Chap 1 Quantitative Chemistry 1 1.2 Formulas3 hours Assessment statementObjTeacher’s notes 1.2.1Define the terms relative atomic mass (A r ) and relative.

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Chap 1 Quantitative Chemistry 1 1.2 Formulas3 hours Assessment statementObjTeacher’s notes 1.2.1Define the terms relative atomic mass (A r ) and relative molecular mass (M r ). 1 1.2.2Calculate the mass of one mole of a species from its formula. 2The term molar mass (in g mol –1 ) will be used. 1.2.3Solve problems involving the relationship between the amount of substance in moles, mass and molar mass. 3 1.2.4Distinguish between the terms empirical formula and molecular formula. 2 1.2.5Determine the empirical formula from the percentage composition or from other experimental data. 3Aim 7: Virtual experiments can be used to demonstrate this. 1.2.6Determine the molecular formula when given both the empirical formula and experimental data. 3

Chemical Formulas Chemical formulas show the ratio of the number of atoms present in a compound or ion. Magnesium chloride MgCl 2 – 2 chloride ions for every 1 magnesium ion (Ionic compound) Acetic Acid CH 3 COOH – 2 atoms carbon, 2 atoms oxygen, 4 atoms hydrogen in one molecule of acetic acid (molecular compound) Ionic compound consist of a metal (positive ion) and non- metal (negative ion). Molecular (covalent) compounds are made of non-metals only. 2

Relative Atomic Mass (A r ) The mass of an atom is compared with that of an atom of carbon-12. (Carbon-12 standard) The relative atomic mass of carbon-12 is taken to be 12. Relative masses are based on elements having isotopes (same identity, different “stuff” within core) For example, naturally occurring chlorine consists of atoms of relative isotopic masses 35 (75%) and 37 (25%). – Its relative atomic mass (A r ) is 35.5. 3

Relative Molecular Mass (M r ) The MASS of a MOLECULE The mass of a molecule is compared with that of an atom of carbon-12. (Carbon-12 standard) A relative molecular mass can be calculated easily by adding together the relative atomic masses of the constituent atoms that have units g mol -1 e. g., ethanol: CH 3 CH 2 OH, has a M r of 46.08 g mol -1 (C 2 + H 6 + O = 24.02 + 6.06 + 16.00 = 46.08) 4

Example: Molecular Mass Find the molecular mass of Glucose C 6 H 12 O 6. Solving Process: – Add the atomic masses of all the atoms in the C 6 H 12 O 6 6 C atoms 6 x 12.0 = 72.0 g mol -1 12 H atoms 12 x 1.0 = 12.0 g mol -1 6 O atoms 6 x 16.0 = 96.0 g mol -1 Formula mass of glucose = 180.0 g mol -1 – This mass represents one mole of glucose. 5

Example: Formula Mass Find the formula mass of Copper (II) phosphate, Cu 3 (PO 4 ) 2. Solving Process: – Add the atomic masses of all the atoms in the Cu 3 (PO 4 ) 2 formula. Remember the subscript applies the entire polyatomic ion. 3 Cu atoms3 x 63.5 = 190.5 g mol -1 2 P atoms2 x 30.9 = 61.8 g mol -1 8 O atoms 8 x 16.0 = 128.0 g mol -1 Formula mass of Cu 3 (PO 4 ) 2 = 380.3 g mol -1 – This mass represents one mole of Copper (II) phosphate. 6

Chap 1 Quantitative Chemistry 7 Assessment statementObjTeacher’s notes 1.1.1Apply the mole concept to substances. 2The mole concept applies to all kinds of particles: atoms, molecules, ions, electrons, formula units, and so on. The amount of substance is measured in moles (mol). The approximate value of Avogadro’s constant (L), 6.02 × 10 23 mol –1, should be known. TOK: Chemistry deals with enormous differences in scale. The magnitude of Avogadro’s constant is beyond the scale of our everyday experience. 1.1.2Determine the number of particles and the amount of substance (in moles). 3Convert between the amount of substance (in moles) and the number of atoms, molecules, ions, electrons and formula units. 1.1 The mole concept and Avogadro’s constant 2 hours

