Empirical and Molecular Formulas

1. Empirical and Molecular Formulas In this presentation you will: • explore the difference between empirical and molecular formulas Next >

2. Introduction There are two different types of formulas for a compound. These are the empirical and molecular formulas. The empirical formula gives the simplest ratio of the number of atoms present in a molecule of a compound. The molecular formula gives the actual number of atoms of each element present in a compound. Next >

3. Empirical Formula Empirical formula = HO Empirical formula = NaCl The empirical formula is often used when dealing with large organic molecules. These may have many hundreds of atoms, or, in ionic or giant covalent lattice structures, millions of atoms, bound together in a fixed ratio. For many simple inorganic substances, the empirical formula is the same as the molecular formula. Next >

4. Example of Empirical and Molecular Formulas Phosphorus(V) oxide exists as P4O10 molecules. This is therefore the molecular formula. This can be simplified to P2O5. This is the empirical formula. Next >

5. Percentage Composition The idea of percentage composition can be used to help us calculate empirical formulas. This is the percentage by mass of a particular element in a chemical compound. The chemical formula of the compound gives the number of molecules of each element in the compound, but not their mass. So, in methane, CH4, there is one molecule of carbon and four molecules of hydrogen. Next >

6. Percentage Composition To calculate the percentage composition: % of element in compound = mass of element in sample of compound × 100 mass of sample of compound However, there is an alternative method: % of element in compound = mass of element in 1 mol of compound × 100 molar mass of compound Using this formula, the percentage composition can be calculated. Next >

7. Percentage Composition Calculate the percentage composition of methane (CH4): % of element in compound = mass of element in 1 mol of compound × 100 molar mass of compound For hydrogen: % of element in compound = mass of hydrogen in 1 mol of CH4× 100 molar mass of CH4 = 4 × 1.01 × 100 = 25.2 to 3 s.f. 16.05 Next >

8. Percentage Composition For carbon: % of element in compound = mass of carbon in 1 mol of CH4× 100 molar mass of CH4 = 12.01 × 100 = 74.83 16.05 Check: 25.2% + 74.83% ≈ 100% Next >

9. Question 1 Find the percentage compositions of PbCl2 given that the relative atomic masses of Pb and Cl are 207.2 and 35.45 respectively. A) Pb: 69.54% Cl: 30.46% B) Pb: 73.99% Cl: 26.01% C) Pb: 74.51% Cl: 25.49% D) Pb: 76.05% Cl: 23.95% Next >

10. Question 1 Find the percentage compositions of PbCl2 given that the relative atomic masses of Pb and Cl are 207.2 and 35.45 respectively. A) Pb: 69.54% Cl: 30.46% B) Pb: 73.99% Cl: 26.01% C) Pb: 74.51% Cl: 25.49% D) Pb: 76.05% Cl: 23.95% Next >

11. Percentage Composition The empirical formula of a compound can be calculated from data which gives the percentage composition, by mass, of each element in the compound. 50 g of unknown substance X 20 g C : 3.3 g H : 26.7 g O = 1.67 C : 3.3 H : 1.67 O The molecular formula can then be deduced if the relative molecular mass is known. = CH2O Next >

12. Calculation of Empirical Formula The empirical formula of a compound can be calculated, with the help of the periodic table, if its mass and the percentages of the components are known. We are given a 100 g sample of a compound containing 40% carbon, 6.7% hydrogen and 53.3% oxygen by mass. The masses of the elements are therefore:C = 40 g, H = 6.7 g and O = 53.3 g. Next >

13. Calculation of Empirical Formula The number of moles, according to the formula mol = mass / Ar is: Carbon: 40 / 12 = 3.3 Hydrogen: 6.7 / 1 = 6.7 Oxygen: 53.3 / 16 = 3.3 To obtain the empirical formula, divide the number of moles of each element by the smallest number and round the results to the nearest integers: 3.3 : 6.7 : 3.3 → 1 : 2 : 1, so the empirical formula is CH2O. Next >

