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THE MOLE

THE MOLE. How many particles are there really? 6.02 x 10 23 "Happy Mole Day to You" Chemistry Song. Mass of a given amount of substance (g). Amount of substance (mol). m. n =. M. Molar Mass (g mol -1 ). The Rule. Worked Example.

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THE MOLE

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  1. THE MOLE How many particles are there really? 6.02 x 1023 "Happy Mole Day to You" Chemistry Song

  2. Mass of a given amount of substance (g) Amount of substance (mol) m n = M Molar Mass (g mol -1) The Rule

  3. Worked Example • Calculate the mass of 0.35 mol of magnesium nitrate (Mg(NO3)2. • What is the formula we use????? • How do we work out M

  4. Solution • m = n x M • (Mg(NO3)2 = 24.3 + 2(14.0 + 3 x 16.0) = 24.3 + 2 x 62.0 = 148.3 g mol-1 So: m(Mg(NO3)2 = n(Mg(NO3)2 x M(Mg(NO3)2 = 0.35 mol x 148.3 g mol-1 = 52 g Why only two significant figures?

  5. Counting by Weighing • If we have a known mass of a substance can we possibly know the number of particles present? • We can by combining both the equations we have learnt so far. • m = n x M and N = n x NA • How do we do this??

  6. Worked Example • a) Calculate the amount (in mol) of CO2 molecules present in 22 g of carbon dioxide. • b) What is the number of molecules present in this mass of CO2.

  7. m n = M m(CO2) n(CO2) = M(CO2) 22 g n(CO2) = 44.0 g mol-1 Solution • Since M(CO2) = 12.0 + 2 x 16.0 = 44.0 g mol-1 = 0.50 mol

  8. Solution Part b) b) Since N = n x NA, where N is the number of specified particles: N(CO2) = n(CO2) x NA = 0.50 x 6.02 x 1023 = 3.0 x 1023 So, 22 g of carbon dioxide contains 3.0 x 1023 ?? Atoms, ions or molecules? Molecules

  9. Percentage Composition • The values for molar masses of elements in compounds can be used to calculate the percentage composition of a compound once its formula is known. • This type of calculation is important in chemistry. • If for example, a company is producing aluminium from alumina (Al2O3), the management will want to know the mass of aluminium that can be extracted given a quantity of alumina.

  10. Mass of element in 1 mole of compound % by mass of the element x 100 = Mass of 1 mole of compound The Rule • Percentage Composition

  11. Worked Example • Calculate the percentage of aluminium in alumina (Al2O3). • Solution: Step 1. Find the molar mass of Al2O3 M(Al2O3) = (2 x 27) + (3 x 16) = 102 g mol-1

  12. Mass of Al in 1 mole of Al2O3 54 x 100 = x 100 = % of Al in Al2O3 102 Mass of 1 mole of Al2O3 Solution • Step 2: Find the percentage of aluminium in the alumina. Since 1 mol Al2O3 contains 2 mol Al atoms: Mass of aluminium in 1 mol (102 g) of Al2O3 = 2 x 27 g = 54 g = 52.9% The company management, therefore, knows that aluminium comprises about 53% by mass of any sample of alumina.

  13. Empirical formulas • The empirical formula of a compound is the formula that gives the simplest whole number ratio, by number of moles, of each element in the compound. • You may have wondered how chemists have determined the formulas of compounds such as water, carbon dioxide, nitrogen dioxide and other compounds.

  14. Empirical Formulas

  15. Empirical Formulas • Empirical formulas are determined experimentally, usually by determining the mass of each element present in a given mass of compound. • To determine the empirical formula of a compound, therefore, an experimentally determined ratio by mass must be converted to a ratio by numbers of atoms. • This is done by calculating the amount (in mol) of each element

  16. Empirical Formulas – the process Step 1: Measure the mass (m) of each element in the compound Step 2: Calculate the amount in mole (n) of each element in the compound Step 3: Calculate the simplest whole number ratio of moles of each element in the compound Step 4: Determine the empirical formula of the compound

  17. Worked example • A compound of carbon and oxygen is found to contain 27.3% carbon and 72.7% oxygen by mass. Calculate the empirical formula of the compound.

  18. 27.3 2.27 72.7 4.54 = 2.27 = 1 = 4.54 = 2 12.0 2.27 16.0 2.27 Solution 27.3 72.7 1 2 The empirical formula of this compound is therefore CO2

  19. Molecular Formula • While an empirical formula gives the simplest whole number ratio of each element in a compound, a molecular formula gives the actual number of atoms in one molecule of the compound. • Note that the empirical formula for a compound can be the same or different from its molecular formula

  20. Molecular Formula The molecular formula is always a whole number multiple of the empirical formula. A molecular formula can be obtained from the empirical formula if the molar mass of a compound is known.

  21. molar mass of compound 78 g mol-1 = = molar mass of one unit 13 g mol-1 Worked example • A compound has the empirical formula CH. The molar mass of this compound is 78 g mol-1. What is the molecular formula of this compound? • Solution The molecule must contain a whole number of (CH) units. The molar mass of a (CH) unit is 13 g mol-1. If the compound has a molar mass of 78 g mol-1, then: The number of CH units in a molecule = 6 The molecular formula of the compound is, therefore, 6 x CH Ie C6H6

  22. End of Outcome 1 • http://www.schooltube.com/video/2370/The-Mole-Worksheet

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