Chemists do not deal with amounts of substance by counting out individual atoms or molecules. Atomic Mass Units (amu) as a Very Small Number It is important to obtain a relationship between mass and the number of particles in a compound. Molecular masses are in atomic mass units, which is a very small unit. An atomic mass unit is only 1.66 x 10 -24 grams. We need to use a larger unit for everyday use such as the gram for laboratory use. We mass quantities of substances on balances. Chap 1.1 Mole Concept & Avogadro’s Constant 8

Avogadro’s Number Chemists have found that 6.022 x 10 23 atoms of Hydrogen have a mass of 1.0079 g. This number 6.022 x 10 23 is called Avogadro’s number. Commonly referred to as the Mole. 9

The Mole Avogadro’s number is an accepted SI standard. The symbol used to represent Avogadro's number is N A. This quantity can be expressed as 6.02205 x 10 23 to be more precise. This number of whatever we are questioning is called one mole (mol) of things. Dozen: 12 of something. Whether feathers or dump trucks. Same number, different masses Mole: just a name representing a number (6.022 x10 23 ) (just like dozen) 10

The Mole (cont.) A mole of feathers and a mole of trucks are the same number – Are they the same mass? One mole of particles (atoms, ions, molecules, electrons, formula units) has a mass (in grams) – Happens to be equivalent in # to that of one particle in atomic mass units. If a mole of any particle has a mass of 4.02 g mol -1, then a single particle has a mass of 4.02 amu (Atomic Mass Units) If you have a single particle with a mass of 53.03 amu, then a mole of the same particles will have a mass of 53.03 g. 11

The Mole Road Map Mass “Stuff” Mole Volume g mole g mole. 6.022 x 10 23 mole 22.4 L mole mole. 22.4 L Stuff: Atoms Molecules Particles Electrons Ions Formula Units For Gases only: @ STP (Standard Temp & Pressure) (0 o C (273 K) & 1 atm) 12

Comparison It is important to note that one mole of atoms contains 6.022 x 10 23 atoms. One mole of molecules contains 6.022 x 10 23 molecules. One mole of ions contains 6.022 x 10 23 ions. N A, therefore can have any of these units (but the SAME number). 13

The Mole Road Map Mass “Stuff” Mole Volume g mole g mole. 6.022 x 10 23 mole 22.4 L mole mole. 22.4 L Stuff: Atoms Molecules Particles Electrons Ions Formula Units For Gases only: @ STP (Standard Temp & Pressure) (0 o C (273 K) & 1 atm) 14

Example: Conversion to Moles 1.20 x 10 25 molecules of NH 3 will be how many moles? Solving Process: – We have molecules and wish to covert to moles. – One mole equals 6.022 x 10 23 molecules. – 1 mol/6.022 x 10 23 molecules – 1.20 x 10 25 molecules NH 3 1 mol NH 3. 6.022 x 10 23 molecules NH 3 = 19.9 mol NH 3 15

Try This! Convert to the indicated units for each question. 1.9.03 x 10 23 atoms bromine to moles of bromine. 2.3.5 mol of sodium to atoms of sodium 3.2 mol of Calcium chloride (CaCl 2 ) to: a)Formula Units of calcium chloride b)Number of calcium ions c)Number of chloride ions 4.4.75 mol of electrons to # of electrons. 5.5.08 x 10 23 molecules of oxygen to moles of oxygen molecules. Moles of oxygen to atoms.* *note: oxygen is a diatomic molecule, O 2. 16

Mole to Mole Calculations 17

Chemical Calculations From the reaction involving nitrogen with hydrogen to form ammonia N 2 + 3H 2  2NH 3 The most important part of this BALANCED equation is that 1 mole of nitrogen reacts with 3 moles of hydrogen to form 2 moles of Ammonia. Based on this information, we can write ratios that relate moles of reactants to each other and to moles of products. 18

