chapter 3: calculations with chemical formulas and equations
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Chapter 3: Calculations with Chemical Formulas and Equations. MASS AND MOLES OF SUBSTANCE 3.1 MOLECULAR WEIGHT AND FORMULA WEIGHT Molecular weight : (MW) sum of the atomic weights of all the atoms in a molecule of the substance. - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 3: Calculations with Chemical Formulas and Equations
MASS AND MOLES OF SUBSTANCE3.1 MOLECULAR WEIGHT AND FORMULA
WEIGHT- Molecular weight: (MW) sum of the atomic
weights of all the atoms in a molecule of the substance.
- H2O = 18.0 amu (2 x 1.0 amu for two H atoms plus 16.0 amu from one O atom)
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Formula weight (FW) – sum of the atomic weights of all atoms in a formula unit of a compound. Whether molecular or not.
Sodium chloride, NaCl (formula unit) formula weight of 58.44 amu (22.99 amu from Na plus 35.45 amu from Cl).
NaCl is ionic (no molecular weight)
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3.2 THE MOLE CONCEPT- Developed to deal with the enormous
numbers of molecules or ions in samples of substances.
- Mole: the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 g of carbon-12.
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The number of atoms in a 12-g sample of carbon-12 is called Avogadro’s Number (NA).
= 6.02 x 10^23A mole of a substance contains Avogadro’s
number of molecules (6.02 x 10^23)Mole (much like dozen or gross) refers to a
particular number of molecules.1 mole of ethanol = 6.02 x 10^23 ethanol
moleculesWhen talking about ionic substances we are
referring to the number of formula units not molecules.
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One mole of sodium carbonate Na2CO3
Contains:6.02 x 10^23 Na2CO3 unitsWhich contains:2 x 6.02 x 10^23 Na+ ions1 x 6.02 x 10^23 CO32- ions
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Specify Mole of oxygen atoms means 6.02 x 10^23 O
atomsMole of oxygen molecules means 6.02 x 10^23
O2 molecules (2 x 6.02 x 10^23 O atoms)
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Molar mass: mass of one mole of the substance.Carbon-12 has a molar mass of exactly 12g/mol
Ethanol, molecular formula C2H5OHMW = 46.1 amuMolar mass = 46.1 g/mol
For all substances, the molar mass in grams per mole is numerically equal to the formula weight in atomic mass units.
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MOLE CALCULATIONS- Conversion from mol to g- Ethanol is 46.1 g/mol1 mol C2H5OH = 46.1 g Cr2H5OHTo convert grams to moles
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• You need to prepare acetic acid from 10.0g of ethanol, C2H5OH. Convert 10.0g C2H5OH to moles C2H5OH by multiplying by the appropriate conversion factor.
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3.3 MASS PERCENTAGES FROM THE FORMULA- Mass percentage of A as the parts of A per
hundred parts of the total, by mass.
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3.4 ELEMENTAL ANALYSIS: PERCENTAGES OF CARBON, HYDROGEN, AND OXYGEN
- To determine the formula of a new compound is to obtain its percentage composition.
- Ex. Burn a sample of the compound of known mass and get CO2 and H2O.
- Next relate the masses of CO2 and H2O to the masses of carbon and hydrogen.
- Then calculate the mass percentages of C and H.- Determine the mass percentage of O by
difference.
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3.5 DETERMINING FORMULAS- Empirical formula (simplest formula) – formula
of a substance written with the smallest integer subscripts.
- For most ionic substances, the empirical formula is the formula of the compound.
- This is often not the case for molecular substances.
- Hydrogen peroxide: molecular formula is H2O2
- The empirical formula (just tells you the ratio of numbers of atoms in the compound) HO
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• Why do ionic compounds usually have the same formula and empirical formula?
• No multiple proportions like molecules.
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• Compounds with different molecular formulas can have the same empirical formula.
• They will also have the same percentage composition.
• Acetylene C2H2
• Benzene C6H6
• Same empirical formula CO, different molecular formulas, different chemical structures.
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To obtain the molecular formula of a substance, you need to know:
1. The percentage composition, from which the empirical formula can be determined
2. Molecular weight
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EMPIRICAL FORMULA FROM THE COMPOSITION- You can find the empirical formula from the
composition of the compound by converting masses of the elements to moles.
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• Ex. 3.10
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MOLECULAR FORMULA FROM EMPIRICAL FORMULA
- Molecular formula of a compound is a multiple of its empirical formula.
- Acetylene C2H2 is equivalent to (CH)2- Benzene C6H6 is equivalent to (CH)6
Molecular weight = n x empirical formula weight
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• N = number of empirical formula units in the molecule.
• Molecular formula is obtained by multiplying the subscripts of the empirical formula by n
n = molecular weight/empirical formula weight
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• After you find the empirical formula, calculate its empirical formula weight.
• From an experimental determination of its molecular weight, you can calculate n and then the molecular formula.
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• Ex. 3.11
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STOICHIOMETRY: QUANTITATIVE RELATIONS IN CHEMICAL REACTIONS
Stoichiometry – the calculation of the quantities of reactants and products involved in a chemical reaction.
Relationship between mass and moles.
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3.6 MOLAR INTERPRETATION OF A CHEMICAL EQUATION
N2 + 3H2 = 2NH3
How much hydrogen is required to give a particular quantity of ammonia?
One N2 One mole of N2 reacts with Three H2 three moles of H2 to give twoTwo NH3 moles of NH3.
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Write pg. 81
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3.7 AMOUNTS OF SUBSTANCES IN A CHEMICAL REACTION
-