 in front of H2 and a coefficient of 2 in front of H2O.

2H2 + O2 2H2O

Finally, indicate the state of the reactants and products

2H2(g) + O2(g) 2H2O(l)

Let us look at another example. If we heat a tin oxide with hydrogen gas to formtin metal and water vapour, how do we write a balanced chemical equation forthis chemical reaction? Let us follow the same steps.

(a) Firstly, write the unbalanced equation.

SnO2 + H2 Sn + H2O




(b) Balance the chemical equation.Identify the elements that are not balanced. If you look carefully, you willnotice that there are two oxygen atoms on the left side of the equation and

only one on the right side. You need to correct this by putting a coefficientof 2 before water:

SnO2 + H2 Sn + H2O

However, you will notice that the hydrogen atoms are now imbalanced asthere are two hydrogen atoms on the left and four hydrogen atoms on theright side of the equation.

You need to add a coefficient of 2 for the hydrogen gas to get four hydrogen

atoms on the right,

SnO2 + H2 Sn + 2 H2O

The equation is now balanced.

SnO2 + 2 H2 Sn + 2 H2O

(c) Next step is to indicate the state of the reactants and products.

SnO2 (s) + 2 H2 (g) Sn (s) + 2 H2O (g)

Finally, you can recheck the balanced equation by making an element inventory, just as the one shown in Table 6.1.

Table 6.1: Element Inventory

Element Before After

Sn 1 1

H 4 4

O 2 2

2

2




RELATIVE ATOMIC MASS AND RELATIVEMOLECULAR MASS

In the previous topic, you have been introduced to the term „mass number of anelement‰. Do you still remember? Let us recap the mass number definition beforewe learn further.

6.2

The mass number of an element refers to the total number ofneutrons and protons in the nucleus of an atom and not thesum of their mass.

SELF-CHECK 6.1

1. Write the chemical formula and chemical equation for thefollowing chemical reactions. Identify the reactants and theproducts of the chemical reactions.

(a) Zinc reacts with chlorine to form zinc chloride; and

(b) Hydrochloric acid reacts with sodium hydroxide to formsodium chloride and water.

2. Write a balanced equation for each of the following:

(a) The reaction between sulphur solid with iron solid to form

solid iron(III) sulphide;(b) The reaction between magnesium metal with hydrochloric

acid to form magnesium chloride solution and hydrogengas;

(c) The reaction between oxygen gas with solid copper metalto form copper(II) oxide solid;

(d) The reaction when hydrogen sulphide gas is bubbledthrough a sodium hydroxide solution to produce sodiumsulphide solution and liquid water.




Do you know that it is impossible to determine the mass of an atom by weighingit? That is why the mass of an atom is compared to the mass of a standard atominstead of weighing it.

Chemists can compare the masses of different atoms using the relative atomicmass scale. This is known as the Ar scale, where r stands for term ‰relative‰.

Initially, the standard atom used for the Ar scale was hydrogen and because it isthe lightest element, it was given an Ar of 1. However, now Carbon-12 is used forcomparison because carbon is easier to handle compared to the gaseoushydrogen. Besides that, it is also because Carbon-12 is an abundant isotope.

Relative atomic mass compares the mass of the atom against the mass of Carbon-

12. The relative atomic mass (Ar) of an element is the average mass of one atom of the element when compared with one-twelfth of the mass of a Carbon-12 atomas shown below.

Average mass of one atom of the element

1/ 12 mass of one Carbon-12 atomelative atomic mass of an element

Figure 6.1 shows you that 12 hydrogen atoms have the same mass as one carbonatom.

Figure 6.1: Relative atomic mass of hydrogenSource: http://www.schoolsnet.com

As you can see, the mass of carbon is 12. Therefore, hydrogen atom has a mass of1, which means its mass is 1/12 to that of Carbon-12.




Keep in mind that the relative atomic mass does not have any unit because therelative atomic mass only shows how many times heavier one atom is comparedto another. You can see this comparison in Table 6.2 as it shows that one atom of

silicon is 28 times heavier than one atom of hydrogen, whereas calcium atoms aretwice as heavy as neon atoms.

Table 6.2: The Approximate Relative Atomic Masses of Some Elements

Element Relative Atomic Mass (Ar)

Hydrogen 1

Carbon 12

Oxygen 16

Neon 20Silicon 28

Calcium 40

Copper 64

The relative molecular mass of a substance is calculated by adding up the relativeatomic masses of all the atoms present in a molecule of the substance.

