1. Chemical Formulas

2. Main objectives for this chapter: Empirical and molecular formulas. Structural formulas. Simple examples Calculations of empirical formulas, given the percentage composition by mass. Calculation of empirical formulas, given the masses of reactants and products.HIGHER LEVEL only Calculation of molecular formulas, given the empirical formulas and the relative molecular masses (examples should include simple biological substances, such as glucose and urea). Percentage composition by mass. Calculations

3. Structural formulas, molecular formulas, empirical formulas

4. Structural Formulas Def: The structural formula of a compound tells you the arrangement of the atoms in a molecule of the compound The structural formula of ethane: HH HH HH The structural formula of ethene: H HH H

5. Molecular Formula Water Carbon dioxide Methane Molecular FormulaStructural FormulaName C H H H H C O O H H O Molecular formula tells you the number of atoms of each element present in a molecule of a compound CH 4 CO 2 H2OH2O

6. Empirical Formula An empirical formula shows only the ratios in which different atoms are present in a molecule of a compound For glucose C 6 H 12 O 6 is the molecular formula The empirical formula is CH 2 O

7. The structural formula of ethane is : The molecular formula is: C 2 H 6 The empirical formula is : CH 3 HH HH HH Once you know the structural formula you can work out the molecular formula and the empirical formula

8. Formula of ionic compounds Ionic compounds are giant structures. butThere can be any number of ions in an ionic crystal - but always a definite ratio of ions. NameRatioFormula Sodium chloride1:1 Magnesium chloride1:2 Aluminium chloride1:3 Aluminium Oxide2:3 + - + - + - - ++ + - + - -- - + + + - + - + - - ++ Sodium chloride A 1:1 ratio NaCl AlCl 3 Al 2 O 3 MgCl 2

9. The structural formula of ethene is: Find the (i) molecular formula (ii) empirical formula H HH H Find the (i) molecular formula= C 2 H 4 (ii) empirical formula= CH 2

10. Calculating Empirical Formulas

11. A compound contains 40% sulfur and 60 % oxygen. What is its empirical formula? Therefore the empirical formula of this compound is SO 3. Element%molesRatio sulfur4040/ 32 = 1.251 oxygen6060/ 16 = 3.753 Method 1- when give & mass of element

12. A compound contains 48.8% carbon and 13.5% hydrogen and 37.7% nitrogen respectively by mass. Determine the empirical formula of the compound. Therefore the empirical formula of this compound is Element%/ ArRatio Carbon48.848.8/ 12 = 4.066666667 1.51 Hydrogen13.513.5 / 1= 13.5 5.04 Nitrogen37.737,5/14= 2.678571429 1 Q282

13. A compound contains 64.9% carbon and 13.5% hydrogen and 21.6 % oxygen respectively by mass. Determine the empirical formula of the compound. Therefore the empirical formula of this compound is C 4 H 10 O Element%/ ArRatio Carbon64.964.9/ 12 = 5.4083 4.006 Hydrogen13.513.5 / 1= 13.5 10 Oxygen21.621.6 /16= 1.35 1 Q284 Finding empirical formula

14. Conservation of Mass in a reaction During chemical reactions the same atoms are present before and after reaction. They have just joined up in different ways. Because of this the total mass of reactants is always equal to the total mass of products. (Law of Conservation of Mass) Reaction but no mass change

15. Mg HCl Gas given off. Mass of chemicals in flask decreases 11.71 Same reaction in sealed container: No change in mass Conservation of Mass

16. You must know the masses of all of the elements in the compound. You might have to work this out... REMEMBER SUM OF MASS OF REACTANTS = SUM OF MASS OF PRODUCTS When 3.175g of copper reacts with chlorine gas 6.725g of copper chloride is formed. Find the empirical formula of the copper chloride Copper + Chlorine gas Copper Chloride 3.175g ? 6.725g Example 3.55g ElementmassmolesRatio Copper3.175g0.051 Chlorine3.55g0.12 Method 2- when given mass of element in compound The empirical formula is CuCl 2

17. When`1.44g of Magnesium was completely burned in oxygen it resulted in the formation of 2.40g of magnesium oxide. Find the empirical formula of magnesium oxide Magnesium + Oxygen Magnesium oxide 1.44g ? 2.40g The empirical formula is MgO Example 0.96g ElementmassmolesRatio Magnesium1.44g0.061 Oxygen0.96g0.061 Method 2- when given mass of element in compound

18. When`2.07g of lead reacts with iodine, 4.61g of lead iodide was formed. Find the empirical formula of lead iodide Lead + Iodine Lead iodide 2.07g ? 4.61g The empirical formula is PbI 2 Q288 2.54 ElementmassmolesRatio Lead2.07g0.011 Iodine2.540.022 Method 2- when given mass of element in compound

