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Chapter 3 - The Mole and Stoichiometry Part 1 - The Mole & Concentration Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 1 / 47

Chapter 3 - The Mole and Stoichiometry - Part 1 - The Mole ... · 1 mole = 6:02 1023 \objects" or \particles" =) 1 mol 6:02 1023 particles or 6:02 1023 particles 1 mol Entering scienti

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  • Chapter 3 - The Mole and StoichiometryPart 1 - The Mole & Concentration

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 1 / 47

  • Section 3.1 - Scientific Notation & Significant Figures

    In science, it is common to come across very large and small numbers. Forexample, a teaspoon of water (about 5 mL) contains about200,000,000,000,000,000,000,000 water molecules.

    Scientists have developed a particular notation for writing very large andvery small numbers called scientific notation.

    Scientific notation simplifies the writing of large and small numbers.

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 2 / 47

  • Scientific Notation

    1 Move the decimal place until you have a number between 1 and 10.

    2 Add a power of ten that corresponds to how many places you movedthe decimal.

    Note:

    A power of ten with a positive exponent, such as 105, means thedecimal was moved to the left.

    A power of ten with a negative exponent, such as 10−5, means thedecimal was moved to the right.

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 3 / 47

  • Examples:

    1) 2 890 000 000

    2) 4 327 500

    3) 0.0000073

    4) 0.00045

    5) 50 000

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 4 / 47

  • Practice:

    1) 60 000 000

    2) 0.000009

    3) 6 340 000

    4) 0.000000607

    5) 0.00027

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 5 / 47

  • Example: Rewrite the following numbers in decimal notation.

    1) 7× 107

    2) 5× 10−4

    3) 1.1× 105

    Practice: Rewrite the following numbers in decimal notation.

    1) 2× 106

    2) 7.53× 10−3

    3) 9.77× 104

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 6 / 47

  • Precision and Accuracy - What’s the Difference?

    Precision is a measure of how closely individual measurements agree withone another.

    Accuracy refers to how closely individual measurements agree with thecorrect or “true” value.

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 7 / 47

  • Precision and Accuracy - What’s the Difference?

    Precision is a measure of how closely individual measurements agree withone another.

    Accuracy refers to how closely individual measurements agree with thecorrect or “true” value.

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 7 / 47

  • Significant Figures

    All digits of a measured quantity are called significant figures. Everymeasurement has uncertainty. Numbers in which there is no uncertaintyare called exact numbers and are uncommon in science.

    What’s the temperature?

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 8 / 47

  • Significant Figures

    All digits of a measured quantity are called significant figures. Everymeasurement has uncertainty. Numbers in which there is no uncertaintyare called exact numbers and are uncommon in science.

    What’s the temperature?

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 8 / 47

  • Determining the Number of Significant Figures:

    1 Zeroes between nonzero digits are always significant.

    Ex: 1005 kg (four sig. figs.); 7.03 cm (three sig. figs.).

    2 Zeroes at the beginning of a number are never significant.

    Ex: 0.02 g (one sig. fig.); 0.0026 cm (two sig. figs.).

    3 Zeroes at the end of a number are significant if the numbercontains a decimal point. If no decimal is present, the trailingzeroes are not significant.

    Ex: 0.0200 g (three sig. figs.); 3.0 cm (two sig. figs.); 10 000 (onesig. fig.).

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 9 / 47

  • Example: Determine the number of significant figures in the followingmeasurements.

    1) 306

    2) 2070

    3) 0.0065

    Practice: Determine the number of significant figures in the followingmeasurements.

    1) 0.350

    2) 3000

    3) 4050

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 10 / 47

  • Significant Figures in Calculations

    Key point: The least certain measurement determines the number ofsignificant figures in the final answer.

    1 Addition & Subtraction: The result has the same number ofdecimal places as the measurement with the fewest decimal places.

    2 Multiplication & Division: The result contains the same number ofsignificant figures as the measurement with the fewest significantfigures. Round off the final result to the correct number of significantfigures.

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 11 / 47

  • Examples: Add, subtract, multiply, or divide the following numbers.

    1) 20.42 + 1.322

    2) 0.08 - 0.0052

    3) 2.3 x 10.12

    4) 0.0003 / 720

    5) (42.023)(12.2)

    6)(22.99

    3.5

    )( 2.037.45

    )

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 12 / 47

  • Combined Calculations Using Significant Figures

    1(1.07− 0.08826)

    0.762

    2 (0.91 + 1.2 + 8.4)(3.70)

    Note: When a calculation involves two or more steps and you writeanswers for intermediate steps, retain at least one non-significant digit forthe intermediate answers.

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 13 / 47

  • Section 3.2 - Dimensional Analysis

    Dimensional analysis is a systematic way of solving numerical problemsthat involve the conversion of units.