The following calculation, which uses ammonia as an example, shows how

relative molecular mass of ammonia is calculated.

Example 6.1How do we calculate the relative molecular mass of ammonia?

Firstly, by now, we know that the molecular formula for ammonia is NH3.

Secondly, based on the periodic table, we know that the relative atomic mass of

H = 1, N = 14

Therefore, the relative molecular mass of ammonia is:

= Ar of nitrogen + (3 Ar of hydrogen)

= 14 + 3(1) = 17

This means that the average mass of one molecule of ammonia is 17 timesheavier than the mass of one-twelfth of a Carbon-12 atom.




THE MOLE CONCEPT

Now, let us move on to the mole concept. Do you notice that in our daily life, weuse the word ‰dozen‰ as a unit that represents the numerical 12, such as a dozenof eggs or a dozen of pencils. How about chemistry?

Well, in chemistry, scientists use a certain unit to present the amount ofsubstance of an atom or a molecule. The measurement unit used in chemistry is

called „mole‰. The following subtopics will elaborate more on mole.

6.3.1 The Mole and Avogadro Constant

First, what does one mole stand for? One mole is defined as the amount ofsubstance that contains as many particles as the number of atoms found in 12 g

of Carbon-12 which is 6.02 1023 particles. Or simply,

Mole (mol) = Amount of substance of an atom/molecule = 6.02 1023 particles

How about Avogadro constant or Avogadro number? Have you heard about it?Avogadro constant or Avogadro number is used to determine the number ofparticles in a substance. Thus, the Avogadro constant (NA) is defined as thenumber of particles in one mole of a substance.

Avogadro constant/Avogadro number (NA) = The number of particles in onemole of a substances

6.3

SELF-CHECK 6.2

1. Define the relative atomic mass of an element.2. Differentiate between relative atomic mass and relative molecular

mass.

3. How many times is krypton atom heavier than a helium atom?(Ar He = 4, Kr = 84)

4. A neon atom is half the mass of a calcium atom. Predict therelative atomic mass of calcium. (Ar Ne = 20)




The particles in the substance can be atoms, molecules or ions. The symbol ofmole is mol. Therefore, the type of particles needs to be carefully specified.

For example:

1 mole of carbon (C) contains 6.02 1023 (C) atoms

1 mole of water (H2O) contains 6.02 1023 water (H20) molecules

1 mole of sodium chloride (NaCl) contains 6.02 1023 sodium chloride(NaCl) formula units

We can find the number of particles in any number of moles of a substance andvice versa using the relationship shown in Figure 6.2.

Figure 6.2: Relationship between moles and particles

Let us look at some examples.

Example 6.2Find the number of atoms in 0.4 mole of iron.

Solution:Iron is an atomic substance.

The number of moles for iron = 0.4 mol

The number of iron atoms = Number of moles NA

= 0.4 6.02 1023

= 2.408

1023

atoms




Example 6.3

How many moles of water contain 3.01 1022 water molecules?

Solution22The number of water molecules 3.01 10 molecules

A

22

23

Number of moleculesNumber of moles of water

N

3.01 10

3.02 10

0.05 mol

Keep in mind that equal numbers of moles of substances always contain the samenumber of particles even though they differ in mass and size. Do you know why?Let us look at Figure 6.3.

In this figure, we can see that both blocks of magnesium and iron have differentmass but the same number of mole. Do they contain the same number of atoms?

Figure 6.3: Blocks of magnesium and iron

Both blocks of magnesium and iron contain 0.5 6.02 1023 atoms. By comparingthe number of moles of the substance, we can compare the number of particles in

a substance.




The molar mass of a substance is known as the mass of one mole of the substancein grams and the unit is grams per mole (g mol1). In our previous lesson, we

learnt that one mole of any substance contains 6.02 1023 particles of the

substance. Therefore, the molar mass of a substance contains 6.02 1023 particlesof the substance.

Molar mass (g mol 1) = The mass of one mole of substance = 6.02 1023 particles

The molar mass of any substance is numerically equal to its relative atomic,molecular or formula mass. Therefore, to measure a mole of atoms of anyelement, what we need to do is to weigh a mass equal to its relative atomic massin grams.

As for molecules, the mass of a mole of molecules of any molecular substance isequal to its relative molecular mass in grams. For ionic substance, the mass of amole of any ionic substance is equal to its relative formula in grams.

You will understand the above explanation better by studying Table 6.3 asshown below.

