1.2 Kaedah Penyiasatan Saintifik

Kaedah Penyiasatan Saintifik
■ Kaedah Penyiasatan Saintifik

►

Kaedah sistematik yang dilakukan menurut peraturan dan prinsip saintifik supaya mendapatkan pengetahuan saintifik.

■ Langkah-langkah kaedah penyiasatan saintifik:

	►	Video ini mengandungi maklumat mengenai langkah-langkah kaedah penyiasatan saintifik:

1. Membuat pemerhatian Mengumpul maklumat, mengenai fenomena, menggunakan lima pancaindera. 2. Membuat inferens Membuat kesimpulan awal, atau penjelasan sementara mengenai fenomena berkenaan berdasarkan pemerhatian. 3. Mengenal pasti masalah Masalah dikenalpasti dengan bertanya soalan, berdasarkan kesimpulan yang dibuat. 4. Membuat hipotesis Membuat pernyataan umum tentang hubungan antara pembolehubah yang dimanipulasi dan pembolehubah bergerak balas untuk menerangkan soalan yang ditanya. 5. Mengenal pasti pembolehubah Mengenal pasti pembolehubah dimanipulasikan, bergerak balas dan dimalarkan untuk menguji hipotesis yang dibuat. 6. Mengawal pembolehubah Menentukan cara mengawal pembolehubah yang dipilih, apa yang perlu mengukur dan bagaimana untuk menjaga pembolehubah yang dimalarkan. 7. Merancang eksperimen Senaraikan semua yang diperlukan untuk menyelesaikan eksperimen. Terangkan langkah-langkah butiran siasatan, bagaimana untuk mengumpulkan data. 8. Mengumpul data Membuat pemerhatian atau pengukuran dan kemudian merekodkan data secara sistematik. 9. Mentafsir data Menyusun dan menganalisis data. Pengiraan, graf atau carta yang disediakan untuk mencari sebarang hubungan antara pembolehubah. 10. Membuat kesimpulan Membentuk satu kenyataan mengenai hasil siasatan yang merumuskan apa yang berlaku dalam eksperimen itu, sama ada hipotesis itu diterima atau ditolak. 11. Menulis laporan Satu laporan lengkap bertulis menurut format kaedah penyiasatan saintifik.

■ Animasi berikut menunjukkan aspek yang diperlukan dalam laporan penyiasatan saintifik.

Aktiviti Makmal 1.2.1: Kaedah Penyiasatan Saintifik

✍ Contoh latihan 1.2(a)
Pilih betul [✔] atau salah [✘] untuk kenyataan-kenyataan berikut.

Pernyataan
Penyiasatan saintifik dilakukan melalui satu siri langkah yang sistematik dikenali sebagai kaedah saintifik.	✔
Semua penyiasatan saintifik yang dijalankan mestilah melibatkan data berangka.	✘
Semua hipotesis dapat dibuktikan benar.	✘
Inferen adalah kesimpulan awal yang dibuat berdasarkan pemerhatian awal.	✔
Pembolehubah yang dimanipulasikan merupakan pembolehubah yang digunakan untuk menguji hipotesis.	✔

✍ Contoh latihan 1.2(b)
Padankan langkah-langkah penyiasatan saintifik berikut berdasarkan penjelasan yang diberikan.

Keterangan	Kaedah penyiasatan saintifik
Proses membentuk kesimpulan yang logik awal, yang mungkin atau mungkin tidak benar, untuk menerangkan proses pemerhatian.	Membuat inferen
Satu proses membuat kenyataan umum yang menyatakan hubungan antara pembolehubah bergerak balas dan pembolehubah yang dimanupulasikan	Membuat hipotesis
Satu langkah dalam penyiasatan saintifik supaya maklumat yang disampaikan dengan cara yang lebih bermakna.	Menganalisa data
Faktor-faktor atau keadaan yang mempengaruhi faktor-faktor lain dalam penyiasatan.	Pembolehubah

✍ Contoh latihan 1.2(c)
Untuk pengaratan berlaku, udara dan air diperlukan. Seorang pelajar menjalankan satu eksperimen untuk mengetahui sama ada udara diperlukan untuk paku besi berkarat. Tentukan pembolehubah yang terlibat dalam eksperimen ini.

