Summary of Chemistry Textbook - Section 4.1 the Mole Concept and Avogadro's Constant

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  Summary of Chemistry Textbook: Section 4.1  –   The Mole Concept and Avogadro’s Constant -   A relative scale   is one in which all measurements are compared to one standard or reference measure -   Relative scales have no units  –  they are merely comparisons of one quantity with another -   Advantage of relative scales is that very large or very small numbers can be compared more easily -   To generate the relative scale of atomic masses, chemists chose the most abundant isotope of the element carbon, the carbon-12 isotope ( 12 C), relative mass of exactly 12 units -   Reasons for choosing carbon:    Carbon is very cheap and is widely available    It is relatively easy to isolate and purify this isotope    Carbon is not toxic in any way -   Mass of 12 units mirrors the mass number of the isotope -   Relative atomic mass will closely parallel the number of these fundamental particles in the nucleus of the element -   Thus, the lightest of all elements, hydrogen, will have a relative atomic mass of close to 1 -   A mole   is defined as the amount of substance that contains as many elementary particles as there are atoms in precisely 12 grams of the carbon-12 isotope -   Actual number of atoms determined as approximately 6.02 x 10 23  -   Number is known as Avogadro’s constant , (symbol, L ) -   It is important to recognise that a mole is simply a number, just as a dozen is equal to 12 or a gross is equal to 144 L  = 6.02 x 10 23  mol -1  -   Atoms are so small that it is impossible to see an individual atom Calculating numbers of particles -   If you are asked to calculate a number of particles, the you are calculating the quantity represented by N  and you will obtain a number with no units -   Asked to calculate an amount   of substance, then you are trying to find a number    of mole  of the substance -   This is the quantity represented by n  and wi ll have the units ‘mol’      Where n = number of mole (amount), N = number of particles and L = Avogadro’s constant   SECTION 4.1 EXERCISES 1.   Calculate the number of mole of the stated particles in 1.00 mole of sulphuric acid. a)   Atoms of sulphur  –   1  b)   Atoms of hydrogen  –  2 c)   Atoms of oxygen    –  4  d)   Total number of atoms  –  7  2.   Calculate the number of mole of the stated particles in each of the following. a)   Magnesium ions in 1.25 mole of magnesium nitrate  –  1.25    b)   Sodium ions in 3.25 mole of sodium sulfate    –  2 x 3.25 = 6.50 mole of sodium ions  3.   Calculate the number of mole of the states particles in each of the following. a)   Carbon atoms in 2.00 mole of propane  –  2.00 x 3 = 6 mole of carbon atoms  b)   Carbon and hydrogen atoms in 1.25 mole of pentane  –  1.25 x 17 = 21.25  4.   Determine the number of particles, present in the following: a)   3.50 mol of ammonia molecules  –  n = 3.50 L = 6.02 x 10 23  Therefore; 3.50 x 6.02 x 10 23  = 2.10 24   b)   3.22 x 10 -3  mol of manganese nitrate n = 3.22 x 10 -3  L = 6.02 x 10 23 Therefore; 3.22 x 10 -3  x 6.02 x 10 23  = 1.94 x 10 21 5.   Calculate the amount of substance in mol present in each of the following. a)   1.44 x 10 24  atoms of neon = 2.39 mol of atoms b)   1000 atoms of mercury = 1.661 x 10 21  mol of atoms
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