Chemical Calculations: The Mole concept and Chemical Formula. Law of Definite Proportions (John Dalton) Chapter 9

Transcription

1 Chapter 9 Chemical Calculations: The Mole concept and Chemical Formula This material is not included in Midterm 1 1 Law of Definite Proportions (John Dalton) Chapter 9 A given compound always contains the same proportion, by mass, of the elements. Decompose samples of a pure compound to give the constituent elements: NO 2 Sample Mass decompose Mass of nitrogen Mass of oxygen A g g g B g g g 2 1 1

2 Percent Composition What is the percent of nitrogen and percent of oxygen contained in each sample of NO 2? Sample A: g g x 100 % = % nitrogen g x 100 % g = % oxygen Sample B: 3 Formula Unit and Formula Mass Formula Unit gives proportions of elements in a pure compound NO 2 CaCl 2 molecular ionic Formula Mass sum of atomic masses of atoms in one formula unit Atomic masses are obtained from the periodic table in amu (atomic mass units) eg: NO 2 atomic mass of N 14.01amu atomic mass of O amu formula mass of NO 2 : (16.00) amu = amu 4 2 2

3 Percent Composition from Formula NO 2 formula mass amu total mass from each element x 100 % formula mass % N 14.01amu / amu x 100% = % % O 2(16.00) amu / amu x 100% = % 5 Percent Composition Sample A g x 100 % g = From decomposition data % nitrogen From formula % g x 100 % g = % oxygen % Sample B g g x 100 % = 30.43% nitrogen 7.383g x 100 % g = % oxygen 6 3 3

4 What is the formula mass of H 2 O? sum of the atomic masses of the total number of atoms represented by the formula 2 x H = 2 (1.008) = amu 1 x O = 1 x = amu formula mass amu = amu 7 What is the formula mass of Ca(NO 3 ) 2? 8 4 4

5 Counting Atoms, Ions and Molecules Consider a reaction to form CO where: 1 atom of C combines with 1 atom of O to give CO, 1 molecule of carbon monoxide 1 pair of C atoms + 1 pair of O atoms gives 1 pair of CO molecules 1 dozen C atoms + 1 dozen O atoms gives 1 dozen CO molecules 1 gross of C atoms + 1 gross of O atoms gives 1 gross CO molecules 1 mole of C atoms + 1 mole of O atoms gives 1 mole of CO molecules 1 mole = x objects Avogadro s number 9 One Mole If one mole of pennies was distributed equally to everyone in the earth s population (~6 billion people), each person would have over a trillion dollars It would take 6 trillion galaxies the size of the Milky Way to provide 6.02 x stars Mole Day 6:02 AM to 6:02 pm on October

6 The Mole 11 Converting between Moles and number of particles Moles of substance Particles of a substance How many atoms are in 1.50 moles of Carbon? 1 mole = 6.02 x atoms 1.50 moles x 6.02 x atoms /mole = 9.03 x atoms How many molecules are in 5.35 moles of CO? 5.35 moles of CO x 6.02 x molecules/mole

7 What is the Mass of one Mole? Molar Mass of an Element The mass in grams that is numerically equivalent to the atomic mass of that element. units of molar mass are: grams/mole atomic mass of 12 C: 12 amu 1 mole of 12 C atoms weighs 12 g/mole atomic mass of C : amu 1 mole of C weighs g/mole 13 Molar Mass of a Compound The mass in grams that is numerically equivalent to the formula mass of that compound. NO 2 formula mass: amu molar mass: g/mole The molar mass of any substance is the mass in grams of one mole of that substance. units of g/mole

8 What is the mass in grams of moles of Na 2 CO 3, an industrial chemical used in the making of glass? To go from moles to grams use molar mass as a conversion factor: Molar mass of Na 2 CO 3 2 x Na 2 (23.00) g/mole 1 x C x O 3 (16.00) g/mole O.250 moles x g/mole = 26.5 g of Na 2 CO 3 15 How many moles of NH 4 Cl are in a 2.75 g sample of NH 4 Cl

9 Substance Dissolved in a Liquid - concentration Molarity (Chapter 13.8 pg ) The number of moles of substance (solute) dissolved in one litre of solution - gives us the concentration Molarity (M) = moles of solute litres of solution 1 Molar solution of CuSO 4 contains 1 mole of CuSO 4 in enough water to make up 1 litre of solution 0.5 Molar solution contains 0.5 moles of CuSO 4 in 1 litre of solution 17 Molarity in Calculations 2.5 moles of NaCl are dissolved in enough water to make 3.0 L of solution. What is the concentration, in molarity, of the solution. 2.5 moles of NaCl 3.0 L of solution = 0.83moles/litre = 0.83 molar = 0.83M 5.5 g of KCl are dissolved in water to make 350 ml of solution. What is the molarity of the final solution? 5.5 g KCl = moles KCl g/mole moles KCl = moles/l = 0.21 moles/litre L