14. Experimental Determination Experimental data can also be used to find the molecular mass of compounds and hence their empirical formula. A useful result of this is the ability to determine the amount of water of crystallization present in a salt sample; this is water bound within the crystal lattice structure of the compound in a solid state. 20.00 g of a sample of iron(II) sulfate, FeSO4·xH2O, is found by analysis to contain 4.03 g of pure iron (Fe). What is the water of crystallization (x) of the sample of iron(II) sulfate? Next >

15. Experimental Determination Mass of anhydrous iron sulfate =mass (Fe) × (Mr (FeSO4)/ Ar (Fe)) Mass of anhydrous iron sulfate =4.03 × (152 / 56) = 10.94 g Mass of water of crystallization =20.00 g – 10.94 g = 9.06 g Moles water = 9.06 / 18 = 0.50 mol; Moles iron = 4.03 / 56 = 0.072 mol Ratio = 0.072 Fe : 0.5 H2O = 1:7 by quantity, so formula is FeSO4·7H2O Next >

16. Question 2 The molar ratio of two elements in a sample of an ionic substance was found to be 2.4 : 4.8. If the two elements in question are Mg and Cl, what is the empirical formula for this magnesium chloride compound? A) MgCl C) MgCl2 B) Mg2Cl D) Mg2Cl4 Next >

17. Question 2 The molar ratio of two elements in a sample of an ionic substance was found to be 2.4 : 4.8. If the two elements in question are Mg and Cl, what is the empirical formula for this magnesium chloride compound? A) MgCl C) MgCl2 B) Mg2Cl D) Mg2Cl4 Next >

18. Question 3 What information does the empirical formula of a compound give? A) The absolute number of atoms in a compound B) The simplest ratio of atoms in a compound C) The number of atoms in one mole of a compound D) The number of moles of a compound in a sample Next >

19. Question 3 What information does the empirical formula of a compound give? A) The absolute number of atoms in a compound B) The simplest ratio of atoms in a compound C) The number of atoms in one mole of a compound D) The number of moles of a compound in a sample Next >

20. Determination of Molecular Formula Once we know the empirical formula, we can go on to calculate the molecular formula. Suppose the empirical formula comes to AB2. We know that the molecular formula must be a whole number multiple of this formula: AB2 or A2B4 or A3B6 and so on. Molecular formula = x(empirical formula) where x is a whole number. Alternatively: Molecular formula mass = x(empirical formula mass) where x is a whole number. Next >

21. Determination of Molecular Formula Molecular formula mass = x(empirical formula mass) where x is a whole number. The empirical formula mass is found by adding together the masses for each of the atoms in the empirical formula. x = molecular formula mass empirical formula mass The molecular formula mass is the molar mass of the compound. x = molar mass   empirical formula mass Dividing the molar mass by the empirical formula mass should give x. Next >

22. Example Determine the molecular formula of the compound with an empirical formula of CH and a formula mass of 78.110 atomic mass units. The empirical formula mass is 12.01 + 1.01 = 13.02. x = 78.11 13.02 x = 6 The molecular formula is C6H6. Next >

23. Question 4 A sample of a compound, with a formula mass of 34.01 atomic mass units, is found to consist of 0.44 g H and 6.92 g O. Find its molecular formula, given the relative atomic mass of hydrogen is 1.01 and of oxygen is 16.00. A) HO B) H2O2 C) H2O D) HO2 Next >

24. Question 4 A sample of a compound, with a formula mass of 34.01 atomic mass units, is found to consist of 0.44 g H and 6.92 g O. Find its molecular formula, given the relative atomic mass of hydrogen is 1.01 and of oxygen is 16.00. A) HO H + O = 17.01 34.01/17.01= 2 ignoring rounding up errors B) H2O2 C) H2O D) HO2 Next >

25. Summary In this presentation you have seen: • the difference between empirical and molecular formulas End >