Mole – Mole Calculations How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen? N 2 + 3H 2  2NH 3 The mole ratio of nitrogen to ammonia is 1:2 meaning for every 1 mole of Nitrogen, 2 moles of ammonia are produced. 19

20 VOLUME MOLES MASS MOLECULES x molar mass  molar mass  6.022 x 10 23  22.4 L x 6.022 x 10 23 x 22.4 L VOLUME MOLES MASS MOLECULES x molar mass  molar mass  6.022 x 10 23  22.4 L x 6.022 x 10 23 x 22.4 L Mole ratio

Mass – Mass Calculations Calculate the number of grams of NH 3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen. Solution: – Balance the equation (if needed) – Grams of H 2 to moles of H 2 to moles of NH 3 to grams of NH 3 21

Try This! 1.How many grams of acetylene are produced by adding water to 5.00 g of CaC 2 ? Acetylene gas (C 2 H 2 ) is produced by adding water to calcium carbide. CaC 2 + 2H 2 O  C 2 H 2 + Ca(OH) 2 2.Using the same equation, determine how many moles of CaC 2 are needed to react completely with 49.0g of water. 22

Calculating Molecules of a Product How many molecules of oxygen are produced when 29.2 g of water is decomposed by electrolysis according to this balanced equation? 2H 2 O (l) → 2H 2 (g) + O 2 (g) 23

Try This! 1.How many molecules of oxygen are produced by the decomposition of 6.54 g of potassium chlorate (KClO 3 )? 2KClO 3 (s) → 2KCl (s) + 3O 2 (g) 2.The last step in the production of nitric acid is the reaction of nitrogen dioxide with water. How many grams of nitrogen dioxide must react with water to produce 5.00 x 10 22 molecules of nitrogen monoxide? 3NO 2 (g) + H 2 O (l) → 2HNO 3 (aq) + NO (g) 24

Example: Conversion to Moles How many moles are represented by 21 g of ammonia, NH 3 ? Solving process: – We have grams and want to convert to moles. – Using the atomic mass from the periodic table, we can calculate the atomic mass of NH 3 (17 g mol -1 ) – 1 mole of NH 3 has a mass of 17 g. – We have a sample of 21 g of NH 3 – Divide our sample by the mass of 1 mole of NH 3 – We end up with 1.24 mol of Ammonia 25

Example: Conversion to Mass What mass is 20 mol of NH 3 ? Solving Process: – Using your periodic table, determine the molecular mass of NH 3 (17.0 g mol -1 ) – 17.0 g for 1 mol of NH 3 – 20 mol NH 3 x 17.0 g NH 3 1 mol NH 3 = 340 g of NH 3 in 20 mol NH 3 26

Example: Conversion to Atoms How many atoms are in a 10.0 g sample of calcium metal? Solving Process: – 1 formula mass of calcium metal is 40.1 g mol -1. – 10 g of calcium is 0.249 mol calcium (10 g/40.1 g per mol) – There are 6.02 x 10 23 (N A ) atoms in one mole – N A (0.249 mol calcium) = # atoms in 10 g of calcium metal = 1.50 x 10 23 atoms in 10 g of calcium 27

Try This! Make the following conversions: 1.7.92 formula units of Water (H 2 O) to moles. 2.2.65 mol of methane (CH 4 ) to molecules. 3.207 g of sodium nitrite (NaNO 3 ) to moles. 4.85.7 g of gold (III) sulfate (AuSO 4 ) to formula units. 5.5.97 x 10 23 formula units of magnesium bicarbonate (Mg(HCO 3 ) 2 )to grams 6.0.458 moles of sodium thiosulfate (Na 2 S 2 O 3 ) to grams 7.1.99 x 10 24 formula units of ammonium phosphate ((NH 4 ) 3 PO 4 )to moles. 8.55 g of Mercury (Hg) to moles. 28

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