19. When`3.94g of hydrated copper (II) sulfate was heated, 2.52g of anhydrous salt remained. Calculate the formula of the hydrated salt. Hydrated copper(II) sulfate Anhydrous copper sulphate + water 3.94g 2.52g The empirical formula is CuSO 4 (H 2 0) 5 Q289 Method 2- when given mass of element in compound 1.42g GroupmassmolesRatio Water1.420.0788888885 Copper sulfate2.520.0157993731

20. 9.76g of a metal forms 20.9g of its oxide whose formula is M 2 0. Calculate the relative molecular mass of the metal. Metal + Oxygen Metal oxide 9.76g 20.90g (1.3925/ 9.76= 7.008976661 Mass of one mole of the metal is 7.009g. This is the relative molecular mass Q290 Method 2- when given mass of element in compound 11.14g GroupmassmolesRatio Metal9.76?2 oxygen11.140.696251 1.3925

21. Calculating Molecular Formulas

22. Calculation of the moleuclar formula: You need: 1. The empirical formula 2. The molecular mass (can work out using the periodic table)

23. Calculating molecular mass Urea is used as a fertiliser and an animal feed. It has a relative molecular mass of 60 and is composed of 46.66% N, 26.66%O, 20%C and 6.66% H. Determine the molecular formula of urea You need: 1. The empirical formula 2. The relative molecular mass 1. The empirical formula Element%molesratio nitrogen46.663.332 oxygen26.661.661 Carbon201.661 Hydrogen6.66 4 N 2 OCH 4 2. Relative molecular mass of urea = 60 3. Empirical formula = molecular formula Mass according to EF = 2(14) +16 +12 +4(1) = 60 Molecular formula N 2 OCH 4

24. 284. Error – see 286 An alcohol was found on analysis to contain 64.9% carbon, 13.5% hydrogen and 21.6% oxygen. If the relative molecular mass of the alcohol is 74 show that the molecular formula is C 4 H 10 O You need: 1. The empirical formula 2. The relative molecular mass 1. The empirical formula Element%molesratio Carbon64.95.4083333334 oxygen21.61.351 Hydrogen13.5 10 C 4 H 10 O 2. Relative molecular mass of urea = 74 3. Empirical formula = molecular formula Mass according to EF = 4(12) +16 +10(1) = 74 C 4 H 10 O

25. 285. Calculating molecular mass An organic acid contain 27.6% carbon, 2.2% hydrogen and 71.1% oxygen. If the relative molecular mass of the alcohol is 90. Determine the molecular formula You need: 1. The empirical formula 2. The relative molecular mass 1. The empirical formula Element%molesratio Carbon27.62.31 oxygen71.14.443752 Hydrogen2.2 1 CO 2 H 2. Relative molecular mass of the organic acid = 90 3. Empirical formula x 2 = molecular formula Mass according to EF = 12 +2(16) +1 = 45 Molecular formula = C 2 O 4 H 2

26. 286. Calculating molecular mass Determine the molecualr formula of a compound whose composition is carbon 64.8%, hydrogen 13.6%and oxygen 21.6% and whose relative molecular mass is 74 You need: 1. The empirical formula 2. The relative molecular mass 1. The empirical formula Element%molesratio Carbon64.85.404 oxygen21.61.351 Hydrogen13.513.610 C 4 H 10 O 2. Relative molecular mass of urea = 74 3. Empirical formula = molecular formula Mass according to EF = 4(12) +16 +10(1) = 74 Molecular formula = C 4 H 10 O

27. Calculating % compositions by mass

28. By the end of today’s class you should be able to: (iii)Calculate the Percentage composition by mass of an element in a compound

29. Percentage composition by mass Mass % of A in compound = Mass of A in compoundx 100 Relative molecular mass of the compound 1

30. What is the percentage by mass of Fe in Fe 2 O 3 ? 2(56) x 100 1601 70% Mass % of A in compound = Mass of A in compoundx 100 Relative molecular mass of the compound 1

31. 291b)What is the percentage by mass of nitrogen in NH 4 NO 3 ? 2(14) x 100 801 35% Mass % of A in compound = Mass of A in compoundx 100 Relative molecular mass of the compound 1

32. 291c) What is the percentage by mass of carbon in methylbenzene ? 7(12) x 100 921 91.3043478% Mass % of A in compound = Mass of A in compoundx 100 Relative molecular mass of the compound 1

33. 291(d)What is the percentage by mass of N in NO 2 ? 14 x 100 46 1 30.4347826% Mass % of A in compound = Mass of A in compoundx 100 Relative molecular mass of the compound 1

34. 291(e)What is the percentage by mass of water in Na 2 CO 3. 10H 2 0 ? 10(18) x 100 286 1 62.9370629% Mass % of A in compound = Mass of A in compoundx 100 Relative molecular mass of the compound 1

35. 291(f)What is the percentage by mass of Fe in Fe 3 O 4. ? 3(56) x 100 232 1 72.4137931% Mass % of A in compound = Mass of A in compoundx 100 Relative molecular mass of the compound 1

