3. Atoms of different elements have different masses. Since atoms are so small it is very difficult to measure their mass. One way is to bombard a sample of an element’s atoms with electrons. This knocks electrons off these atoms and leaves them with a positive charge.

4. The positive ions are then fired through a magnetic field. The less massive ions are deflected more than the more massive ions. By comparing the amount of deflection a relative mass can be determined.

7. The more massive the _______(more/less) it is deflected. less

8. Any atom’s relative mass can be determined using this method called mass spectrometry. A scale based on the mass of a C-12 atom was arbitrarily established. The standard is defined as 1/12 the mass of a C-12 atom. 1  = 1/12 mass of a C-12 atom The average mass of all isotopes of a given element are listed on the periodic table.

9. This value is based on the relative abundances of all natural isotopes for any given element. For example the element chlorine consists of a mixture of 75.53% Cl-35 and 24.47% Cl-37. Any sample of chlorine from any source will consist of these two isotopes in this proportion. The average mass of a chlorine atom is calculated by (0.7553 X 35  ) + (0.2447 x 37  ) = 35.49  Try pg. 166 # 10,11 in the Nelson Textbook

10. In the real world single atoms are too small to concern us. However if the mass of 1 oxygen atom is 16  and the mass of 1 hydrogen atom is 1  then the mass of 100 atoms of O is 1600 , the mass of 100 atoms of H is 100  Regardless of the number of O and H atoms if an equal number of each is compared the ratio of O to H is always 16 to 1. This can be turned around to get a more valuable statement.

11. If samples of oxygen and hydrogen in a mass ratio of 16 to 1 are taken there are an equal number of atoms in each sample, regardless of the units used. Thus 16 lb of oxygen contains the same number of atoms as 1 lb of hydrogen. If grams (g) are used for atomic masses instead of  the number of atoms involved is called Avogadro’s Number. 12.011 g of C has Avogadro’s number of atoms as does 14 g of N, or 4 g of He, or 19 g of F, or 26.98 g of Al, or 6.94 g of Li.

12. Oxygen atomsHydrogen atoms Equal numbers of oxygen and hydrogen atoms oxygen is 16x’s more massive

13. The quantity of an element having a weight in grams numerically equal to that element’s atomic mass always contains Avogadro’s number of atoms. The value of Avogadro’s number has been determined by crystallographic and electrochemical techniques as 6.023 x 10 23. The magnitude of this number is difficult to imagine. If Avogadro’s number of 1 cm 3 ice cubes were spread out over all of Canada it would cover the entire country to a depth of more than 30 km.

14. Here is Avogadro’s Number 602 300 000 000 000 000 000 000. This quantity is also called a mole

15. The quantity of a substance that contains Avogadro’s number of the units indicated by the formula of the substance is called a mole. (mol) The term can be applied to a variety of items: collections of atoms, ions, electrons, molecules or chemical bonds. A mole of calcium chloride, CaCl 2 for instance, contains 1 mol of Ca, 2 mol of Cl; the mass of 1 mol of CaCl 2 can be calculated as: 40.078 g + (35.4527 g x 2) = 110.983 g

16. Calculate the molar mass of magnesium sulfate 1 mol of MgSO 4 contains 1 mol of Mg 1 mol of S 4 mol of O (1 x 24.305 g + 1 x 32.066 g + 4 x 15.9994 g) = 120.369 g

17. Calculate the molar mass of barium phosphite 1 mol of Ba 3 (PO 3 ) 2 contains 3 mol of Ba 2 mol of P 6 mol of O (3 x 137.33 g + 2 x 30.974 g + 6 x 15.9994 g) = 569.93 g

18. If 1.0 mole of barium phosphite has a mass of 569.93 g how many moles are there in 1139.86 g of Ba 3 (PO 3 ) 2 ? 1139.86 g / 569.9 g/mol = 2.0 mol This approach gives us a simple formula to use when calculating the number of moles of a substance. Number of moles = mass Molar mass n= m M

19. How many moles are there in 40 g of hydrogen fluoride? The formula is HF n= m M n= 40 g 1 g/mol + 19 g/mol 40 g 20 g/mol = =2.0 mol

20. What is the mass of 2.5 mol of carbon dioxide? The formula is CO 2 n= m M To find m get rid of the fraction multiply both sides by the bottom of the fraction, M M M m = n x M m = 2.5 mol x 44 g/mol m = 1.1 x 10 2 g

21. If 4.5 mol of a substance has a mass of 207 g what is this substance’s molar mass? n= m M get rid of the fraction multiply both sides by the bottom of the fraction, M M M m = n x MTo solve for M divide both sides by n nn M = 207 g / 4.5 mol = 46 g/mol