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Mass Numbers & Masses of atoms and moles Cont.

Mass Numbers & Masses of atoms and moles Cont. Learning Outcomes 1) Masses of atoms and molecules of compounds can be compared using atomic mass. 2) A mole of an element or a compound is the relative formula mass expressed in grams. The mole.

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Mass Numbers & Masses of atoms and moles Cont.

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  1. Mass Numbers & Masses of atoms and moles Cont. Learning Outcomes 1) Masses of atoms and molecules of compounds can be compared using atomic mass. 2) A mole of an element or a compound is the relative formula mass expressed in grams.

  2. The mole The relative atomic or formula mass in grams is one mole

  3. How many atoms in a mole? In one mole of a substance there is always 602000000000000000000000 atoms, ions or molecules. This is called Avogadro’s number. Don’t worry you don’t need to learn it, just remember there is always the same number of ‘bits’ if you have equal moles

  4. Mole calculations Number of moles = mass/ Ar Number of moles = mass/ RAM Try the gdoc Moles 2 and see how far you can get

  5. Mole calculations Moles Number of moles = mass/ Ar Number of moles = mass/ RAM Mass Atomic or molecular mass

  6. Number of Moles - Elements For an element: • Number of moles = mass . • relative atomic mass Example: How many moles of sodium atoms are in 69g of sodium? Sodium -Ar= 23 • Number of moles = mass . • relative atomic mass • Number of moles = 69 . • 23 Number of moles = 3 moles of sodium atoms in 69g of Na

  7. Number of Moles - Elements For an element: • Number of moles = mass . • relative atomic mass Example: What mass would 4 moles of sodium atoms have? Sodium -Ar= 23 mass = number of moles x relative atomic mass mass = 4 x 23 mass = 92g

  8. Number of Moles - Elements For an element: • Number of moles = mass . • relative atomic mass Example: If 7 moles of sodium atoms have a mass of 161g. What is the relative atomic mass of sodium? • Relative atomic mass = mass . • number of moles • Relative atomic mass = 161 . • 7 Relative atomic mass = 23

  9. Number of Moles - Compounds For a compound: • Number of moles = mass . • relative formula mass Example: How many moles of water molecules are in 90g of water? Water -Ar= 18 • Number of moles = mass . • relative formula mass • Number of moles = 90 . • 18 Number of moles = 5 moles of water molecules in 90g of H2O

  10. Number of Moles - Compounds For a compound: • Number of moles = mass . • relative formula mass Example: What mass would 6 moles of water molecules have? Water -Ar= 18 mass = number of moles x relative formula mass mass = 6 x 18 mass = 108g

  11. Number of Moles - Compounds For a compound: • Number of moles = mass . • relative formula mass Example: If 10 moles of water molecules have a mass of 180g. What is the relative formula mass of water? • Relative formula mass = mass . • number of moles • Relative formula mass = 180 . • 10 Relative formula mass = 18

  12. % Formulae Learning Outcomes • We can calculate the % of mass of an element in a compound.

  13. Percentage Composition • It is sometimes useful to know how much of a compound is made up of some particular element. • This is called the percentage composition by mass. % Z = (Number of atoms of Z) x (atomic Mass of Z) x 100% Formula Mass of the compound E.g. % of oxygen in carbon dioxide (Atomic Masses: C=12. O=16) Formula = Number oxygen atoms = Atomic Mass of O = 16 Formula Mass CO2 = % oxygen = CO2 2 12 +(2x16)=44 (2 x 16 / 44) x 100% = 72.7%

  14. Activity – copy and complete % Z = (Number of atoms of Z) x (atomic Mass of Z) x100% Formula Mass of the compound • Calculate the percentage of oxygen in the compounds shown below 16 24+16=40 (1x16/40)x100%=40% (2x39)+16 =94 16 (1x16/94)x100%=17% 23+16+1=40 16 (1x16/40)x100%=40% 32+(2x16)=64 16 (2x16/64)x100%=50%

  15. Formulae • When a new compound is discovered we have to deduce its formula. • This always involves getting data about the masses of elements that are combined together. • What we have to do is work back from this data to calculate the number of atoms of each element and then calculate the ratio. • We call this simplest version of the formula the empirical formula of a compound. • In order to do this we find the number of moles of each element (divide the mass of each atom by its atomic mass). • The calculation is best done in 5 stages:

  16. For example: • We found 3.2g of copper reacted with 0.8g of oxygen. What is the formula of the oxide of copper that was formed? (Atomic Mass, Ar Cu=64: O=16) 0.8 3.2 0.8/16 =0.05 3.2/64 =0.05 1:1 CuO

  17. For example: • We found 5.5g of manganese reacted with 3.2g of oxygen. What is the formula of the oxide of manganese formed? (Atomic Mass, Ar Mn=55: O=16) 3.2 5.5 3.2/16 =0.20 5.5/55 =0.10 1:2 MnO2

  18. For example: • A chloride of silicon was found to have the following % composition by mass: Silicon 16.5%: Chlorine 83.5% (Atomic Mass, Ar Si=28: Cl=35.5) 83.5 16.5 83.5/35.5 =2.35 16.5/28 =0.59 Cl÷Si = (2.35 ÷ 0.59) = (3.98) Ratio of Cl:Si =4:1 Divide biggest by smallest SiCl4

  19. Activity • Calculate the formula of the compounds formed when the following masses of elements react completely ( use the 5 stage method): (Atomic Mass, Ar Si=28: Cl=35.5) FeCl3 KBr PCl5 CH4 MgO

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