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Chemical Formulas

Chemical Formulas. Main objectives for this chapter:. Empirical and molecular formulas . Structural formulas. Simple examples Calculations of empirical formulas, given the percentage composition by mass.

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Chemical Formulas

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  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. H H H H H H Structural Formulas Gives the arrangement of the atoms in a molecule of the compound The structural formula of ethane: The structural formula of ethene: H H H H

  5. Name Structural Formula Molecular Formula Methane H H C H H Carbon dioxide O O C Water H O H Molecular Formula • Molecular formula tells you the number of atoms of each element present in a molecule of a compound CH4 CO2 H2O

  6. Sodium chloride A 1:1 ratio + + - - - - + + + + - - - + - + + - - + + + - + - + - Formula of ionic compounds • Ionic compounds are giant structures. • There can be any number of ions in an ionic crystal - but always a definite ratio of ions. NaCl MgCl2 AlCl3 Al2O3

  7. Empirical Formula • Gives only the ratios in which different atoms are present in a molecule of a compound For glucose C6H12O6 is the molecular formula The empirical formula is CH2O

  8. H H H H H H Once you know the structural formula you can work out the molecular formula and the empirical formula • The structural formula of ethane is : • The molecular formula is: C2H6 • The empirical formula is : CH3

  9. The structural formula of ethene is: H H • Find the (i) molecular formula (ii) empirical formula H H • Find the (i) molecular formula= C2H4 (ii) empirical formula= CH2

  10. Check your learning • Define • Structural formula • Molecular formula • Emperical formula

  11. Objectives for today • Calculations of empirical formulas: (i) Given the % mass of elements present in the compound (ii) given the masses of reactants and products.HIGHER LEVEL (ii) Calculation of molecular formulas

  12. Calculating Empirical Formulas

  13. Method 1- when given % mass of element • A compound contains 40% sulfur and 60 % oxygen. What is its empirical formula? • Therefore the empirical formula of this compound is SO3.

  14. Q282 • 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

  15. Q284 Finding empirical formula • 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 C4H10O

  16. 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

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

  18. Method 2- when given mass of element in compound 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 Example • 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 3.55g The empirical formula is CuCl2

  19. Method 2- when given mass of element in compound Example • 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 0.96g

  20. Method 2- when given mass of element in compound Q288 • 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 PbI2 2.54

  21. Method 2- when given mass of element in compound Q289 • 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 CuSO4 (H20)5 1.42g

  22. Q290 Method 2- when given mass of element in compound • 9.76g of a metal forms 20.9g of its oxide whose formula is M20. 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 11.14g 1.3925

  23. Calculating Molecular Formulas

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

  25. You need: 1. The empirical formula 2. The relative molecular mass 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 Mass according to EF = 2(14) +16 +12 +4(1) = 60 1. The empirical formula N2OCH4 2. Relative molecular mass of urea = 60 3. Empirical formula = molecular formula Molecular formula N2OCH4

  26. You need: 1. The empirical formula 2. The relative molecular mass 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 C4H10O Mass according to EF = 4(12) +16 +10(1) = 74 1. The empirical formula C4H10 O 2. Relative molecular mass of urea = 74 3. Empirical formula = molecular formula C4H10 O

  27. You need: 1. The empirical formula 2. The relative molecular mass 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 Mass according to EF = 12 +2(16) +1 = 45 1. The empirical formula CO2 H 2. Relative molecular mass of the organic acid = 90 3. Empirical formula x 2 = molecular formula Molecular formula = C2O4 H2

  28. You need: 1. The empirical formula 2. The relative molecular mass 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 Mass according to EF = 4(12) +16 +10(1) = 74 1. The empirical formula C4H10 O 2. Relative molecular mass of urea = 74 3. Empirical formula = molecular formula Molecular formula = C4H10 O

  29. Calculating % compositions by mass

  30. 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

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

  32. What is the percentage by mass of Fe in Fe2O3 ? 2(56) x 100 160 1 70% Mass % of A in compound = Mass of A in compound x 100 Relative molecular mass of the compound 1

  33. 291b)What is the percentage by mass of nitrogen in NH4NO3 ? 2(14) x 100 80 1 35% Mass % of A in compound = Mass of A in compound x 100 Relative molecular mass of the compound 1

  34. 291c) What is the percentage by mass of carbon in methylbenzene ? 7(12) x 100 92 1 91.3043478% Mass % of A in compound = Mass of A in compound x 100 Relative molecular mass of the compound 1

  35. 291(d)What is the percentage by mass of N in NO2 ? 14 x 100 46 1 30.4347826% Mass % of A in compound = Mass of A in compound x 100 Relative molecular mass of the compound 1

  36. 291(e)What is the percentage by mass of water in Na2 CO3. 10H20 ? 10(18) x 100 286 1 62.9370629% Mass % of A in compound = Mass of A in compound x 100 Relative molecular mass of the compound 1

  37. 291(f)What is the percentage by mass of Fe in Fe3 O4.? 3(56) x 100 232 1 72.4137931% Mass % of A in compound = Mass of A in compound x 100 Relative molecular mass of the compound 1

  38. Extra questions not on worksheet

  39. Urea has an empirical formula of CON2H4 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 CON2H4 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 = CON2H4

  40. Glucose has an empirical formula of CH2O and a relative molecular mass of 180. Find its molecular formula. • If the atoms in each molecule actually present were CH2Othen 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 CH2O! • 30 x n = 180 • n = 6 • So the molecular formula = C6H12O6

  41. Q1. Heptane has an empirical formula of C7H16 and a relative molecular mass of 100. Find its molecular formula. • If the atoms in each molecule actually present were C7H16 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 = C7H16

  42. Q2. Butanoic acid has an empirical formula of C2H4O and a relative molecular mass of 88. Find its molecular formula. • If the atoms in each molecule actually present were C2H4O2 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 = C4H8O2

  43. 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.

  44. First you need to find the empirical formula:

  45. First you need to find the empirical formula: Moles present = mass in grams Ar

  46. First you need to find the empirical formula: To get a simple ratio divide each number by the smallest number present!

  47. First you need to find the empirical formula: empirical formula = CH2O

  48. the relative molecular mass of fructose is 180, the empirical formula is CH2O - find its molecular formula. • If the atoms in each molecule actually present were CH2Othen 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 = C6H12O6

  49. What is the percentage composition by mass of carbon present in ethanol ( C2H5OH )? • 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 C2H5OH = 12 +12+ 1+1+1+1+1+16+1 = 46 • Percentage of carbon by mass in ethanol = 24 x 100 = 52.17% 46

  50. What is the percentage composition by mass of nitrogen present in (NH4)2HPO4 ? • (NH4)2HPO4 = N2H9PO4 • 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 N2H9PO4: (14 x 2)+ (1 x 9)+ 40 +(16 x4) = 109g • Percentage of nitrogen by mass in N2H9PO4 28 x 100 = 25.68% 109

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