Table 6.3: Molar Mass of Substances

Substance Relative Mass Molar Mass (g mol-1)

Magnesium, M 24 24

Methane, CH4 12 + 4(1) = 16 16

Sodium chloride, NaCl 23 + 35.5 = 58.5 58.5

Based on Table 6.3, we can deduce that the mass of any number of mole ofa substance or vice versa can be calculated using the following relationship(Figure 6.4).

Figure 6.4: Relationship between number of moles and mass in grams

Now, let us look at an example on how to calculate moles of atoms.




Example 6.4How many moles of atoms does 136.9 g of iron metal contain?

Solution:We know that the relative atomic mass of iron is 55.85.

Thus, the molar mass of iron is 55.85 g mol1.

1 mol of iron atoms = 55.85 g of iron

136.9 g of iron = 1 mol of iron atoms/55.85 g of iron 136.9 g of iron

= 2.451 mol of iron atoms

How many atoms are there in 2.451 mol of iron?

We know that one mole of iron contains 6.02 1023 iron atoms.

Thus, 2.451 mol of iron = 2.451 6.02 1023 iron atoms

= 1.476 1024 iron atoms

1. How many atoms are there in 0.5 mole of Al?

2. Calculate the number of moles of S in 1.8 1024 S atoms.

3. An aspirin has the molecular formula of C9H8O4 (relative

atomic mass of H = 1, C = 12, O = 16; NA = 6.02 1023)

(a) Find the relative molecular mass of aspirin;

(b) Number of moles of H in aspirin; and

(c) Number of moles of O in aspirin.

ACTIVITY 6.1




6.3.2 Moles of Gases

Do you know that under the same temperature and pressure, equal volumes of

all gases contain the same number of particles? We will investigate this throughthe concept of molar volume.

Firstly, what does molar volume of a gas stand for? The molar volume of a gas isdefined as the volume of one mole of the gas. Thus, the molar volume is also the

volume occupied by 6.02 1023 particles of gas.

The molar volume of any gas is 22.4 dm3 mol1 at the standard temperature

of 0C and pressure of 1 atmosphere (STP) or 24 dm3 mol1 at room conditions

(25 C and the pressure of 1 atmosphere). This means that one mole of hydrogen

gas occupies the same volume as in one mole of oxygen at STP, which is22.4 dm3.

The relationship shows in Figure 6.5 describes that the volume of a gas can beconverted to the number of moles and vice versa.

Figure 6.5: Relationship between number of moles and volume of gas

6.3.3 Moles and Solutions

Now, let us learn about moles and solutions. Firstly, let us recap the solutiondefinition. A solution is a mixture formed by dissolving a solute in a solvent. Canyou think of any example? An example is a salt solution. This salt solution is

formed by dissolving salt such as sodium chloride (solute) in water (solvent).

Different concentrations of sodium chloride solution can be prepared by varyingthe amount of sodium chloride in a fixed volume of water. Therefore, theconcentration of a solution refers to the quantity of solute in a given volume ofsolution, which is normally 1dm3 of solution.




Hence the concentration of a solution can be expressed in moles per dm3, whichis also known as molarity or molar concentration. What does it mean bymolarity? Molarity is the number of moles of solute present in 1dm3 of solution,

or can be written as:

33

Moles of solute (mol)Molarity (mol dm )

Volume of solution (dm )

For example, hydrochloric acid with the molarity of 0.50 mol dm3 is prepared bydissolving 0.5 mol of hydrochloric acid in 1dm3 of solution.

EMPIRICAL FORMULA AND MOLECULARFORMULA

Let us start this subtopic by looking at the solutions shown in Figure 6.6 whichcan be found in your Chemistry laboratory. Look at the labels on the bottles. Doyou notice that besides the name of the solutions, their chemical formulas,NH4OH and CH3COOH, are also written on the label?

Figure 6.6: Chemical formulaeSource: https://science7acidbase.wikispaces.com

6.4




So, what is the meaning of a chemical formula?

The chemical formula of an element may represent its atoms or their molecules assome elements exist naturally as molecules.

Let us look at Figure 6.7 which shows the chemical formula of oxygen gas, whichis an element.

Figure 6.7: Chemical formula of an element

Keep in mind that if the substance is a compound, then the chemical formula fora compound has all the elements that are present in the compound and thenumber of atoms of each element.

Now, let us look at Figure 6.8, which shows the chemical formula of a compound water.