Faktor	Jenis pembolehubah
Panjang paku besi	Pembolehubah yang dimalarkan
Kehadiran atau ketiadaan udara	Pembolehubah yang dimanupulasikan
Pengaratan paku besi	Pembolehubah yang bergerak balas

← 1.1 Kimia dan Kepentingannya

Senarai Video →

⇲ Dapatkan latihan (objektif and subjektif), secara percuma untuk OS Android.

Bab 1: Pengenalan kepada Kimia

Bab 1: Pengenalan kepada Kimia

Isi Kandungan
► 1.1 Kimia dan Kepentingannya
► 1.2 Kaedah Saintifik
► Senarai Video

1.1 Kimia dan Kepentingannya

Maksud kimia
■ Kimia

	►	Kajian saintifik terhadap sifat, komposisi dan struktur bahan-bahan yang terdapat di sekeliling kita.
	►	Perubahan yang berlaku apabila bahan-bahan ini berinteraksi antara satu sama lain.

■ Kimia dan penggunaannya

	►	Kebanyakan aktiviti dalam kehidupan seharian melibatkan tindak balas kimia.
	►	Bahan kimia yang digunakan dalam kehidupan harian yang dihasilkan oleh industri adalah berasaskan kimia.
	Bahan kimia Kegunaan Simen Digunakan sebagai bahan binaan bangunan. Garam Digunakan dalam masakan. Digunakan sebagai bahan pengawet makanan untuk membolehkan makanan bertahan lebih lama. Cat Digunakan untuk melindungi permukaan. Ejen pembersihan Digunakan sebagai agen pencuci untuk menanggalkan kesan kotoran. Ubat Digunakan untuk mengubati penyakit.

Kerjaya yang melibatkan bidang kimia

Kerjaya Penerangan Ahli ekologi Untuk mengkaji sifat bahan kimia dalam alam persekitaran. Jurutera genetik Mengkaji komposisi genetik dalam kromosom dan kesan kejuruteraan terhadap komposisi kromosom. Geokimia/Geologi Untuk mengkaji sifat bahan kimia di bumi. Ahli farmakologi Mengkaji dan mensintesis ubat-ubatan baru serta  kesan mereka ke atas pelbagai penyakit.

Industri di Malaysia

Bidang Penerangan Industri petroleum dan gas asli Menyediakan sumber bahan api untuk kenderaan dan tukang. Industri kimia Membuat bahan-bahan petrokimia seperti polivinil klorida atau PVC dan objek plastik seperti baldi dan komponen elektronik. Industri kimia pertanian Membuat racun perosak dan baja. Industri makanan Menghasilkan bahan tambahan makanan seperti perisa makanan. Menghasilkan agen pengawet seperti sodium benzoate. Industri farmasi Menghasilkan dan memasarkan ubat berlesen untuk digunakan sebagai ubat-ubatan. Industri elektronik Menghasilkan mikrocip daripada semikonduktor seperti silikon dan seramik. Menghasilkan superkonduktor untuk memproses maklumat dengan lebih cepat.

✍ Contoh latihan 1.1(a)
Isikan tempat kosong dengan jawapan yang tepat.

(a)	__________Kimia adalah satu bidang kajian yang berkaitan struktur, komposisi, sifat dan interaksi jirim.
(b)	__________Ahli farmasi memerlukan kedua-dua bahan kimia dan ilmu perubatan untuk menemui ubat-ubatan baru bagi menyembuhkan penyakit.
(c)	__________Geokimia terlibat dalam penyelidikan kimia di dalam kerak bumi.
(d)	__________Industri berasaskan kimia menghasilkan pelbagai produk yang berguna untuk pengguna dan menyumbang kepada pembangunan ekonomi negara.

↶ SPM Kimia

1.2 Kaedah Penyiasatan Saintifik →

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2.1.1 Laboratory Activity : Investigating the diffusion of particles in a gas, liquid and solid

Laboratory Activity 2.1.1: Investigating the diffusion of particles in a gas, liquid and solid

Aim: Investigating the diffusion of particles in a gas, liquid and solid
Hypothesis : Diffusion takes place in a gas, liquid and solid. The rates of diffusion in a gas, liquid and solid are in decreasing order.
Problem statement: Investigating the diffusion of particles in a gas, liquid and solid
Variable:

	»	Fixed variable: Temperature
	»	Manipulated variable: Medium of diffusion
	»	Responding variable: Rate of diffusion

Material: » Liquid bromine » Potassium permanganate crystals » Water » Hot jelly solution

Apparatus: » Two gas jars with plastic covers » Petri dish » Teat pipette » Spatula » Rubber stopper » Boiling tube

