1 / 22

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.

monty
Télécharger la présentation

THE MOLE

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related