    The strategy:

    (((((

    Given unit × desired unit���

    ��given unit= desired unit

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 14 / 47

  • Example: Use dimensional analysis to perform the following conversions.

    1) 35 centimetres to metres

    2) 2180 metres to kilometres

    3) 565 900 seconds to hours

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 15 / 47

  • Practice: Use dimensional analysis to perform the following conversions.

    1) 180 grams to kilograms

    2) 1.25 litres to millilitres

    3) 45 minutes to days

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 16 / 47

  • Converting Units Using Two or More Conversion Factors

    1) The average speed of a nitrogen molecule in air at 25◦C is 515 metresper second (m/s). Convert this speed to kilometres per hour.

    2) Convert 20 feet per second to miles per hour. [1 mi = 5280 ft]

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 17 / 47

  • Section 3.3 - Particles, Volume, and the Mole

    In chemistry, the counting unit for numbers of atoms, ions, or molecules ina laboratory-size sample is called the mole (abbreviated “mol”).

    One mole is the amount of matter that contains as many objects as thenumber of atoms in exactly 12 g of pure 12C.

    From experiments, scientists have determined this number to be6.022 142 1× 1023 ≈ 6.02× 1023.

    Avogadro’s number: 6.02× 1023

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 18 / 47

  • A mole of atoms, a mole of molecules, or a mole of anything else allcontain Avogadro’s number of objects:

    1 mol C atoms = 6.02× 1023 C atoms

    1 mol H2O molecules = 6.02× 1023 H2O molecules

    1 mol NO –3 ions = 6.02× 1023 NO–

    3 ions

    Referencing the size of Avogadro’s number: Placing Avogadro’snumber of pennies side by side in a straight line would encircle the Earth300 trillion times.

    Watch: “Professor Martyn Poliakoff on the Mole”

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 19 / 47

    https://www.youtube.com/watch?v=2dzS_LXvYA0

  • Example: Name the object involved in the following measurements.

    1 mol Fe = 6.02× 1023 of Fe

    1 mol Na+ = 6.02× 1023 of Na+

    1 mol H2 = 6.02× 1023 of H2

    1 mol NaCl = 6.02× 1023 of NaCl

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 20 / 47

  • Here’s the end goal:

    Mass (g)

    ↓↑

    Particles � Mole � Volume (L)

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 21 / 47

  • Particles and Moles

    1 mole = 6.02× 1023 “objects” or “particles”

    =⇒ 1 mol6.02× 1023 particles

    or6.02× 1023 particles

    1 mol

    Entering scientific notation on a calculator: Find the EE or EXPbutton.

    6.02× 1023 = 6.02 EE 23

    = 6.02 EXP 23

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 22 / 47

  • Example:

    1) How many atoms are in 3.50 mol of silver?

    2) If you had 5.34× 1023 atoms of gold, how many moles do you have?

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 23 / 47

  • Practice:

    1) How many atoms are in 2.0 mol of copper?

    2) If you had 7.42× 1023 atoms of carbon, how many moles do you have?

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 24 / 47

  • Volume of a Mole

    Standard conditions: Set of conditions established to allow easycomparisons between sets of data.

    Standard Temperature and Pressure (STP):

    Temperature = 0◦C; Pressure = 100.00 kPa

    Standard Ambient Temperature and Pressure (SATP):Temperature = 25◦C; Pressure = 100.00 kPa

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 25 / 47

  • At Standard Temperature and Pressure (STP):

    1 mole of gas = 22.4 L

    Conversion factor:

    1 mol

    22.4 Lor

    22.4 L

    1 mol

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 26 / 47

  • Examples:

    1) At STP, convert 10 L of CH4 to moles.

    2) At STP, convert 2.4 mol of CO2 to L.

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 27 / 47

  • Practice:

    1) At STP, convert 5.55 L of O2 to moles.

    2) At STP, convert 4.2 mol of N2O3 to L.

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 28 / 47

  • Section 3.4 - Molar Mass

    The atomic mass of an element is given in atomic mass units (amu).Atomic masses of atoms are relative values.

    Carbon = 12.01 amu; Magnesium = 24.31 amu

    Therefore, magnesium is twice as heavy as carbon.

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 29 / 47

  • Recall: A mole is always the same number (6.02× 1023). However, 1-molsamples of different substances have different masses.

    Example:

    1 mol C =⇒ 12 amu (rounded)1 mol Mg =⇒ 24 amu (rounded)

    By definition, one mole of carbon has a mass of 12 g. Therefore, one moleof magnesium must have a mass of 24 g.

    Important Facts

    A mole is the amount of particles in exactly 12 g of 12C. Thisnumber is Avogadro’s number, which is 6.02× 1023.The unit for molar mass is grams per mole (g/mol).