Procedures:

	►	(A) Diffusion in a gas 1. The animation below shows the arrangement and the results of the experiment. 2. A few drops of liquid bromine are dropped into a gas jar using a teat pipette. 3. The gas jar is covered with a gas jar cover. 4. Another gas jar is placed upside down on top of the first jar. 5. The cover is removed and any colour change is recorded. 6. The time taken for the brown bromine vapour to spread into the second gas jar is recorded.
	►	(B) Diffusion in a liquid 1. The animation below shows the arrangement and the results of the experiment. 2. A petri dish is filled with water. 3. A few potassium permanganate crystals are placed at the bottom of the water using a spatula. 4. Any colour change is recorded. 5. The time taken for the purple permanganate ions to spread throughout the water is recorded.
	►	(C) Diffusion in a solid 1. The animation below shows the arrangement and the results of the experiment. 2. Some freshly cooked jelly solution is placed into a boiling tube until it is almost full. 3. The jelly is allowed to set. 4. A small potassium permanganate crystal is placed on top of the jelly. 5. The boiling tube is then stoppered using a rubber stopper. 6. Any colour change is recorded. 7. The time taken for the purple permanganate ions to spread throughout the solid jelly is recorded.

Discussion:

	►	Diffusion has taken place in the gas (air in experiment A), liquid (water in experiment B) and solid (jelly in experiment C).
	►	The rates of diffusion of the particles in the solid, liquid and gaseous states differ. It is highest in gases, lower in liquids and the lowest in solids.
	►	This shows that there are more and bigger spaces between particles in the gas. The spaces between liquid particles are smaller. The particles in the solid state are very close with little space between them.
	►	The occurrence of diffusion proves that matter (bromine and potassium permanganate) consist of particles in constant motion
	►	The diffusion experiments show that particle posses the kinetic energy and are in constant motion.

Conclusion:

	►	Bromine is made of tiny and discrete particles.
	►	Potassium permanganate is made up of tiny and discrete particles (ions).

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1.2 Scientific Method

Scientific Investigation Methods
■ Scientific investigation methods

►

Systematic method that performed following rules and scientific principles to obtain scientific knowledge.

■ Steps of Scientific investigation method:

	►	This video contains information about steps of scientific investigation method:

1. Making observation Gathering information, about a phenomenon, using our five senses. 2. Making inference Making early conclusion, or tentative explanation, about the phenomenon, based on the observation. 3. Identify problem Problem is identified by asking question, based on the inference made. 4. Making hypothesis Making a general statement on the relationship between a manipulated variable and a responding variable in order to explain the question asked. 5. Identifying variables Identifying the manipulated, responding and fixed variables of an experiment to test the hypothesis made. 6. Controlling variables Deciding how to manipulate the chosen variable, what to measure and how to keep the fixed variables constant. 7. Planning an experiment List everything needed to complete the experiment. Describe details steps of investigations, how to collect the data. 8. Collecting data Making observations or measurement and then recording the data systematically. 9. Interpreting data Organising and analysing data. Calculations, graph or charts may be drawn to look for any relationship between the variables. 10. Making conclusion Forming a statement on the outcome of an investigation that sums up what happened in the experiment, whether the hypothesis was accepted or rejected. 11. Writing report A complete report is written according to the format of the scientific investigation method.

■ The animation below shows the items to be included in a scientific investigation report.

Laboratory Activity 1.2.1: Scientific Investigation method

✍ Worked-example 1.2(a)
Choose true [✔] or false [✘] for the statement given.

Statement
Scientific investigation is done through a series of systematic steps called scientific methods.	✔
All scientific investigations conducted must involve numerical data.	✘
All hypothesis can be proven true.	✘
An inference is the initial conclusion that is made based on the beginning observation.	✔
Variables that are manipulated are variables that are used to test the hypothesis.	✔

✍ Worked-example 1.2(b)
Match the following scientific investigation steps based on the explanation given.

Explanation	Scientific investigation method
A process of forming an initial logical conclusion, which may or may not be true, to explain a process or observation.	Making inference
A process of making a general statement that states the correlation between the responding variable and the manipulated variable.	Making hypothesis
A step in a scientific investigation in which information is presented in a more meaningful way.	Analizing data
Factors or conditions which influence other factors in an investigation.	Variables

✍ Worked-example 1.2(c)
For rusting to take place, air and water must be present. A students carry out an experiment to find out whether air is required for an iron nail to rust. Determine the variable involved in this experiment.

