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06 Empirical Formulae

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Anthony Hardwicke
Anthony Hardwicke

Worksheet for practicing working out empirical formulae. Pupils will need a data sheet or a list of relative atomic masses to be able to complete the questions.

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6. Empirical Formulae What is an Empirical Formula? You already know that the molecular formula of a molecule represents the number of each kind of atom in the molecule. So, for example, the molecular formula of ethene is C2H4, because there are two carbon atoms and four hydrogen atoms in the molecule. But the empirical formula of a substance is defined as the simplest ratio of atoms in a compound, so the empirical formula of ethene is just CH2. 1. Write down the molecular and empirical formulae of the following molecules: molecular formula empirical formula There are not very many cases of molecules in which the molecular formula and the empirical formula are different. There is one in the table above. 2. Can you think of any more examples? (Clue: think of some hydrocarbons) Example One What is the empirical formula of a compound containing 0.15 mol of zinc and 0.30 mol of chlorine? Zn Cl moles 0.15 mol 0.30 mol ratio of moles 1 : 2 empirical formula ZnCl2 3. Write down the empirical formula of the compounds obtained when: a) 0.2 mol of carbon and 0.8 mol of hydrogen are combined b) 2 mol of oxygen and 2 mol of hydrogen are combined c) 0.65 mol of oxygen and 1.30 mol of hydrogen are combined d) 4.4 mol of barium and 8.8 mol of chlorine are combined e) 1.8 mol of aluminium and 2.7 mol of oxygen are combined Quantitative Analysis You are probably wondering what empirical formulae are for. They are interesting from a historical point of view. From 1860, when a list of relative atomic masses was agreed upon, to about 1960, when NMR spectroscopy became widely available, chemists used empirical formulae calculations to work out the formula of unknown substances. Modern chemists still
need to know how to do empirical formulae calculations, even if there are easier ways to determine formulae experimentally. The conventional Example 2 Thomas Edison weighed out 3.84g of copper and found that it way to set out empirical reacted with 0.48g of oxygen. What was the empirical formula formulae is as a of the oxide of copper that he was investigating? Notice that we have divided by 16, which table. The table has a column for is the RAM of oxygen atoms (not the RMM each element and we write the Cu O of O2 molecules). This is because we symbol and then the mass of each mass 3.83g 0.48g need to find out the ratio of copper one. moles 3.83 0.48 = 0.0598 mol = 0.03 mol atoms to oxygen We then divide 64 16 atoms, in order to obtain the empirical the mass of each elements by the ratio of moles 2 : 1 formula. element¡¯s RAM. We are essentially empirical formula Cu2O Finally we have to write the two amounts as a simple ratio. If you can¡¯t ¡®see¡¯ the simple ratio working out the number of moles straight away, choose the smallest of the two of each element, Don¡¯t be put off if the ratio isn¡¯t exact. It amounts and divide both amounts by this because moles = is OK to round up or down, if your ratio smallest amount. This doesn¡¯t always get you mass / RAM. is close to a whole number. straight to the answer, but it often helps. 4. Find the empirical formulae of the d) 0.60g of magnesium was heated with compounds formed when the following sulfur until further reaction occurred. quantities react: The mass of magnesium sulfide formed a) 4.32g of silver and 1.42g of chlorine was 1.40g. b) 2.5g of calcium and 1.5g of carbon e) 10.0g of carbon was reacted with c) 0.3g of carbon and 0.8g of oxygen hydrogen to give a compound that d) 0.7g of iron and 3.0g of bromine weighed 12.5g. e) 1.6g of calcium and 2.84g of chlorine f) 2.07g of sodium burnt completely in f) 3.2g of oxygen and 5.5g of manganese oxygen to give 2.79g of sodium oxide. g) 2.54g of iodine and 0.8g of oxygen g) 44.0g of manganese reacted completely h) 2.1g of silicon and 2.66g of chlorine to give 88.8g of manganese oxide. i) 5.0g of arsenic and 0.2g of hydrogen h) After reacting 1.5g of aluminium with j) 4.00g of magnesium and 1.56g of fluorine gas, 4.67g of aluminium fluoride nitrogen was produced. i) 5.00g of chromium underwent complete 5. Calculate the empirical formulae of the combustion in oxygen to give an oxide compounds formed in the following that weighed 7.31g experiments: j) 1.54g of an oxide of nitrogen was found a) 3.60g of magnesium was heated with to contain 0.40g of nitrogen. bromine until further reaction occurred. The mass of magnesium Answers 1. NH3, NH3, C3H8, C3H8, C2H6O, C2H6O, C4H10, C2H5 bromide formed was 27.60g. b) 8.00g of a gaseous hydrocarbon was 3. a) CH4, b) HO, c) H2O, d) BaCl2, e) Al2O3 reduced to give 6.00g of carbon. 4. a) AgCl, b) CaC2, c) CO2, d) FeBr3, e) CaCl2, f) MnO2, g) c) When 15.61g of an oxide of lead was I2O5, h) SiCl4, i) AsH3, j) Mg2N3 completely reduced, 14.49g of lead was produced. 5. a) MgBr2, b) CH4, c) PbO, d) MgS, e) CH3, f) Na2O, g) Mn2O7, h) AlF3, i) Cr2O3, j) N2O5