Figure 6.8: Chemical formula of a compound

A chemical formula is a representation of a chemical substance using

symbols; letters for atoms and subscript numbers for the number ofatoms of each element that are present in the substance.




In the following subtopics, we will look at the two types of formula that exist empirical formula and molecular f ormula.

SELF-CHECK 6.3

Based on Figure 6.6, explain the elements and number of atoms of eachelement found in NH4OH and CH3COOH.

1. Find the chemical formula of the compound shown in the

following table; and

2. Find the meaning of the chemical formula of the compound.

CompoundChemicalFormula

Meaning of Chemical Formula

Elements PresentRatio of Number ofAtoms in Elements

Ammonia NH3 Nitrogen, hydrogen N:H = 1:3

Sulphuric acid

Zinc hydroxide

Magnesiumnitrate

Sodiumhydroxide

ACTIVITY 6.2




6.4.1 Empirical Formula of Compound

Let us learn about the empirical formula of a compound by looking at this

example. Can you recall the formula for glucose? The formula for a glucosemolecule is C6H12O6. The ratio of carbon to hydrogen to oxygen atom in theglucose molecule is 6: 12: 6. This ratio can further be simplified to 1: 2: 1. Withthis ratio, you can simplify C6H12O6 and write it as CH2O.

Based on the previous discussion, we learnt that CH2O is the empirical formulaof a glucose molecule. So, what can you conclude about the empirical formula?The empirical formula of a compound gives us the simplest whole number ratioof atoms of each element in the compound. Thus, the simplest whole numberratio of atoms of glucose is CH2O.

6.4.2 Molecular Formula of Compound

How about molecular formula? The molecular formula is the actual number ofatoms of each element that are present in a molecule of the compound. Based onthe previous example, C6H12O6 is the molecular formula for a glucose molecule.

In short, the subscript numbers of a molecular formula are simple multiples ofthe empirical formula as shown below:

Molecular formula = (Empirical formula)n

n = Positive integer

SELF-CHECK 6.4

In your own words, differentiate between empirical formula andmolecular formula. Use some examples to support your explanation.




6.4.3 Calculating the Formula of a Compound

How do we calculate the formula of a compound? Keep in mind that there areoccasions when the information on the empirical formula of a compound isrequired. This is how we can determine the empirical formula of a compound;you just need to follow these simple steps:

(a) Firstly, the mass of each element in the compound must be determined;

(b) Then, the mass of each element must be converted to the number of molesof atoms; and

(c) Finally, the simplest ratio of moles of the elements must be determined.

1. Fill the blanks in the table below.

(a) Write the empirical formula of the following compounds based on the given molecular formula; and

(b) Find the value of „n‰.

Compound MolecularFormula

EmpiricalFormula

n

Water H2O

Methane CH4

Hydrogen Peroxide H2O2

Ethane C2H6

Octane C8H18

Caffeine C8H10 N4O2

2. Compare and contrast the empirical formula and molecularformula by filling the table below with the correct statement.

Molecular Formula Empirical Formula

Differences

Similarities

ACTIVITY 6.3




Let us look at the two examples below to learn how the empirical formula of acompound can be determined by using the steps listed just now.

Example 6.5Copper(II) iodide was found to contain 20.13% copper by mass.

Find its empirical formula. (Relative atomic mass: Cu, 64; I, 127.)

Solution:To find the solution:

(a) Find the Mass of Each ElementBased on the percentage, we know that 100 grams of the substance contain:

Copper = 20.13

Iodine = 79.87 (100 20.13)

(b) Convert the Mass to the Number of Moles

Number of moles of copper atoms = 20.13/64 = 0.315

Number of moles of iodine atoms = 79.87/127 = 0.629

(c) Find the Simplest Ratio of Moles of the Elements

Simplest ratio of the moles of copper = 0.315/0.315 = 1Simplest ratio of moles of iodine = 0.629/0.315 = 2

Thus, one mole of copper atoms combines with two moles of iodine atoms.Hence, the empirical formula of copper(II) iodide is CuI2.

Example 6.6A compound contains 8.63% hydrogen, 68.54% carbon and 22.83% oxygen. ThecompoundÊs molecular weight is approximately 140 g/mol.