36. Extra questions not on worksheet

37. Urea has an empirical formula of CON 2 H 4 and a relative molecular mass of 60. Find its molecular formula. The empirical formula of urea is a simple whole number ratio of the atoms from each element that are present. If the atoms in each molecule actually present were CON 2 H 4 then the molecular mass of urea would be: 12 + 16+ 14 +14+ 1+1+1+1 = 60 We are told the molecular mass of urea is 60. molecular formula = empirical formula! Molecular formula = CON 2 H 4

38. Glucose has an empirical formula of CH 2 O and a relative molecular mass of 180. Find its molecular formula. If the atoms in each molecule actually present were CH2O then the molecular mass of glucose would be: 12+ 1+1+16= 30 But we are told the molecular mass of glucose is 180. So the molecular formula is not CH 2 O ! 30 x n = 180 n = 6 So the molecular formula = C 6 H 12 O 6

39. If the atoms in each molecule actually present were C 7 H 16 then the molecular mass of heptane would be: (12 X 7) + (1 X 16)= 100 We are told the molecular mass of Heptane is 100. Molecular formula = empirical formula Molecular formula of Heptane = C 7 H 16 Q1. Heptane has an empirical formula of C 7 H 16 and a relative molecular mass of 100. Find its molecular formula.

40. If the atoms in each molecule actually present were C 2 H 4 O 2 then the molecular mass of Butanoic acid would be: 12 + 12 + 1+1+ 1+1 + 16 = 44 We are told the molecular mass of Butanoic acid is 88. Molecular formula x 2 = empirical formula Molecular formula of Butanoic acid = C 4 H 8 O 2 Q2. Butanoic acid has an empirical formula of C2H4O and a relative molecular mass of 88. Find its molecular formula.

41. Q3. Fructose has the following composition by mass – 40% carbon, 6.66% hydrogen, 53.33% oxygen. If the relative molecular mass of fructose is 180 find its molecular formula.

42. Element present Mass present in 100g of fructose Moles present in 100g of fructose Molar ratio Simplest ratio Simplest whole number ratio Carbon40g Hydrogen6.66g Oxygen53.33g First you need to find the empirical formula:

43. Element present Mass present in 100g of fructose Moles present in 100g of fructose Molar ratio Simplest ratio Simplest whole number ratio Carbon40g40 = 3.33 12 Hydrogen6.66g6.66 = 6.67 1 Oxygen53.33g53.33 = 3.33 16 First you need to find the empirical formula: Moles present = mass in grams Ar

44. Element present Mass present in 100g of fructose Moles present in 100g of fructose Molar ratio Simplest ratio Simplest whole number ratio Carbon40g40 = 3.33 12 3.33 =1 3.33 Hydrogen6.66g6.66 = 6.67 1 6.67 =2 3.33 Oxygen53.33g53.33 = 3.33 16 3.33 =1 3.33 First you need to find the empirical formula: To get a simple ratio divide each number by the smallest number present!

45. Element present Mass present in 100g of fructose Moles present in 100g of fructose Molar ratio Simplest ratio Simplest whole number ratio Carbon40g40 = 3.33 12 3.33 =1 3.33 11 Hydrogen6.66g6.66 = 6.67 1 6.67 =2 3.33 22 Oxygen53.33g53.33 = 3.33 16 3.33 =1 3.33 11 First you need to find the empirical formula: empirical formula = CH 2 O

46. the relative molecular mass of fructose is 180, the empirical formula is CH 2 O - find its molecular formula. If the atoms in each molecule actually present were CH 2 O then the molecular mass of fructose would be: 12 + 1+1+ 16 = 30 We are told the molecular mass of Fructose is 180. Molecular formula x 6 = empirical formula Molecular formula of Fructose = C 6 H 12 O 6

47. What is the percentage composition by mass of carbon present in ethanol ( C 2 H 5 OH )? Moles of carbon present : 2 Mass of carbon present : Ar x number of moles = mass present in grams 12 x 2 = 24g Total mass of C 2 H 5 OH = 12 +12+ 1+1+1+1+1+16+1 = 46 Percentage of carbon by mass in ethanol = 24 x 100 = 52.17% 46

48. What is the percentage composition by mass of nitrogen present in (NH 4 ) 2 HPO 4 ? (NH 4 ) 2 HPO 4 = N 2 H 9 PO 4 Moles of nitrogen present : 2 Mass of nitrogen present : Ar x number of moles = mass present in grams 14 x 2 = 28g Total mass of N 2 H 9 PO 4: (14 x 2)+ (1 x 9)+ 40 +(16 x4) = 109g Percentage of nitrogen by mass in N 2 H 9 PO 4 28 x 100 = 25.68% 109

49. What is the percentage composition by mass of sodium in sodium hydroxide NaOH? Moles of sodium present : 1 Mass of sodium present : Ar x number of moles = mass present in grams 23 x 1 = 23g Total mass of NaOH = 23 + 16 +1 = 40 Percentage of sodium by mass in sodium hydroxide= 23 x 100 = 57.5% 40