THE MOLE CONCEPT II Quantifying the World of the Very Small

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Programme Chemistry for Engineers Prepared by M. J.

Department of Pure and Applied Sciences Portmore

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1 THE MOLE CONCEPT II Quantifying the World of the Very Small ADEng. Programme Chemistry for Engineers Prepared by M. J. McNeil, MPhil. Department of Pure and Applied Sciences Portmore Community College Main Campus SAVE PAPER AND INK!!! When you print out the notes on PowerPoint, print "Handouts" instead of "Slides" in the print setup. Also, turn off the backgrounds(tools>options>print>unchec k "Background Printing")! ROCK N MOLE

2 LECTURE OBJECTIVES PART TWO Moles and The Chemical Equations. Percentage Composition of elements in substances. Calculate empirical and molecular formulae, using combustion data or composition by mass. 2

3 THE MOLES AND CHEMICAL EQUATIONS Reacting Ratios - show the mole-tomole ratio between two of the substances in a balanced equation. Use the coefficients of two substances in the equation. The balanced chemical equation below can be interpreted in numbers of molecules, but generally chemists interpret equations as mole-to-mole relationships. Stoichiometry - the calculation of the quantities of reactants and products in a chemical reaction. Based on the balanced chemical equation and on the relationship between for e.g. mass and moles. For example, the Haber process for producing ammonia involves the reaction of hydrogen and nitrogen. N2(g) 3 H2(g) 2 NH3(g) 3

4 MOLAR INTERPRETATION OF A CHEMICAL EQUATION This balanced chemical equation shows that one mole of N 2 reacts with 3 moles of H 2 to produce 2 moles of NH 3. N (g) 3H (g) 2 NH (g) molecule N molecules H 2 2 molecules NH 3 1mol N2 3 mol H2 2 mol NH3 In terms of mass, the substances in the equation would be. Because moles can be converted to mass, you can also give a mass interpretation of a chemical equation. 4

5 MOLES AND MASS OF A CHEMICAL EQUATION Suppose we wished to determine the number of moles of NH 3 we could obtain from 4.8 mol H 2. N (g) 3H (g) 2 NH (g) Amounts of substances in a chemical reaction by mass. How many grams of HCl are required to react with 5.00 grams manganese (IV) oxide according to this equation? Because the coefficients in the balanced equation represent moleto-mole ratios, the calculation is simple. 4 HCl(aq) MnO2 (s) H O(l) MnCl (aq) Cl (g)

6 MASS OF A CHEMICAL REACTION The reaction between H 2 and O 2 produces 13.1 g of water. How many grams of O 2 reacted? Write the equation H 2 (g) + O 2 (g) H 2 O (g) Balance the equation 2 H 2 (g) + O 2 (g) 2H 2 O (g) The volume of gases or amount of substances can also be determined from a chemical equation. 1. Magnesium burns in air to form magnesium oxide. What mass of magnesium oxide would be formed if 6 g of magnesium was burned in plenty (excess) of oxygen. 2. Calcium cyanamide, CaCN 2 reacts with water to form calcium carbonate and ammonia gas. What mass of calcium carbonate will be formed if 20 g of calcium cyanamide reacts with excess water? 3. How many molecules of oxygen are produced from the decomposition of 12 g of water into its elements. 6

CALCULATING THE PERECENTAGES OF ELEMENTS IN A COMPOND What is a compound? Compounds has a definite composition and ratio (combining capacity). For e.g. Fe(II) vs Fe(III) oxides.

7 CALCULATING THE PERECENTAGES OF ELEMENTS IN A COMPOND What is a compound? Compounds has a definite composition and ratio (combining capacity). For e.g. Fe(II) vs Fe(III) oxides. Any other Scientists often want to find the percentage of an element in a compound. Usually this is to find out which compounds are better value for money. The percentage composition ifs the composition of each element in a compound given in percentages. Percent by mass We define the mass percentage of A as the parts of A per hundred parts of the total, by mass. That is, mass of "A" in whole mass %"A" 100% mass of the whole 7

8 MASS PERCENTAGES FROM FORMULA Let s calculate the percent composition of butane, C 4 H 10. First we need to molecular mass of butane. 1. What is the percent carbon in C 5 H 8 NO 4 (the glutamic acid used to make MSG (monosodium glutamate), a compound used to flavor foods and tenderize meats? Find out next the percent compositon by mass of each constituent element in the compound. 2. TNT (trinitrotoluene) is a white crystalline substance that explodes at 240 C. Calculate the percent composition of TNT, C 7 H 5 (NO 2 ) 3. What is the % composition of C 6 H 12 O 6? % composition is 40.0% C, 6.7% H, and 53.3% O 3. A chemical engineer has 50 tonnes of haematite, [iron(iii) oxide] and 50 tonnes of magnetite [iron(ii) oxide]. Which of the two ores contains the more iron? 8

LAW OF DEFINITE COMPOSITION The law of definite composition states that Compounds always contain the same elements in a constant proportion by mass. Sodium chloride is always 39.3% sodium and 60.