10 Calculations Involving Moles Moles of substance Avogadro s number particles/mole Particles of a substance Grams of substance molar mass g/mole Moles of substance Volume of a solution molarity moles/l Moles of substance 19 Example How many moles of sulfuric acid, H 2 SO 4, are contained in 0.80 L of a M solution of the acid? volume = 0.80 L molarity, M = moles/l # moles = 0.80 L x moles L = moles (2 sig figs allowed)

11 Chemical Formula of Compounds Formula give the relative numbers of atoms or moles of each element in a formula unit - always a whole number ratio (the law of definite proportions). 1 molecule of SO 2 1 atom of S for every 2 atoms of O 1 mole of SO 2 : 1 mole S atoms for every 2 moles of O atoms If we know, or can determine, the relative number of moles of each element in a compound we can determine a formula for the compound. 21 Chemical Formula of Compounds Empirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms presents. Ionic formula are always empirical formula Molecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound

12 Calculating Empirical Formula From experimental data: composition mass of individual elements % composition of individual elements eg: decomposition of a known mass of sample to determine grams of each element. 23 Formula A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34 g O. Determine a formula for this substance. We require mole ratios so must convert grams to moles moles of N = 2.34g of N g/mole = moles of N moles of O = 5.34 g = moles of O g/mole N O = NO Formula: N0.167O

13 Example: A sample of a liquid compound contains g of C and 6.06g of H. What is the chemical formula of the compound? 25 Decomposition by Combustion compounds containing only carbon and hydrogen burn in oxygen to produce CO 2 and H 2 O calculate the mass of C from mass of CO 2 calculate the mass of H from mass of H 2 O

14 Sample Calculation A sample of a pure substance containing only carbon and hydrogen weighs g. When burned in oxygen, g of CO 2 and g of H 2 O were produced. What is the empirical formula of the sample compound? moles of C in the compound from moles of CO 2 (mass / molar mass) g CO 2 = mol of CO g/mol of CO mol CO 2 x 1 mol C = mol of C 1 mol CO 2 27 moles of H in the compound from moles of H 2 O moles of H 2 O = mass of H 2 O molar mass of H 2 O = g H 2 O = mol H 2 O g /mol H 2 O moles of H = moles of H 2 O x 2 mol H mol H 2 O = moles of H Formula based on relative number of moles: C H Simplest whole number ratio CH 3 Empirical Formula

15 Combustion Products: The following compounds are produced from each of the listed elements when samples containing those elements are burned in oxygen: C gives CO 2 H gives H 2 O N gives NO or NO 2 S gives SO 2 29 Calculating Empirical formula The combustion of g of a substance that contains only C, H, and O results in the recovery of g of CO 2 and g of H 2 O. What is the empirical formula of this substance? -determine the moles of C based on moles of CO 2 formed -determine the moles of H based on moles of H 2 O formed moles of O: -oxygen in the CO 2 and the H 2 O comes from both the O in the sample and the O in the air

16 Determining amount of oxygen from combustion data Mass of sample = mass of O + mass of C + mass of H from moles of C calculate mass of C in the sample from moles of H calculate mass of H in the sample mass of O = mass of sample mass of (C+ H) -determine the moles of oxygen from mass of oxygen -determine the empirical formula knowing moles of C, H, O 31 Empirical Formula from % Composition A substance has the following composition by mass: % Na ; % B ; % H What is the empirical formula of the substance? Consider a sample size of 100 grams: This will contain grams of sodium, grams of B and grams H. Determine the number of moles of each Determine the simplest whole number ratio

17 To obtain an Empirical Formula 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 * Be careful! Do not round off numbers prematurely 33 Calculation of the Molecular Formula molecular formula = n(empirical formula) molar mass = n(empirical formula mass) A compound has an empirical formula of NO 2. The colourless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance? empirical formula mass: (16.00) = g/mol n = molar mass = 92.0 g/mol emp. f. mass g/mol n = 2 2(NO 2 ) = N 2 O

18 Example An unknown compound has the empirical formula CH 2 O A sample of 0.25 moles weighs grams. What is the molecular formula of this compound?