c4e237864d12cdb0be66e0e86247e4cben4.pptx

Mole concept it's explanation and calculations on mole concept including concentration.

1. MOLE CONCEPT

2. HOW WE MEASURE ENTITIES? MASS- GRAMS VOLUME – LITRES DISTANCE -METRES

3. How to measure atoms or molecules of such smaller magnitude? Just as we denote 1 Dozen for 12 items, it can be anything 12 eggs-1 Dozen 12 banana-1 Dozen 12 bangles- 1 Dozen Similarly mole is used to count entities on the microscopic level ( i.e. atoms/molecules/ions/ particles etc.)

4. Define moles One mole is the amount of substance that contains as many entities or particles as there are present in 12 g of C-12 atom which is exactly equal to 6.02 x 10 23 particles. Here 6.02 x 10 23 is the Avogadro number or Avogadro constant

5. Why C-12 is used as relative atomic mass(atomic weight) Atomic weights are an average of the relative masses of all of the isotopes of the given element. The number of C-12 atoms in exactly 12.00 g of C-12 is 6.02 X 10 23 . So C-12 is used as relative atomic mass in calculation of Avogadro number

6. Relative atomic mass The ratio of average of mass of an atom of the element to the 1/12 th mass of the carbon atom is defined as the relative atomic mass of any element. Example Mass of one N atom=14x1/12 (mass of 1 carbon -12 atom) Mass of one O atom=16x1/12(mass of 1 carbon-12 atom) Atomic mass unit: 1/12 x mass of 1 carbon-12 atom

7. What is Avogadro number /Avogadro constant? Avogadro number is the number of constituent particles usually atoms/molecules present in 1 mole of the substance which is equals to the 6.02 x 10 23 Example- 1 mol of hydrogen atoms = 6.02 x 10 23 atoms 1 mol of water molecules= 6.02 x 10 23 molecules 1 mol of sodium chloride = 6.02 x 10 23 formula units

8. How we calculated Avogadro constant? Mass of a C-12 atom=1.99x 10 -23 g ( determined by mass spectrometer) Knowing that one mole of the carbon atom weighs 12g, the number of atoms in it is equal to 12g / mol 1.99x10 -23 g/atom 6.02 x 10 23 atoms

9. The mole concept for (A) elements, (B) compounds, and (C) molecular substances. A mole contains 6.02 X 10 23 particles. Since every mole contains the same number of particles, the ratio of the mass of any two moles is the same as the ratio of the masses of individual particles making up the two moles.

10. Molar mass Molar mass is defined as the mass of 1 mol of substance in grams is numerically equal to atomic mass. molar mass of water= 18.02g/ mol Molar mass of sodiumm chloride =58.5g/ mol Molar mass in grams is numerically equal to atomic/molecular/formula mass in u.

11. Molecular mass Molecular mass is defined as the sum of the atomic masses of all the elements present in a compound. Example H 2 O= 2x1+1x16= 18 H 3 PO 4 = 3X1+1X31+4X16= 98 Problems: Calculate the molecular mass of glucose and hydrogen peroxide.

12. How to calculate moles? Calculate the number of sodium atoms in 0.120 mol Na? 0.120 mol Na x 6.02 x 10 23 atoms Na = 7.22 x 10 22 1 mol Na

13. Calculate the number of moles of potassium in 1.25 x 10 21 atoms of K. 1.25 x 10 21 atoms K x 1 mol K = 2.08 x 10 -3 mol K 6.02 x 10 23 atoms K

14. Problems What is the mass in grams of 2.01 x 10 22 atoms of sulfur? How many O 2 molecules are present in 0.470g of oxygen gas?