Variables	Types of variable
The length of iron nail	Constant variable
Presence or absence of air	Manipulated variable
Rusting of iron nail	Responding variable

← 1.1 Chemistry and Its Importance

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1.2.1 Laboratory Activity : Scientific Investigation method

Laboratory Activity 1.2.1: Scientific Investigation method

Aim: Investigating the effect of the temperature of water on the solubility of sugar
Problem statement: Does the amount of sugar that dissolve in water increase when the temperature of water increase?
Hypothesis: The higher the temperature of the water, the greater the mass of sugar that dissolve in it.
Variable:

	»	Fixed variable :	Volume of water and size of sugar
	»	Manipulated variable :	Temperature of water
	»	Responding variable:	Amount of sugar that dissolves at different temperatures

Material: » Sugar » Water

Apparatus: » 100 ml measuring cylinder » 250 ml beaker » Electronic balance » Bunsen burner » Tripod stand » Wire gauze » Spatula » Thermometer » Glass rod

Procedures:

	►	The animation below shows the arrangement and the results of the experiment.
	►	1. 100 cm³ of water is measured using a measuring cylinder and is poured into a 250 cm³ beaker. 2. The temperature of the water is recorded using a thermometer. 3. 100 ml beaker is filled with sugar. The beaker and its contents are then weighed and recorded as a gram. 4. The water is added a little at a time to the 100 cm³ of water in the beaker using a spatula. The mixture is then stirred using a glass rod. 5. The process is continued until no sugar can further dissolve in the water. 6. The beaker and its sugar content is weighed again and recorded as b gram. 7. The amount of sugar that dissolved in the water at room temperature is (a - b) gram. 8. The experiment is repeated by heating the water to temperatures of 40 ºC, 50 ºC, 60 ºC and 70 ºC respectively.

Results:

	►	Temperature (ºC) Room temperature 40 50 60 70 Initial mass of beaker and its contents(g) a b c d e Final mass of beaker and its contents(g) b c d e f Mass of sugar dissolve(g) a-b b-c c-d d-e d-f
	►	A graph of the mass of sugar dissolved against temperature is plotted.

Conclusion: The amount of sugar that dissolves in the water increases when the temperature of the water increases. The hypothesis is accepted.

↶ 1.2 Scientific Method

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3.3 Number of Moles and Mass

Number of moles and mass
■ Definition : Molar mass: The mass of a substance that contains one mole of the substance.

	►	Molar mass of substance = mass of 1 mole of substance
	►	Molar mass of substance = mass of substances for $6.02\times {10}^{23}$ particles.

■ Relationship between the number of moles, the mass molar and the mass of a substance.

	►	Molar mass of a substance is equivalent to their relative atomic mass or relative molecular mass or relative formula mass written in the unit of gram.
	►	For example: ○ The relative atomic mass of nitrogen, N = 14, thus the molar mass of nitrogen atom is 14g. ○ The relative molecular mass of nitrogen, ${\text{N}}_2$ = 14 x 2 = 28, thus the molar mass of a molecule = 28g.

■ Summary of relationship between the number of moles, the mass molar and the mass of a substance.

✍ Worked-example 3.3(a)
Calculate the number of particle in

(a) Calculate the mass of 0.5 mole chlorine ${\text{Cl}}_2$ [RAM : Cl = 35.5]

Solution The mass of 0.5 mole Chlorine ${\text{Cl}}_2$ Number of moles $\begin{array}{c}\text{=}\frac{\text{Mass of Chlorine}}{\text{Relative molecular mass}}\end{array}$ $\begin{array}{c}\text{0.5 =}\frac{\text{Mass of Chlorine}}{2.0\times 35.5}\end{array}$ Mass of Chlorine = $0.5\times 71$ Mass of Chlorine = 35.5g

(b) Find the number of moles in 32g Copper, Cu. [RAM : Cu = 64]

Solution Number of moles $\begin{array}{c}\text{=}\frac{\text{Mass of Copper}}{\text{Relative molecular mass}}\end{array}$ Number of moles $\begin{array}{c}\text{=}\frac{32}{64}\end{array}$ Number of moles = 0.5mol

✍ Worked-example 3.3(b)
Calculate the number of particles in 20g Methane, ${\text{CH}}_4$ , [RAM: H = 1, C = 12]