9 LAW OF DEFINITE COMPOSITION The law of definite composition states that Compounds always contain the same elements in a constant proportion by mass. Sodium chloride is always 39.3% sodium and 60.7% chlorine by mass, no matter what its source. A drop of water, a glass of water, a lake of water all contain hydrogen and oxygen in the same percent by mass. Water is always 11.2% hydrogen and 88.8% oxygen by mass. The percent composition of a compound leads directly to empirical formula calculations. 9

or Mass of Each Element sddjpcjq Empirical (Simplest or

10 THE MOLE AND THE CHEMICAL FORMULAE Using moles to determine the formula of compounds Chemical Analysis (% Composition or Mass of Each Element sddjpcjq Empirical (Simplest or Reduced) Formula Molecular (Actual) Formula (Using Molecular Mass)

McNelium (Mc) 4Mc + O 2 2Mc 2 O 2Mc + O 2 2McO Combustion Analysis: the technique of finding the mass composition of an unknown organic sample (X) by examining the products of its combustion.

11 McNelium (Mc) 4Mc + O 2 2Mc 2 O 2Mc + O 2 2McO Combustion Analysis: the technique of finding the mass composition of an unknown organic sample (X) by examining the products of its combustion. The relative number of atoms of Mc and O in each oxide can be determine by reducing each separately with injected hydrogen (must be dried by conc. H 2 SO 4 ) (H 2 + O 2 H 2 O), resulting in metallic Mc, the mass of which can be found. X (weighed sample) + O 2 CO 2 + H 2 O (GC Analysis) Which element(s) has a similar combining capacity? When the elements present in a compound have been qualitatively identified, it is possible to find by quantitative analysis, the proportions by mass of each element. Considering Mc. Mc + H 2 Mc + H 2 O From these data, the empirical formula and molecular formulas can be found. 11

12 COMBUSTION OF MCNELIUM In an experiment to determine the formation of an oxide of Mcnelium, the following data was obtained. Mass of oxide = g Mass of copper = g Mass of O removed = g Determining the empirical formula. Elements Mc : O Masses g : g No. of moles g : g 63.5 g/mol 10g/mol = 0.05 = 0.05 Divide by smallest # 0.05 : 0.05 of moles Simple mole ratio 1 : 1 Empirical formula is McO 12

13 TYPES OF FORMULAS EMPIRICAL FORMULA Simplest formula The lowest whole number ratio of elements in a compound. smallest integer (whole number) subscripts The empirical formula of a compound is the simplest whole number ratio of ions in a formula unit or element in a compound. Compounds existing as formula units (ionic compounds) are always referred to as empirical formulas. A compound s empirical formula can be determined from its percent composition. MOLECULAR FORMULA Actual formula The molecular formula is the actual ratio of elements in a compound. The formulas of compounds existing as molecules (covalent compounds) are ALWAYS referred to as molecular formulas. A compound s molecular formula is determined from its molar mass and empirical formula. Both empirical and molecular formulas can be the same e.g. water 13

14 COMPOUND EMPIRICAL FORMULA MOLECULAR FORMULA INTEGER MULTIPLIER acetylene CH C 2 H 2 2 benzene CH C 6 H 6 6 hydrogen peroxide HO H 2 O 2 2 butyl alcohol C 4 H 10 O C 4 H 10 O 1 Ethyl ether C 4 H 10 O C 4 H 10 O 1 Pentane C 5 H 12 C 5 H 12 1 A simple analogy may help to illustrate these two types of formulas. In PCC, the ratio of men to women may be 1 : 2 (empirical formula), but the actual number of men to women may be 400 : 800 (molecular formula) 14

Divide each by the smallest number of moles to obtain the simplest whole number ratio. 4.

