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

Mr. Palmarin Chapter 3 - The Mole and Stoichiometry 1 / 50

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.


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.


Examples:

1) 2 890 000 000

While the decimal is not visible, it’s actually present on the far right:2890000000.

Now, move the decimal 9 places to the left. Since you moved it left,

the exponent is positive. 2.89× 109

2) 4 327 500

Identical to 1). Simply move the decimal left until you’re left with a

number between 1 and 10. 4.3275× 106

3) 0.0000073

To move the decimal to arrive at a number between 1 and 10, you’llhave to move it to the right (6 places). Because you moved it right,

the exponent is negative. 7.3× 10−6

4) 0.00045 4.5× 10−4


Practice: [See Section 3.1 Video]

1) 60 000 000

2) 0.000009

3) 6 340 000

4) 0.000000607

5) 0.00027


Example: Rewrite the following numbers in decimal notation.

1) 7× 107

To go the reverse direction, simply move the decimal right or leftdepending on the sign of the exponent. Here, the exponent is positive,so we move the decimal to the right (remember, we’re workingbackwards). 70 000 000

2) 5× 10−4

The exponent is negative, so we move the decimal to the left. Youshould expect a small number to be formed when the exponent isnegative. 0.0005

3) 1.1× 105

Move the decmial 5 to the right. 110 000


Practice: Rewrite the following numbers in decimal notation.[See Section 3.1 Video]

1) 2× 106

2) 7.53× 10−3

3) 9.77× 104


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.


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?


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.).


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

1) 306

Locate the zeroes (there’s only one here). Is that zero significant?

Reference rule 1. Yes it is significant. 3 s.f.

2) 2070

There’s one zero between to numbers, so it’s significant. The trailingzero is not significant because a decimal point is not present. 3 s.f.

3) 0.0065

Zeroes out front never are significant. 2 s.f.


Practice: Determine the number of significant figures in the followingmeasurements. [See Section 3.1 Video]

1) 0.350

2) 3000

3) 4050


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.


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

1) 20.42 + 1.322

Use Rule 1 (least number of decimals). 21.74

2) 0.08 - 0.0052

Use Rule 1 again. 0.07

3) 2.3 x 10.12

Use Rule 2 (report your answers to the lowest number of s.f. in thequestion). 23

4) 0.0003 / 720

Use Rule 2 again. Round to 1 s.f. Your calculator may convert this toscientific notation. In scientific notation, you only count s.f. for the

decimal portion. 4× 10−7

5) (42.023)(12.2) 512

6)(22.99

3.5

)( 2.0

37.45

)0.35


Practice: Add, subtract, multiply, or divide the following numbers.

1) 15.5 + 2.35

2) 0.016 - 0.22

3) 42.023 x 12.2

4) 23.1 / 12


Combined Calculations Using Significant Figures

1)(1.07− 0.08826)

0.762

0.98174

0.762= 1.3 (see Section 3.1 Video for explanation)

2) (0.91 + 1.2 + 8.4)(3.70)

(10.51)(3.70) = 38.887 = 38.9

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.


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


Example: Use dimensional analysis to perform the following conversions.

1) 35 centimetres to metres

35��cm

(1 m

100��cm

)= 0.35 m

Don’t underestimate this technique. It must be used for the followingmaterial. Become familiar with converting all units in this manner - itwill be worth it in the end.

2) 2180 metres to kilometres

2180��m

(1 km

1000��m

)= 2.18 km

3) 565 900 seconds to hours

565 900�s

(1��min

60�s

)(1 h

60��min

)= 157.2 h

Note: This conversion is done in one big step. This is critical. Doingconversions in multiple steps may allow rounding errors to creep in.


Practice: Use dimensional analysis to perform the following conversions.[See Section 3.2 Video]

1) 180 grams to kilograms

2) 1.25 litres to millilitres

3) 45 minutes to days


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.(

515��m

1�s

)(1 km

1000��m

)(60�s

1��min

)(60��min

1 h

)= 1854 km/h = 1850 km/h

Notice how the answer is rounded to 3 s.f. This is because the originalvalue used (515) has 3 s.f. Always round your answers to the lowestnumber of significant figures stated in the question that is used inthe calculation.