Solution Number of moles $\begin{array}{c}\text{=}\frac{\text{Mass of Methane}}{\text{Relative molecular mass}}\end{array}$ Number of moles$\begin{array}{c}\text{=}\frac{\text{20}}{12+\mathrm{(4}\times \mathrm{1)}}\end{array}$ Number of moles = 1.25mol Number of particles = Number of moles $\times$ Avogrado constant Number of particles = $1.25\times 6.02\times {10}^{23}$ Number of particles = $7.75\times {10}^{23}$

← 3.2 Number of Moles and the Number of Particles

3.4 Number of Moles and Its Molar Volume →

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3.2 The Mole and the Number of Particles

The number of moles and the number of particles
■ Definition : Mole: One mole is the amount of any substance which contains the same number of particles as there are in 12g of Carbon-12, that is $6.02\times {10}^{23}$ or the Avogrado constant.
■ Relationship between number of moles and number of particles.

	►	1 mole of any substance contains $6.02\times {10}^{23}$ particle of that substances (atoms, molecules or ions).
	►	For atom : ○ 1 mole of lithium (Li) : $6.02\times {10}^{23}$ atoms of lithium ○ 1 mole of carbon (C) : $6.02\times {10}^{23}$ atoms of carbon

►

For molecule: ○ 1 mole of hydrogen ( ${\text{H}}_2$ ) : $2\times 6.02\times {10}^{23}$ atoms of hydrogen or $6.02\times {10}^{23}$ molecules of hydrogen. ○ 1 mole of carbon dioxide ( ${\text{CO}}_2$ ) : $3\times 6.02\times {10}^{23}$ atom or $6.02\times {10}^{23}$ molecules of carbon dioxide.

►

For ion: ○ 1 mole of sodium chloride (NaCl) : $6.02\times {10}^{23}$ ions ${\mathrm{Na}}^{\mathrm{+}}$ and $6.02\times {10}^{23}$ ions ${\mathrm{Cl}}^{\mathrm{-}}$ ○ 1 mole of zinc bromide ( ${\text{ZnBr}}_2$ ) $6.02\times {10}^{23}$ ions ${\mathrm{Zn}}^{\mathrm{+}}$ and $2\times 6.02\times {10}^{23}$ ions ${\mathrm{Br}}^{\mathrm{-}}$

■ Summary of relationship between number of moles and number of particles.

✍ Worked-example 3.2(a)
Calculate the number of particle in

(a) 0.5 mole of Copper

Solution Number of particle in 0.5 mole of copper $\text{=}0.5\times 6.02\times {10}^{23}$ $\text{=}3.01\times {10}^{23}$ atom

(b) 0.25 mole Lead(II) ions, ${\mathrm{Pb}}^{\mathrm{2+}}$

Solution Number of particle in 0.25 mole of Lead(II) ions, $\text{=}0.25\times 6.02\times {10}^{23}$ $\text{=}1.505\times {10}^{23}$ ion ${\mathrm{Pb}}^{\mathrm{2+}}$

(c) 3.0 moles Carbon Dioxide,

Solution Number of particle in 3.0 moles of Carbon Dioxide $\text{=}3.0\times 6.02\times {10}^{23}$ $\text{=}1.806\times {10}^{24}$ molecule ${\mathrm{CO}}_2$

✍ Worked-example 3.2(b)
Calculate the number of moles of the following substances

(a) $1.505\times {10}^{24}$ atom of Silver, Ag

Solution Number of moles $\begin{array}{c}\text{=}\frac{1.505\times {10}^{24}}{6.02\times {10}^{23}}\end{array}$ = 2.5 mol

(b) $3.01\times {10}^{22}$ Oxide ions, ${\mathrm{O}}^{\mathrm{2-}}$

Solution Number of moles $\begin{array}{c}\text{=}\frac{3.01\times {10}^{22}}{6.02\times {10}^{23}}\end{array}$ = 0.5 mol

(c) $4.515\times {10}^{24}$ Nitrogen molecules, ${\mathrm{N}}_2$

Solution Number of moles $\begin{array}{c}\text{=}\frac{4.515\times {10}^{24}}{6.02\times {10}^{23}}\end{array}$ = 7.5 mol of molecules ${\mathrm{N}}_2$

✍ Worked-example 3.2(c)

