The Mole Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second?

Slides:



Advertisements
Similar presentations
The Mole Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second?
Advertisements

The Mole Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second?
The Mole A counting unit Similar to a dozen, except instead of 12, it’s 602 billion trillion 602,000,000,000,000,000,000, X (in scientific.
The Mole Concept Goal: To develop the concept of the mole as a useful measurement and to apply this in calculations involving mass and volume.
Empirical and Molecular Formulas
Chapter 8 Chemical Composition Chemistry B2A. Atomic mass unit (amu) = × g Atomic Weight Atoms are so tiny. We use a new unit of mass:
The Mole: A Shortcut for Chemists S-C-8-1_The Mole Presentation Source:
The Mole. Not the type of mole we are talking about.
Chapter 7 Chemical Quantities Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings.
The Mole Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second?
Unit 9 part 1: The Mole Chemical Measurements Mole Conversions Empirical & Molecular Formulas.
The Mole Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second?
The Mole Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second?
Proportions/Rates/Fractions  1,500 pages in 12 days = ______ pages in 3 days  18 calls in 6 hours = ______ calls in 11 hours.
Empirical Formula Molecular Formula
Chemical Calculations Mole to Mass, Mass to Moles.
Section 12.2: Using Moles (part 3). Mass Percent Steps: 1) Calculate mass of each element 2) Calculate total mass 3) Divide mass of element/ mass of compound.
Chemical Formulas NaCl H2OH2O C 6 H 12 O 6 NaHCO 3.
1.1 The Mole Concept 1.2 Formulas. Assessment Objectives Apply the mole concept to substances Determine the number of particles and the amount.
The Mole Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second?
Percent Composition.  We go to school for 180 days a year. What % of the year are we in school? Similar to finding % in any other situation.
The Mole Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second?
Chemistry 200 Fundamentals D Chemical Composition.
NaCl H2O C6H12O6 Chemical Formulas NaHCO3.
The Mole.
Chapter 7 “Chemical Formulas and Chemical Compounds”
Determining the Empirical Formula for a Compound
Stoichiometry Chapter 3
Measuring matter The mole
Empirical Formula.
The Mole Concept.
Empirical Formulas.
Molecular mass and formula mass
NaCl H2O C6H12O6 Chemical Formulas NaHCO3.
Chemical Quantities Part I
Unit 4: Formula Stoichiometry
Chapter 10 “Chemical Quantities”
The Mole Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second?
III. Formula Calculations
Dr. Chirie Sumanasekera
Chapter 10 – The Mole.
Empirical Formula Molecular Formula
Moles and Mass -The molar mass of a compound (aka molecular mass or molecular weight) is the sum of the atomic masses of each atom in the compound -Molar.
Moles.
Simplest Chemical formula for a compound
The Mole Chapter 10.1.
Mole Calculations 2.
Chemistry 100 Chapter 6 Chemical Composition.
Molecular Formula number and type of atoms covalent compounds
Chapter 10 “Chemical Quantities”
Chapter 9 “Chemical Quantities”
4/5/16 Today I will explain the mole and use it in calculations
Compounds and Molecules
From percentage to formula
Empirical Formulas and Mole Ratios
Chapters 10 Chemical Quantities.
The Mole Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second?
Empirical and Molecular Formulas
Mass Relationships in Chemical Reactions: STOICHIOMETRY
Empirical and Molecular Formulas
mole (symbolized mol) = 6.02 x particles
The Mole: A Shortcut for Chemists
The Mole.
From percentage to formula
Stoichiometry.
The Mole L.
Empirical Formula of a Compound
Empirical and Molecular Formulas
III. Formula Calculations (p )
Chapter 7- Sec. 3 and 4 “Chemical Formulas and Chemical Compounds”
Presentation transcript:

The Mole Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second?

Background: atomic masses Look at the “atomic masses” on the periodic table. What do these represent? E.g. the atomic mass of C is 12 (atomic # is 6) We know there are 6 protons and 6 neutrons Protons and neutrons have roughly the same mass. So, C weighs 12 u (atomic mass units). What is the actual mass of a C atom? Answer: approx. 2 x 10-23 grams (protons and neutrons each weigh about 1.7 x10-24 grams) Two problems Atomic masses do not convert easily to grams They can’t be weighed (they are too small)

The Mole With these problems, why use atomic mass at all? Masses give information about # of p+, n0, e– It is useful to know relative mass E.g. Q - What ratio is needed to make H2O? A - 2:1 by atoms, but 2:16 by mass It is useful to associate atomic mass with a mass in grams. It has been found that 1 g H, 12 g C, or 23 g Na have 6.02 x 1023 atoms 6.02 x 1023 is a “mole” or “Avogadro’s number” “mol” is used in equations, “mole” is used in writing; one gram = 1 g, one mole = 1 mol. Read 4.3 (167-9). Stop after text beside fig 2. Do Q1-6. Challenge: 1st slide (use reasonable units)

