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CALCULATION of EMPIRICAL FORMULA Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations 5. Simple empirical formula and formula mass from reacting masses or % composition (easy start, no moles involved !) Quantitative Chemistry calculations online Help for problem solving in doing empirical formula calculations, using experiment data, making predictions. Practice revision questions on empirical formula from reacting masses or % composition by mass of a compound. This page describes and explains, with fully worked out examples, how to work out the empirical formula of a compound. The empirical formula of a compound is defined and explained. Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB courses Spotted any careless error? EMAIL query?comment or request a type of GCSE calculation not covered?  Self-assessment Quizzes: type in answer for F and H  or  multiple choice for F and H

5. Empirical formula and formula mass from reacting masses (easy start, no moles!) The EMPIRICAL FORMULA of a compound can be worked out by knowing the exact masses of the elements that combine to form a given mass of a compound. The empirical formula of a compound is the simplest whole number ratio of atoms present in a compound. (see section 3. for some simpler examples). Here the word 'empirical' means from experimental data. Do not confuse with molecular formula which depicts the actual total numbers of each atom in a molecule. The molecular formula and empirical formula can be different or the same. They are the same if the molecular formula cannot be simplified on a whole number basis. Examples where molecular formula = empirical formula e.g. for sodium sulfate Na2SO4  and  propane C3H8 You cannot simplify the atomic ratios 2 : 1 : 4 or 3 : 8 to smaller whole number (integer) ratios Examples of where molecular formula and empirical formula are different e.g. butane molecular formula C4H10, empirical formula C2H5 numerically, the empirical formula of butane is 'half' of its molecular formula 4 : 10 ==> 2 : 5 glucose molecular formula C6H12O6, empirical formula CH2O numerically, the empirical formula of glucose is '1/6th' of the full molecular formula 6 : 12 : 6 ==> 1 : 2 : 1 Where the empirical formula and molecular formula are different, you need extra information to deduce the molecular formula from the empirical formula (see link below). This page is only concerned with calculating empirical formula. For more advanced students see Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition) The following examples illustrate the ideas using numbers more easily appreciated than in real experiments. In real laboratory experiments only a fraction of a gram or a few grams of elements would be used, and a more 'tricky' mole calculation method is required than shown here (dealt with later for higher students in section 8). However the examples below show in principal how formulae are worked out from experiments. Any calculation method must take into account the different relative atomic masses of the elements in order to get to the actual ratio of the atoms in the formula. For example, just because 10g of X combines with 20g of Y, it does not mean that the formula of the compound is XY2 ! If you divide the mass of each element by its atomic mass, you actually get the atomic ratio. Empirical formula calculation Example 5.1 The compound formed between lead and sulfur It is found that 207g of lead combined with 32g of sulphur to form 239g of lead sulphide. From the data work out the formula of lead sulphide. (Relative atomic masses: Pb = 207 and S = 32) In this case it easy to see that by the atomic mass ratio, 239 splits on a 1 to 1 basis of 1 atom of lead to 1 atom of sulphur (1 x 207 to 1 x 32 by mass) so the formula is simply PbS You can set out the calculation in a simple table format, in this case the numbers are very easy to deal with! RATIOS ... lead (Ar = 207) sulphur S (Ar = 32) Comments and tips Reacting mass 207g 32g not the real atom ratio atom ratio from mass / atomic mass values 207/207 = 1 32/32 = 1 work out the simplest whole number ratio simplest whole number atom ratio by trial & error 1 1 therefore the integer simplest ratio of 1 : 1 gives the empirical formula for lead sulphide as PbS - Empirical formula calculation Example 5.2 The empirical formula of a lead oxide It is found that 207g of lead combined with oxygen to form 239g of a lead oxide. From the data work out the formula of the lead oxide. (Relative atomic masses: Pb = 207 and O = 16) In this case, you first have to work out the amount of oxygen combined with the lead. By simple logic from the law of conservation of mass, this is 239 - 207 = 32g In atomic ratio terms, the 207 is equivalent to 1 atom of lead and the 32 is equivalent to 2 atoms of oxygen (1 x 207 to 2 x 16), so the formula is simply PbO2 Note: The mass of oxygen combined with the lead is deduced by subtracting the original mass of lead from final total mass of lead oxide. RATIOS ... lead (Ar = 207) oxygen O (Ar = 16) Comments and tips Reacting mass 207g 239-207 = 32g not the real atom ratio atom ratio from mass / atomic mass values 207/207 = 1 32/16 = 2 work out the simplest whole number ratio simplest whole number atom ratio by trial & error 1 2 therefore the simplest whole number ratio of 1 : 2 gives the empirical formula for this lead oxide as PbO2 Its actually called lead(IV) oxide - Empirical formula calculation Example 5.3 The empirical formula of aluminium sulfide It is found that 54g of aluminium forms 150g of aluminium sulphide. Work out the formula of aluminium sulphide. (Relative atomic masses: Al = 27 and S = 32). Amount of sulphur combined with the aluminium = 150 - 54 = 96g By atomic ratio, the 54 of aluminium is equivalent to 2 atoms of aluminium and the 96 of sulphur is equivalent to 3 atoms of sulphur. Therefore the atomic