	(a) Find the number of atoms in 1.5 moles of ${\mathrm{Cl}}_2$ Solution $\begin{array}{c}\text{1.5 =}\frac{\text{Number of molecules}}{6.02\times {10}^{23}}\end{array}$ Number of molecules = $1.5\times 6.02\times {10}^{23}$ Number of molecules = $9.03\times {10}^{23}$ Number of atom = $2\times 9.03\times {10}^{23}$ Number of atom = $1.806\times {10}^{24}$ atom
	(b) Find the number of ions in 0.5 mole of Magnesium Chloride, ${\mathrm{MgCl}}_2$ Solution $\begin{array}{c}\text{0.5 =}\frac{\text{Number of particles}}{6.02\times {10}^{23}}\end{array}$ Number of particles = $0.5\times 6.02\times {10}^{23}$ Number of particles = $3.01\times {10}^{23}$ Number of ions = $3\times 3.01\times {10}^{23}$ Number of ions = $9.01\times {10}^{23}$

← 3.1 Relative atom Mass and Relative molecular Mass

3.3 Number of Moles and Mass →

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3.1 Relative Atomic Mass and Relative Molecular Mass

Relative atomic mass
■ Relative atomic mass based on Carbon-12 scale [modern standard]

	►	Definition : The number of times the mass of atom is greater than $\frac{1}{12}$ of the mass of a carbon-12 atom. Relative atomic mass $=\frac{\text{Mass of atom of substance}}{\frac{1}{12}\times \text{mass of carbon-12 atom} }$
	►	For example : A magnesium atom is 24 times greater than $\frac{1}{12}$ of the mass of a carbon-12 atom, thus the relative atomic mass of magnesium is 24. $\text{Relative atomic mass of magnesium}$ $\text{=}24\times \frac{1}{12}\times \text{r.a.m of carbon-12}$ $\text{=}24\times \frac{1}{12}\times 12$ $\text{=}24$
	►	The animation below shows the idea of relative atomic mass in details.

■ Relative atomic mass of several elements :

Element Symbol Relative Atomic Mass Hydrogen H 1 Helium He 4 Carbon C 12 Nitrogen N 14 Oxygen O 16 Sodium Na 23

Relative Molecular Mass
■ Relative molecular mass based on Carbon-12 scale

	►	Definition : The number of times the mass of a molecule is greater than $\frac{1}{12}$ of the mass of a carbon-12 atom. Relative molecular mass $=\frac{\text{Mass of molecule of substance}}{\frac{1}{12}\times \text{mass of carbon-12 atom} }$
	►	For example : One nitrogen molecule (N2)is 28 times greater than $\frac{1}{12}$ of a carbon-12 atom, thus, the relative molecular mass of nitrogen is 28.

■ Calculation of relative molecular mass :

►

Adding the relative atomic masses of all the atoms of the elements forming the molecule.

■ This video gives the idea about relative atomic mass and relative molecular mass.

✍ Worked-example 3.1(a) Calculate the relative molecular mass for the following elements: (Relative atomic mass : H:1; C:12; N:14; O:16) Solution Element Molecular Formula Relative Molecular Mass Nitrogen N2 $14\times 2\text{=}28$ Water H2O $\mathrm{(1}\times \mathrm{2)}+16\text{=}18$ Glucose C6H12O6 $\mathrm{(12}\times \mathrm{6)}+\mathrm{(1}\times \mathrm{2)}+\mathrm{(16}\times \mathrm{6)}\text{=}180$

Relative Formula Mass
■ Relative molecular mass based on Carbon-12 scale

►

Definition : The number of times the mass of an ionic substances is greater than $\frac{1}{12}$ of the mass of a carbon-12 atom. Relative formula mass $=\frac{\text{Mass of an ionic substance}}{\frac{1}{12}\times \text{mass of carbon-12 atom} }$

■ Calculation of relative formula mass :

►

Adding the relative atomic masses of all the atoms of the elements forming the substances.

✍ Worked-example 3.1(b) Calculate the relative formula mass for the following substances: (Relative atomic mass : H:1; C:12; O:16, Mg:24, S:32, Cl:35.5, Ca:40, Cu:64) Solution: Element Compound Formula Relative Formula Mass Copper Chloride CuCl2 $64+\mathrm{(35.5}\times \mathrm{2)}\text{=}135$ Calcium Carbonate CaCO3 $40+12+\mathrm{(16}\times \mathrm{3)}\text{=}100$ Magnesium Sulphate Dehydrate MgSO4.7H2O $24+32+\mathrm{(16}\times \mathrm{4)}+\mathrm{7\{(1}\times \mathrm{2)}+\mathrm{16\}}\text{=}246$

↶ SPM Chemistry

3.2 Number of Moles and the Number of Particles →

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