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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