15 EMPIRICAL FORMULA DETERMINATION 1. Determine the mass in grams of each element present, if necessary. 2. Calculate the number of moles of each element. 3. Divide each by the smallest number of moles to obtain the simplest whole number ratio. 4. If whole numbers are not obtained* in step 3), multiply through by the smallest number that will give all whole numbers. Do not round off numbers prematurely. Use appropriate sig. fig. 15

16 EMPIRICAL FORMULA FROM COMPOSITION DATA We can calculate the empirical formula of a compound using: - composition data (g) - Percentage data (%) A g sample of radium metal was heated to produce g of radium oxide. What is the empirical formula? USING COMPOSITION DATA We can determine the mole ratio of each element from the mass to determine the formula of radium oxide, Ra? O?. We have g Ra and = g O. 16

17 EMPIRICAL FORMULA FROM PERCENT DATA 100 grams of sample. Benzene is 92.2% carbon and 7.83% hydrogen, what is the empirical formula. If we assume 100 grams of sample, we have 92.2 g carbon and 7.83 g hydrogen. To work out the empirical formula, the total mass of the compound is assumed to be 100 g, and each percentage is turned into a mass in grams. Show all Working here. 17

Determine the empirical formula of Acetaminophen, the active ingredient in Tylenol, from elemental analysis: C 63.65%, H 6.00%, N 9.

18 DETERMINING FORMULAS FROM ORGANIC COMPOUNDS Glucose has the molecular formula C 6 H 12 O 6. What is its empirical formula, and what is the percentage composition of glucose? E.F. = smallest whole number ratio, CH 2 O 1. Determine the empirical formula of Acetaminophen, the active ingredient in Tylenol, from elemental analysis: C 63.65%, H 6.00%, N 9.27%, O 21.17%. 2. Benzoic acid is a white, crystalline powder used as a food preservative. The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. What is its empirical formula? 18

19 COMBUSTION ANALYSIS - ORGANICS This analysis provides the Empirical Formula. If a second technique provides the molecular weight, then the molecular formula may be deduced. Calculate the empirical formula weight. Find the number of formula units by dividing the known molecular weight by the formula weight. X + O 2 CO 2 + H 2 O 1. Find the mass of C and H in the sample. 2. Find the actual number of moles of C and H in the sample. 3. Find the relative number of moles of C and H in whole numbers 4. Write the empirical formula for the unknown compound. X is usually substituted with C x H y Multiply the number of atoms in the empirical formula by the number of formula units. Try to balance the equation algebraically. 19

20 MOLECULAR FORMULA DETERMINATION Determining the molecular formula from the empirical formula. The molecular formula should be a multiple of the empirical formula (since both have the same percent composition). To determine the molecular formula, we MUST KNOW the molecular weight of the compound. Determining the molecular formula from the empirical formula. For example, suppose the empirical formula of a compound is CH 2 O and its molecular weight is 60.0 g/mol. The molar weight of the empirical formula (the empirical weight) is only 30.0 g/mol. This would imply that the molecular formula is actually the empirical formula doubled, or C 2 H 4 O 2 20

21 MOLECULAR FORMULA PROBLEMS The molecular weight of glucose is 180 g/mole and its empirical formula is CH 2 O. Deduce the molecular formula. What is the Empirical Formula if the % composition is 40.0% C, 6.7% H, and 53.3% O? 1. Formula weight for CH 2 O is g/mole 2. # of formula units = 180/30.03 = 6 3. Molecular formula = C 6 H 12 O 6 E.F. AND M.F RELATIONSHIP Molecular formula subscripts = n x E.F. subscripts, where n = 1, 2, 3 In other words, you multiply the empirical value by a whole number factor to obtain the M. F. The Empirical Formula is CH 2 O (MW = ) If MW of the real formula is , what is the actual formula? ( )/(30.026) = 6 CH 2 O x 6 = C 6 H 12 O 6 21

22 RESOURCES VIDEOS Chemguy videos (google AP chemistry and chemguy videos; use in sequence) (good, short, video lectures that are appropriate for regular, honors, or AP chemistry) chem Guy video for mole 1 _in_order Mole 2 chem guy: aynext=1&playnext_from=pl&index=19 Mole 3 (Junior) : ynext=1&playnext_from=pl&index=20 Stoichiometry video with worked example for predicting amount of product from 2.6 mol of one of reactant: see Empirical formula for AP chem with chem. Guy: 22

23 CONCLUSIONS The percent composition of a substance is the mass percent of each element in that substance. The empirical formula of a substance is the simplest whole number ratio of the elements in the formula. The molecular formula is a multiple of the empirical formula. 23

24 Good luck! Remember the best practice is practice! THE MOLE CONCEPT PART THREE IS NEXT 24

Calculations with Chemical Formulas and Equations

Calculations with Chemical Formulas and Equations Mass and Moles of a Substance Chemistry requires a method for determining the numbers of molecules in a given mass of a substance. This allows the chemist