2) Convert 20 feet per second to miles per hour. [1 mi = 5280 ft][See Section 3.2 Video]


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


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”


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

Example: Name the object involved in the following measurements (eitheratoms, ions, or molecules).

1) 1 mol Fe = 6.02× 1023 of Fe

2) 1 mol Na+ = 6.02× 1023 of Na+

3) 1 mol H2 = 6.02× 1023 of H2

4) 1 mol NaCl = 6.02× 1023 of NaCl

Answers: 1) atoms 2) ions 3) molecules 4) molecules


Here’s the end goal:

Mass (g)

↓↑

Particles � Mole � Volume (L)

We wish to be able to convert easily from particles to volume to mass tomoles. Notice how the mole is at the centre of all conversions.


Particles and Moles

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

=⇒ 1 mol

6.02× 1023 particlesor

6.02× 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


Example:

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

3.50��mol Ag

(6.02× 1023 atoms

1��mol

)= 2.11× 1024 atoms Ag

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

5.34× 1023 ��atoms Au

(1 mol

6.02× 1023��atoms

)= 0.887 mol Au

Notice how each answer is rounded to the correct number ofsignificant figures. That is, the least number of s.f. in each question thatis involved in the calculations.



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?


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


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


Examples:

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

10 �L

(1 mol

22.4 �L

)= 0.446 mol = 0.4 mol CH4 (1 s.f.)

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

2.4��mol

(22.4 L

1��mol

)= 53.76 L = 54 L CO2 (2 s.f.)



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

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


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.


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).


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


Examples: Find the molar mass of the following elements.

1) oxygen

Very simple. Reference the atomic weights from the periodic table. Do

this for each. 16.00 g/mol

2) silver 107.87 g/mol

3) uranium 238.03 g/mol

4) hydrogen 1.008 g/mol


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

Follow the steps above. Carbon’s mass is 12.01 g, chlorine’s mass is35.45 g. We have 1 carbon and 4 chlorines. Therefore, the calculation

is: 12.01 + 4(35.45) = 153.81 g/mol

2) C6H12O6 6(12.01) + 12(1.008) + 6(16.00) = 180.16 g/mol


Practice: Find the molar mass of the following compounds to twodecimals. [See Section 3.4 Video]

1) CO2

2) Ca(NO3)2

3) silver carbonate


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.


Examples:

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

35 �g

(1 mol

107.87 �g

)= 0.32 mol Ag

Note: Notice how the conversion factor was the molar mass of silver:107.87 g/mol. You can flip it depending on what you wish to cancel.

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

18.5��mol

(40.08 g

1��mol

)= 741 g Ca

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

486 �g

(1 mol

180.16 �g

)= 2.70 mol C6H12O6

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

12.05��mol

(63.018 g

1��mol

)= 759.4 g HNO3



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?


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.


Examples:

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

3.50× 1023��particles

(1��mol


)(22.4 L

1��mol

)= 13.0 L iodine

Use the diagram in the above slide. Put your finger on the unit thatyou start with (in this case “particles”). Then follow the path towhere to want to go (in this case “litres”). The diagram informs youthat you need 2 conversions. Both should be familiar to you already.

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


(1��mol


)(22.99 g

1��mol

)= 57.3 g

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

30.0 �L

(1��mol

22.4 �L

)(6.02× 1023 molecules

1��mol

)= 8.06× 1023 molecules H2



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).


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


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

(1) (2)


Examples:

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

M =mol

L=⇒ M =

1.05 mol

2.00 L= 0.525 M (3 s.f.)

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

M =mol

L=⇒ mol = (M)(L) = (0.5M)(2.75 L) = 1.4 mol (2 s.f.)

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

=⇒ L =mol

M=⇒ L =

35.5 mol

2.0M= 18 L (2 s.f.)



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?


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


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?

Use the formula: C1V1=C2V2

C1 = 12.0 M V1 = ?C2 = 2.0 M V2 = 100.0 mL

Now plug the values into the formula and solve for V1.

(12.0M)(V1) =(2.0M)(100.0 mL)

V1 =(2.0��M)(100.0 mL)

12.0��MV1 =17 mL

Note: As usual, we round our answer to the lowest number of significantfigures in the question (2 s.f. in this case).



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?


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