ratio is 2 to 3, so the formula of aluminium sulphide is Al2S3 RATIOS ... aluminium (Ar = 27) sulfur S (Ar = 32) Comments and tips Reacting mass 54g 150-54 = 96g not the real atom ratio atom ratio from mass / atomic mass values 54/27 96/32 work out the simplest whole number ratio simplest whole number atom ratio by trial & error 2 3 therefore the simplest integer ratio of 2 : 3 gives the empirical formula for aluminium sulphide as Al2S3 - Empirical formula calculation Example 5.4 From now on, questions just using the table method to work out empirical formula from more awkward numbers! In this case a compound formed between copper and chlorine. A compound of copper contained 47.4% copper and 52.6% chlorine. The atomic masses are: Cu = 64 and Cl = 35.5 Think of the percentages as masses in grams to solve the empirical formula problem. RATIOS ... Cu Cl Comments and tips Reacting mass 47.4 52.6 not the real atom ratio atom ratio from mass / atomic mass values 47.4/64 = 0.74 52.6/35.5 = 1.48 work out the simplest whole number ratio simplest whole number atom ratio by trial & error 0.74/0.74 = 1.0 1.48/0.74 = 2.0 therefore the simplest whole number ratio of 1 : 2 gives the empirical formula for copper chloride as CuCl2 Its actually called copper(II) chloride - Empirical formula calculation Example 5.5 The empirical formula of a compound of carbon and chlorine It was found that 0.39 g of carbon was combined with 4.61g of chlorine. Atomic masses: C = 12 and Cl = 35.5 RATIOS ... C Cl Comments and tips Reacting mass 0.39 4.61 not the real atom ratio atom ratio from mass / atomic mass values 0.39/12 = 0.0325 4.61/35.5 = 0.130 work out the simplest whole number ratio simplest whole number atom ratio by trial & error 0.0325/0.0325 = 1.0 0.130/0.0325 = 4.0 therefore the simplest ratio of 1 : 4 gives the empirical formula is CCl4 Its actually called tetrachloromethane - Empirical formula calculation Example 5.6 The formula of a hydrocarbon. It was that 0.75g of carbon was combined with 0.25g of hydrogen. Atomic masses: C = 12 and H = 1 Calculate the empirical formula of the hydrocarbon RATIOS ... C H Comments and tips Reacting mass 0.75g 0.25g not the real atom ratio atom ratio from mass / atomic mass values 0.75/12 = 0.0625 0.25/1 = 0.25 work out the simplest whole number ratio simplest whole number atom ratio by trial & error 0.0625/0.0625 = 1.0 0.25/0.0625 = 4.0 therefore the simplest ratio gives the empirical formula for the hydrocarbon = 1 : 4, so formula is CH4 This is the simplest hydrocarbon molecule called methane (main constituent in natural gas) - Empirical formula calculation Example 5.7 The analysis of sodium sulfate, calculating its empirical formula from the % composition by mass. On analysis, the salt sodium sulfate was found to contain 32.4% sodium, 22.5% sulfur and 45.1% oxygen. Atomic masses: Na = 23, S = 32 and O = 16 Calculate the empirical formula of sodium sulfate RATIOS ... Na S O Comments and tips Reacting mass 32.4 22.5 45.1 not the real atom ratio atom ratio from mass / atomic mass values 32.4/23 = 1.41 22.5/32 = 0.70 45.1/16 = 2.82 work out the simplest whole number ratio, in this case you have to make a reasonable judgement as to the values of the integers simplest whole number atom ratio by trial & error 1.41/7 = 2.01 ~2.0 0.70/0.70 = 1.0 2.82/0.70 = 4.03 ~4.0 the simplest ratio gives the empirical formula for sodium sulfate = 2 : 1 : 4, formula is Na2SO4 - Empirical formula calculation Example 5.8 The formula of a hydrocarbon. Analysis of hydrocarbon showed it consisted of 83.3% carbon and 16.7% hydrogen. Atomic masses: C = 12 and H = 1 Calculate the empirical formula of the hydrocarbon (just think of it as 83.3g C combined with 16.7g H) RATIOS ... C H Comments and tips Reacting mass 83.3 16.7 not the real atom ratio atom ratio from mass / atomic mass values 83.3/12 = 6.94 16.7/1 = 16.7 work out the simplest whole number ratioIn this case from the 1 2.4 to the 5:12 ratio, you have to multiply the 2.4 up until you get a whole number, x2, x3 and x4 don't work, but x5 does! simplest whole number atom ratio by trial & error 6.94/6.94 = 1.0 16.7/6.94 = 2.4 This is a bit awkward! 1.0 x 5 = 5.0 2.4 x 5 = 12.0 therefore the simplest ratio gives the empirical formula for the hydrocarbon = 5 : 12, so formula is C5H12 This is the simplest hydrocarbon molecule called pentane - - See section 8. for more empirical/molecular formula calculations involving moles. Self-assessment Quizzes [emp] type in answer for F and H  or  multiple choice for F and H OTHER CALCULATION PAGES What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass Calculating relative formula/molecular mass of a compound or element molecule Law of Conservation of Mass and simple reacting mass calculations Composition by percentage mass of elements in a compound Empirical formula & formula mass of a compound from reacting masses (easy start, not using moles) (this page) Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination Reacting masses, concentration of solution and volumetric titration calculations (NOT using moles) Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass) Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition) Moles and the molar volume of a gas, Avogadro's Law Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants-products) Molarity, volumes and solution concentrations (and diagrams of apparatus) How to do acid-alkali titration calculations, diagrams of apparatus, details of procedures Electrolysis products calculations (negative cathode and positive anode products) Other calculations e.g. % purity, % percentage & theoretical yield, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy 14.1 % purity of a product 14.2a % reaction yield 14.2b atom economy 14.3 dilution of solutions 14.4 water of crystallisation calculation  14.5 how much of a reactant is needed? 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