15. Mole calculations Calculate the number of Magnesium and Chlorine ions present in 0.450 mol of MgCl 2 . # Mg 2+ ions: 0.450 mol MgCl 2 x 6.02 x 10 23 formula units x 1 Mg 2+ 1 mole 1 formula unit # Mg 2+ ions = 2.71 x 10 23 # Cl - ions: 0.450 mol MgCl 2 x 6.02 x 10 23 formula units x 2 Cl - 1 mole 1 formula unit # Cl - ions = 5.42 x 10 23

16. Percentage composition Percentage composition is defined as the percentage by mass of each element contributed in the compound. Mass % of an element= Mass of an element in the compound x100 Molar mass of the compound Example: Calculate the percentage of Nitrogen, by mass in Ca (NO 3 ) 2 : Formula Weight = 1 x Ca + 2 x N + 6 x O = 1 x 40.1 + 2 x 14.0 + 6 x 16.0 = 164.1 amu % N = (2)(14.0) = 17.1% 164.1

17. Empirical formula Empirical formula represents the simplest whole number ratio of all the atoms present in a compound What is different from the molecular formula? Molecular formula represents the actual number of different atoms present in a compound. For example glucose molecular formula is C 6 H 12 O 6 but the empirical formula deduced from it after taking its whole number ratio is CH 2 O

18. Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g so 38.67 g C x 1mol C = 3.220 mole C 12.01 gC 16.22 g H x 1mol H = 16.09 mole H 1.01 gH 45.11 g N x 1mol N = 3.219 mole N 14.01 gN The ratio is 3.220 mol C = 1 mol C 3.219 mol N 1 mol N The ratio is 16.09 mol H = 5 mol H 3.219 mol N 1 mol N C 1 H 5 N 1

19. P roblems A compound is 43.64 % P and 56.36 % O. What is the empirical formula? Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? The percent composition of a solid is known to be 68.4% Ba, 10.3% P, and 21.3% O. What is the empirical formula of the compound.

20. Q . What is the percent composition of sodium hydrogen carbonate? The formula for sodium hydrogen carbonate is NaHCO 3 . The Na:H:C:O mole ratio is 1:1:1:3. Convert this mole ratio into a mass ratio by assuming there is a 1 mole sample present. Answer = 23g Na, 1.0g H, 12g C, 48 g O To determine the percent composition, divide the mass of each element present by the total mass of the compound and multiply by 100. Total mass of 1 mole of NaHCO 3 = 84 g Answer = 27% Na, 1.2% H, 14% C, 57% O

21. A compound contains 4.07% hydrogen, 24.27%carbon,71.65%chlorine, its molar mass is 98.9g. What are its empirical formula and molecular formula . Steps involved in the solution Conversion of mass per cent to grams Convert into number of moles of each element. Divide the mole value obtained above by the smallest number Write empirical formula by mentioning the numbers after writing the symbols of respective elements Writing molecular formula

22. STOICHIOMETRY AND STOICHIOMETRIC CALCULATIONS Stoichiometry deals with the calculation of masses of the reactant and the products involved in a chemical reaction. But before calculation we need to know the exact amount of all the masses of reactant and product taking part in any chemical reaction for which we need to balance the chemical equation. For example a balanced equation for this reaction is: CH 4 (g)+2O 2 (g) CO 2 (g)+ 2H 2 O(g) The coefficients 2 for O 2 and H 2 O are called stoichiometric coefficients

23. Calculate the amount of water (g) produced by the combustion of 16g of methane . The balanced equation for combustion of the methane is : CH 4(g) +20 2(g) CO 2(g) +2H 2 O(g) 1. 16 g of CH 4 corresponds to one mole From above equation , 1mol of CH 4 gives 2 mol of H 2 O 2 mol of water H 2 O = 2x(2+16) 2x18= 36 g 1 mol of H 2 O =18 g H 2 0= 18g H 2 = 1 1mol H 2 Hence 2 mol H 2 0X 18g H 2 1 mol H 2 O 2X18 g H 2 O= 36 g H 2 O

24. Limiting reagent The reactants which get consumed and limits the amount of product formed is therefore called the limiting reagent. For example in the reaction N 2 (g)+ 3H 2 (g) 2NH 3 (g) In the reaction dihydrogen is the limiting reagent because the amount of product formed ammonia is mostly depends on the availability of the dihydrogen

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