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06 Empirical Formulae

  • 1. 6. Empirical Formulae What is an Empirical Formula? You already know that the molecular formula of a molecule represents the number of each kind of atom in the molecule. So, for example, the molecular formula of ethene is C2H4, because there are two carbon atoms and four hydrogen atoms in the molecule. But the empirical formula of a substance is defined as the simplest ratio of atoms in a compound, so the empirical formula of ethene is just CH2. 1. Write down the molecular and empirical formulae of the following molecules: molecular formula empirical formula There are not very many cases of molecules in which the molecular formula and the empirical formula are different. There is one in the table above. 2. Can you think of any more examples? (Clue: think of some hydrocarbons) Example One What is the empirical formula of a compound containing 0.15 mol of zinc and 0.30 mol of chlorine? Zn Cl moles 0.15 mol 0.30 mol ratio of moles 1 : 2 empirical formula ZnCl2 3. Write down the empirical formula of the compounds obtained when: a) 0.2 mol of carbon and 0.8 mol of hydrogen are combined b) 2 mol of oxygen and 2 mol of hydrogen are combined c) 0.65 mol of oxygen and 1.30 mol of hydrogen are combined d) 4.4 mol of barium and 8.8 mol of chlorine are combined e) 1.8 mol of aluminium and 2.7 mol of oxygen are combined Quantitative Analysis You are probably wondering what empirical formulae are for. They are interesting from a historical point of view. From 1860, when a list of relative atomic masses was agreed upon, to about 1960, when NMR spectroscopy became widely available, chemists used empirical formulae calculations to work out the formula of unknown substances. Modern chemists still
  • 2. need to know how to do empirical formulae calculations, even if there are easier ways to determine formulae experimentally. The conventional Example 2 Thomas Edison weighed out 3.84g of copper and found that it way to set out empirical reacted with 0.48g of oxygen. What was the empirical formula formulae is as a of the oxide of copper that he was investigating? Notice that we have divided by 16, which table. The table has a column for is the RAM of oxygen atoms (not the RMM each element and we write the Cu O of O2 molecules). This is because we symbol and then the mass of each mass 3.83g 0.48g need to find out the ratio of copper one. moles 3.83 0.48 = 0.0598 mol = 0.03 mol atoms to oxygen We then divide 64 16 atoms, in order to obtain the empirical the mass of each elements by the ratio of moles 2 : 1 formula. element¡¯s RAM. We are essentially empirical formula Cu2O Finally we have to write the two amounts as a simple ratio. If you can¡¯t ¡®see¡¯ the simple ratio working out the number of moles straight away, choose the smallest of the two of each element, Don¡¯t be put off if the ratio isn¡¯t exact. It amounts and divide both amounts by this because moles = is OK to round up or down, if your ratio smallest amount. This doesn¡¯t always get you mass / RAM. is close to a whole number. straight to the answer, but it often helps. 4. Find the empirical formulae of the d) 0.60g of magnesium was heated with compounds formed when the following sulfur until further reaction occurred. quantities react: The mass of magnesium sulfide formed a) 4.32g of silver and 1.42g of chlorine was 1.40g. b) 2.5g of calcium and 1.5g of carbon e) 10.0g of carbon was reacted with c) 0.3g of carbon and 0.8g of oxygen hydrogen to give a compound that d) 0.7g of iron and 3.0g of bromine weighed 12.5g. e) 1.6g of calcium and 2.84g of chlorine f) 2.07g of sodium burnt completely in f) 3.2g of oxygen and 5.5g of manganese oxygen to give 2.79g of sodium oxide. g) 2.54g of iodine and 0.8g of oxygen g) 44.0g of manganese reacted completely h) 2.1g of silicon and 2.66g of chlorine to give 88.8g of manganese oxide. i) 5.0g of arsenic and 0.2g of hydrogen h) After reacting 1.5g of aluminium with j) 4.00g of magnesium and 1.56g of fluorine gas, 4.67g of aluminium fluoride nitrogen was produced. i) 5.00g of chromium underwent complete 5. Calculate the empirical formulae of the combustion in oxygen to give an oxide compounds formed in the following that weighed 7.31g experiments: j) 1.54g of an oxide of nitrogen was found a) 3.60g of magnesium was heated with to contain 0.40g of nitrogen. bromine until further reaction occurred. The mass of magnesium Answers 1. NH3, NH3, C3H8, C3H8, C2H6O, C2H6O, C4H10, C2H5 bromide formed was 27.60g. b) 8.00g of a gaseous hydrocarbon was 3. a) CH4, b) HO, c) H2O, d) BaCl2, e) Al2O3 reduced to give 6.00g of carbon. 4. a) AgCl, b) CaC2, c) CO2, d) FeBr3, e) CaCl2, f) MnO2, g) c) When 15.61g of an oxide of lead was I2O5, h) SiCl4, i) AsH3, j) Mg2N3 completely reduced, 14.49g of lead was produced. 5. a) MgBr2, b) CH4, c) PbO, d) MgS, e) CH3, f) Na2O, g) Mn2O7, h) AlF3, i) Cr2O3, j) N2O5