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 30 / 47

  • General rule: The atomic weight of an element in atomic mass units(amu) is numerically equal to the mass in grams of 1 mol of that element.

    Examples:

    Cl has an atomic weight of 35.5 amu =⇒ 1 mol Cl has a mass of 35.5 g

    Au has an atomic weight of 197 amu =⇒ 1 mol Au has a mass of 197.0 g

    ∴ Cl has a mass of 35.5 g/mol and Au has a mass of 197.0 g/mol

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 31 / 47

  • Examples: Find the molar mass of the following elements.

    1) oxygen

    2) silver

    3) uranium

    4) hydrogen

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 32 / 47

  • How would you find the molar mass of 1 mole of a compound?

    1 Count the number of atoms of each element.

    2 Add the atomic masses of each atom together.

    Examples: Find the molar mass of the following compounds to twodecimals.

    1) CCl4

    2) C6H12O6

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 33 / 47

  • Practice: Find the molar mass of the following compounds to twodecimals.

    1) CO2

    2) Ca(NO3)2

    3) silver carbonate

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 34 / 47

  • Converting Moles to Mass and Mass to Moles

    Recall: The molar mass of a substance is always in g/mol.

    Conversion factors:

    ? g

    1 moleor

    1 mole

    ? g

    Note: The “?” is unknown because each substance has a different molarmass.

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 35 / 47

  • Examples:

    1) How many moles are in 35 g of Ag?

    2) What is the mass (in grams) of 18.5 mol of Ca?

    3) How many moles are in 486 g of C6H12O6?

    4) What is the mass (in grams) of 12.05 mol of HNO3?

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 36 / 47

  • Practice:

    1) How many moles are in 28.5 g of CuCl2

    2) How many moles are in 72 g of Na3PO4

    3) What is the mass (in grams) of 8.6 mol of NaCl?

    4) What is the mass (in grams) of 15.0 mol of Cu?

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 37 / 47

  • Section 3.5 - Multi-step Calculations

    The mole is at the center of all calculations. From the mole, we canconvert to grams, particles, or litres.

    Mass (g)

    ↓↑

    Particles � Mole � Volume (L)

    The goal is to complete all the calculations in one big step usingdimensional analysis.

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 38 / 47

  • Examples:

    1) Convert 3.50× 1023 particles of iodine to litres.

    2) Convert 1.50× 1024 particles of sodium to grams.

    3) How many molecules are present (at STP) in a 30.0 L container ofhydrogen gas?

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 39 / 47

  • Practice:

    1) Convert 58 L Cl2 to particles.

    2) Convert 105 g of sodium to particles.

    3) Convert 55 g of oxygen to litres.

    4) Convert 123 L Mg(OH)2 to milligrams (1000 mg = 1 g).

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 40 / 47

  • Section 3.6 - Molarity

    The concentration (c) of a solution is a measure of the amount of solutethat is dissolved in a given quantity of solvent.

    Molarity is the most common and useful measure of concentration inchemistry.

    Molarity =mol

    Lor M =

    mol

    L

    Common notation:24.0 mol

    L= 24.0 M

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 41 / 47

  • Which solution is more concentrated? i.e. Which beaker has a highermolarity?

    (1) (2)

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 42 / 47

  • Examples:

    1) If 1.05 mol of hydrochloric acid is dissolved in 2.00 L of water, what isthe concentration?

    2) How many moles of sodium hydroxide are in 2.75 L of a 0.5 M solution?

    3) How many litres of water are needed to make a 2.0 M solution that has35.5 mol of potassium hydroxide in it?

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 43 / 47

  • Practice:

    1 If you took 35 g of NaOH and dissolved it in 500.0 mL of water, what isthe concentration of your solution?

    2 How many grams of glucose (C6H12O6) are required to produce500.0 mL of a 3.0 M glucose solution?

    3 How many litres of a 2.0 M solution can be prepared from 40.5 g CuCl2?

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 44 / 47

  • Section 3.7 - Dilutions

    A solution is diluted whenever a solvent is added. The additional solvent(usually water) lowers the concentration. We can calculate theconcentration of a diluted solution using the following equation:

    C1V1=C2V2

    Where,C1 - original concentration of the solution

    V1 - original volume of solution

    C2 - new concentration of solution after dilution

    V2 - new volume of solution after dilution

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 45 / 47

  • Example: A stock solution of hydrochloric acid is 12.0 M. How much ofthe stock solution do you need to use if you want to make 100.0 mL of a2.0 M solution of HCl?

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 46 / 47

  • Practice:

    1) How many millilitres of 5.5 M NaOH are needed to prepare 300.0 mLof 1.2 M NaOH?

    2) What would be the concentration of a solution made by adding 250 mLof water to 45.0 mL of 4.2 M KOH?

    Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 47 / 47