Find its empirical formula and its molecular formula.(Relative atomic mass: H, 1.008; C, 12.011; O, 16)




Solution:To find the solution:

(a) Find the Mass of Each Element Based on the percentage, we know that 100 grams of the substance contain:

Carbon : 68.54 grams

Hydrogen : 8.63 grams

Oxygen : 22.83 grams

(b) Convert the Mass to the Number of Moles

Carbon : 68.54/12.011 = 5.71 mol

Hydrogen : 8.63/1.008 = 8.56 molOxygen : 22.83/16.00 = 1.43 mol

(c) Find the Simplest Ratio of Moles of the Elements

Carbon : 5.71 1.43 = 3.99

Hydrogen : 8.56 1.43 = 5.99

Oxygen : 1.43 1.43 = 1.00

Thus, the empirical formula of the compound is C4H6O.

(d) Calculate the value of n

(C4H6O)n = 140 where n represents an integer

[(12 4) + (6 1) + (16)]n = 140

70n = 140

n = 2

(C4H6O)2 = C8H12O2

Therefore, the molecular formula of the compound is C8H12O2




1. Find the relative atomic mass of the elements. 2. Apply the steps shown in the rhymes in order.

3. Derive the empirical formula of the compound.

(a) A compound is found to have (by mass) 48.38% carbonand 8.12% hydrogen, while the rest is oxygen. Find itsempirical formula.

(b) A compound is found to have 46.67% nitrogen, 6.70%hydrogen, 19.98% carbon and 26.65% oxygen. Find itsempirical formula.

ACTIVITY 6.4

SELF-CHECK 6.5

1. Compare the mass of an oxygen atom with that of a sulphur atom(relative atomic mass: O, 16; S, 32).

2. Sulphur and oxygen can form sulphur dioxide and sulphurtrioxide.

(a) Write the formula of both compounds;

(b) Calculate the relative molecular masses of sulphur dioxideand sulphur trioxide; and

(c) How many times heavier is a molecule of sulphur dioxidecompared to a molecule of sulphur trioxide?

3. A substance has a molar mass of 34 gmol1

. What is the mass of2.5 moles of the substance?

4. Compounds containing sulphur and oxygen are serious airpollutants; they cause acid rain. It is revealed that a purecompound contains 50.1% sulphur and 49.9% oxygen by mass.What is the simplest formula (empirical formula) for thiscompound?




There are three steps to follow when writing and balancing a chemicalequation:

Write down the unbalanced chemical equation;

Balance the equation by adjusting the coefficients in front of the formula.Check that the equation is balanced; and

Indicate the symbols for the state of the reactants and products.

The relative atomic mass (Ar) of an element is the average mass of one atomof the element when compared with one-twelfth of the mass of a Carbon-12atom.

The relative molecular mass of a substance is the average mass of a moleculeof the substance when compared with one-twelfth of the mass of one Carbon-12 atom.

A mole is defined as the amount of substance which contains as manyparticles as the number of atoms found in a 12 g of Carbon-12, which is

6.02 1023 particles.

The Avogadro constant (NA) is defined as the number of particles in onemole of a substance.

The molar volume of a gas is defined as the volume of one mole of the gas.

The concentration of a solution can be expressed in moles per dm3, which isalso known as molarity or molar concentration.

The empirical formula of a compound gives the simplest whole number ratioof atoms of each element in the compound.

The molecular formula is the actual number of atoms of each element that arepresent in a molecule of a compound.

In a molecular formula, the subscript numbers are simple multiples of theempirical formula.

The molar mass of a substance is defined as the mass of one mole of thesubstance in grams.




There are three steps to determine the empirical formula of a compound:

Find the mass of each element in the compound;

Convert the mass to the number of moles of atoms; and Find the simplest ratio of moles of the elements.

Avogadro constant

Chemical formula

Empirical formula

Molar mass

Mole

Molecular formula

Relative atomic mass

Relative molecular mass

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Conoley, C., & Hills, P. (2002). Chemistry (2nd ed.). London: Harper-Collins.

Hewitt, P. G. (1998). Conceptual physics (8th ed.). Massachusetts: Addison-Wesley.

Kementerian Pendidikan Malaysia Bahagian Pendidikan Guru. (1995). Bukusumber pengajaran pembelajaran sains sekolah rendah: Strategi pengajarandan pembelajaran sains. Kuala Lumpur: Kementerian Pendidikan Malaysia.

Ralph, A. B. (2003). Fundamentals of chemistry . New Jersey: Prentice Hall.

Whitten, K. W., Davis, R. E., Peck, M. L., & Stanley, G. G. (2010). Chemistry(9th ed.). Belmont: Brooks/Cole.

topic 6 chemical calculations

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