The Mole: Answers A mole is a number (like a dozen). Having this number of atoms allows us to easily convert atomic masses to molar masses. 6.02 x 1023 602 000 000 000 000 000 000 000 3.00 x 6.02 x 1023 = 18.06 x 1023 or 1.81 x 1024 (note: there are 3 moles of atoms in one mole of CO2 molecules. In other words, there are 5.42 x 1024 atoms in 3.00 mol CO2) 3.01 x 1023 a) 1.43 kg  12 = 0.119 kg per orange b) 1.01 g  6.02 x 1023 = 1.68 x 10 –24 g

A: It would take 19 million years Mollionaire Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second? A: $ 6.02 x 1023 / $1 000 000 000 = 6.02 x 1014 payments = 6.02 x 1014 seconds 6.02 x 1014 seconds / 60 = 1.003 x 1013 minutes 1.003 x 1013 minutes / 60 = 1.672 x 1011 hours 1.672 x 1011 hours / 24 = 6.968 x 109 days 6.968 x 109 days / 365.25 = 1.908 x 107 years A: It would take 19 million years

Comparing sugar (C12H22O11) & H2O Same 1 gram each 1 mol each volume? No, they have dif. densities. No, molecules have dif. sizes. mass? Yes, that’s what grams are. No, molecules have dif. masses # of moles? No, they have dif. molar masses Yes. # of molecules? No, they have dif. molar masses Yes (6.02 x 1023 in each) # of atoms? No, sugar has more (45:3 ratio) No

Molar mass The mass of one mole is called “molar mass” E.g. 1 mol Li = 6.94 g Li This is expressed as 6.94 g/mol What are the following molar masses? S SO2 Cu3(BO3)2 32.06 g/mol 64.06 g/mol 308.27 g/mol Calculate molar masses (to 2 decimal places) CaCl2 (NH4)2CO3 O2 Pb3(PO4)2 C6H12O6 Cu x 3 = 63.55 x 3 = 190.65 B x 2 = 10.81 x 2 = 21.62 O x 6 = 16.00 x 6 = 96.00 308.27

Molar mass The mass of one mole is called “molar mass” E.g. 1 mol Li = 6.94 g Li This is expressed as 6.94 g/mol What are the following molar masses? S SO2 Cu3(BO3)2 32.06 g/mol 64.06 g/mol 308.27 g/mol Calculate molar masses (to 2 decimal places) CaCl2 (NH4)2CO3 O2 Pb3(PO4)2 C6H12O6 110.98 g/mol (Ca x 1, Cl x 2) 96.11 g/mol (N x 2, H x 8, C x 1, O x 3) 32.00 g/mol (O x 2) 811.54 g/mol (Pb x 3, P x 2, O x 8) 180.18 g/mol (C x 6, H x 12, O x 6)

Converting between grams and moles If we are given the # of grams of a compound we can determine the # of moles, & vise-versa In order to convert from one to the other you must first calculate molar mass g = mol x g/mol mol = g  g/mol This can be represented in an “equation triangle” g mol g/mol g= g/mol x mol 0.25 HCl 53.15 H2SO4 3.55 NaCl 1.27 Cu Equation mol (n) g g/mol Formula 36.46 9.1 98.08 0.5419 mol= g  g/mol 58.44 207 g= g/mol x mol 63.55 0.0200 mol= g  g/mol

Simplest and molecular formulae Consider NaCl (ionic) vs. H2O2 (covalent) Cl Na Cl H O Na H O H O Cl Cl Na Na Chemical formulas are either “simplest” (a.k.a. “empirical”) or “molecular”. Ionic compounds are always expressed as simplest formulas. Covalent compounds can either be molecular formulas (I.e. H2O2) or simplest (e.g. HO) Q - Write simplest formulas for propene (C3H6), C2H2, glucose (C6H12O6), octane (C8H14) Q - Identify these as simplest formula, molecular formula, or both H2O, C4H10, CH, NaCl

For more lessons, visit www.chalkbored.com Answers For more lessons, visit www.chalkbored.com Q - Write simplest formulas for propene (C3H6), C2H2, glucose (C6H12O6), octane (C8H14) Q - Identify these as simplest formula, molecular formula, or both H2O, C4H10, CH, NaCl A - CH2 A - H2O is both simplest and molecular C4H10 is molecular (C2H5 would be simplest) CH is simplest (not molecular since CH can’t form a molecule - recall Lewis diagrams) NaCl is simplest (it’s ionic, thus it doesn’t form molecules; it has no molecular formula) CH CH2O C4H7