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Percentage Composition by mass and empirical formula.

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Presentation on theme: "Percentage Composition by mass and empirical formula."— Presentation transcript:

1 Percentage Composition by mass and empirical formula

2 % composition by mass = ∑ Ar (atoms/ions) % composition by mass = ∑ Ar (atoms/ions)Mr To find the % of copper in copper oxide, CuO RFM of CuO = 64 + 16 = 80 Of this 64 is Cu % of copper = 64/80 = 80%

3 Example 2 Example 2 To find the % composition in ammonium nitrate,NH 4 NO 3 RFM of NH 4 NO 3 = 14+4+14+(16*3) = 80 Of this (2*14) is nitrogen % of N = 28/80 = 35%

4 Find the percentage composition of C in butanol, C 4 H 9 OH Find the percentage composition of C in butanol, C 4 H 9 OH

5 And another one – And another one – Find the percentage composition of Sulphur in anhydrous copper sulphate, CuSO 4

6 Last one (a tough problem) Last one (a tough problem) Calculate the percentage composition of water in hydrated magnesium sulphate, MgSO 4.7H 2 O

7 Empirical formula Empirical formula – simplest ratio of atoms of each element in a compound Empirical formula – simplest ratio of atoms of each element in a compound Molecular formula – actual number of atoms Molecular formula – actual number of atoms Ionic compounds – always use the empirical formula since they consist of giant ionic lattices in which ions are present in a fixed ratio Ionic compounds – always use the empirical formula since they consist of giant ionic lattices in which ions are present in a fixed ratio Simple covalent – use molecular formula Simple covalent – use molecular formula

8 Lets take a simple formula, CuO Lets take a simple formula, CuO This means – 1 atom of copper reacts with 1 atom of oxygen It is better to read this as – 1mole of Copper atoms reacts with 1 mole of oxygen atoms Even better – 64g of Cu reacts with 16g of oxygen

9 Example 1 Example 1 After completing a practical, it has been found that 2.4g of Mg has reacted with 1.6g of oxygen. MgO Combining masses2.4g1.6g No of moles of atoms2.4/241.6/16 = 0.1= 0.1 Ratio of moles 11 Simplest formulaMgO

10 Example 2 Example 2 A compound containing 4.6g Na, 2.8g N, 9.6g O. Find the empirical formula. Na N O Combining masses g4.6 2.8 9.6 No of moles of atoms4.6/23 2.8/14 9.6/16 = 0.2 = 0.2 = 0.6 Ratio of moles 1 1 3 Simplest formulaNaNO 3

11 Questions 1. Find the empirical formula of a compound containing 85.7% C, 14.3% H by mass? (These percentage figures apply to any masses, so chose 100g. That is – 85.7g C, 14.3g H)

12 Now read pages 64 – 65 of your AS text book and do questions 1 – 5 on these pages. Now read pages 64 – 65 of your AS text book and do questions 1 – 5 on these pages. The answers are on pages 250 – 251. Check your answers with these but no cheating!!!!! The answers are on pages 250 – 251. Check your answers with these but no cheating!!!!!


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