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JF Tutorial: Mole CalculationsJF Tutorial: Mole CalculationsShane PlunkettShane Plunkettplunkes@tcd.ieplunkes@tcd.ie

1. Some Mathematical Functions2. What is a mole?• Avogadro’s Number• Converting between moles and mass• Calculating mass % from a chemical formula• Determining empirical and molecular formulae from mass

Recommended reading• T.R. Dickson, Introduction to Chemistry, 8th Ed., Wiley, Chapters 2 & 4• M.S. Silberberg, Chemistry, The Molecular Nature of Matter and Change, 3rd Ed., Chapter 3• P. Atkins & L. Jones, Molecules, Matter and Change, 3rd Ed., Chapter 2• Multiple choice tests: http://www.mhhe.com/silberberg3

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Carrying out Carrying out CalculationsCalculations

In chemistry, must deal with several mathematical functions. In chemistry, must deal with several mathematical functions.

1. Scientific Notation• Makes it easier to deal with large numbers, especially

concentrations• Written as A ×10b, where A is a decimal number and b is a whole

numberExample: Avogadro’s number

602 213 670 000 000 000 000 000It is very inconvenient to write this. Instead, use scientific notation:

6.022 × 1023

Calculators:•Sharp & Casio Type in 6.022

Press the exponential function [EXP]Key in 23

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QuestionsQuestions

(a) 784000000(a) 784000000

(b) 0.00023(b) 0.00023

(c) 9220000(c) 9220000

(d) 0.000000015(d) 0.000000015

How would you write the following:

7.84 × 108

2.3 × 10-4

9.22 × 106

1.5 × 10-8

(a) (1.38 × 104) × (8.21 × 106)

Calculate the following:

1.13 × 1011

(b) (8.56 × 10-8) × (2.39 × 104) 2.05 × 10-3

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Common Decimal PrefixesCommon Decimal PrefixesPrefixPrefix SymbSymb

ololNumberNumber WordWord Exponential Exponential

NotationNotationteratera TT 1,000,000,000,001,000,000,000,00

00trilliontrillion 10101212

gigagiga GG 1,000,000,0001,000,000,000 billionbillion 101099

MegaMega MM 1,000,0001,000,000 millionmillion 101066

kilokilo kk 1,0001,000 thousandthousand 101033

hectohecto hh 100100 hundredhundred 101022

decadeca dada 1010 tenten 101011

decideci dd 0.10.1 tenthtenth 1010-1-1

centicenti cc 0.010.01 hundredthhundredth 1010-2-2

millimilli mm 0.0010.001 thousandtthousandthh

1010-3-3

micromicro 0.0000010.000001 millionthmillionth 1010-6-6

nanonano nn 0.0000000010.000000001 billionthbillionth 1010-9-9

picopico pp 0.0000000000010.000000000001 trillionthtrillionth 1010-12-12

femtofemto ff 0.00000000000000.00000000000000101

quadrillionquadrillionthth

1010-15-15

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2. 2. LogarithmsLogarithms Makes dealing with a wide range of numbers more convenient, Makes dealing with a wide range of numbers more convenient,

especially pHespecially pH Two types: common logarithms and natural logarithmsTwo types: common logarithms and natural logarithms

Common Logarithms

• Common log of x is denoted log x

• gives the power to which 10 must be raised to equal x

• 10n = x

• written as: log10x = n (base 10 is not always specified) Example:

The common log of 1000 is 3, i.e. 10 must be raised to the power of 3 to get 1000

Written as: log101000 = 3 103 = 1000

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CalculatorsCalculators

Sharp:Sharp: Press the [LOG] functionPress the [LOG] functionType the numberType the numberHit answerHit answer

Casio:Casio: Key in the numberKey in the numberPress the [LOG] functionPress the [LOG] function

Questions

(a) 10

(b) 1,000,000

(c) 0.001

(d) 853

1

6

-3

2.931

Calculate the common logarithms of the following:

log 10

log 1000000

log 0.001

log 853

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Natural logarithmsNatural logarithms

Natural log of Natural log of xx is denoted ln is denoted ln xx the difference here is, instead of base 10, we have base the difference here is, instead of base 10, we have base ee

(where (where ee = 2.71828) = 2.71828) Gives the power to which Gives the power to which ee must be raised to equal must be raised to equal xx lnx lnx oror log logeex = n x = n oror e enn = x = x

Example

The natural log of 10 is 2.303, i.e. e must be raised to the power of 2.303 to get 10

Calculators

• Sharp: Press the [ln] function

Enter the number and hit answer

• Casio: Enter the number

Press the [ln] function

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QuestionsQuestions

What is the natural log of:What is the natural log of:

(a) 50

(b) 1.25 × 105

(c) 2.36 × 10-3

(d) 8.98 × 1013

3.91

11.74

-6.05

32.13

ln 50

ln 1.25x105

ln 2.36x10-3

ln 8.98x1013

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3. 3. GraphsGraphs

Experimental data often represented in graph form, especially in Experimental data often represented in graph form, especially in straight linesstraight lines

Equation of straight line given byEquation of straight line given byy = mx + cy = mx + c

where where x and y are the axes valuesx and y are the axes valuesm is the slope of the graphm is the slope of the graphc is the intercept of the plotc is the intercept of the plot

y- axis

x-axis

Slope

Intercept

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Sign of slope tells you the direction of the lineSign of slope tells you the direction of the line Magnitude of slope tells you steepness of lineMagnitude of slope tells you steepness of line Slope found by taking two x values and the two corresponding y values Slope found by taking two x values and the two corresponding y values

and substituting these into the following relation:and substituting these into the following relation:

xy

xxyym

12

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Example

Given the (x, y) coordinates (2, 4) and (5, 9), find the slope of the line containing these two points.

x1 = 2 y1 = 4 x2 = 5 y2 = 9

Sub into above relation: m = 9 – 4 = 5 = 1.675 – 2 3

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4. Quadratic 4. Quadratic EquationsEquations

May be encountered when dealing with concentrations May be encountered when dealing with concentrations Involve Involve xx22 ( (xx-squared terms)-squared terms) Take the form aTake the form axx22 + b + bxx + c = 0 + c = 0 Can be solved by:Can be solved by:

aacbbx

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• this expression finds the roots or the solution for x of the quadratic equation

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Find the roots of the equation xFind the roots of the equation x22 – 6x + 8 = 0 – 6x + 8 = 0

)1(2)8)(1(4)6()6( 2

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Example:

a = 1 b = -6 c = 8

x =

x =

x = =2

26

Therefore, x = 2

26 or2

26 = 4 = 2

aaxx22 + b + bxx + c = 0 + c = 0

aacbbx

242

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QuestionQuestionYou have been asked to calculate the concentration of [HYou have been asked to calculate the concentration of [H33OO++] ions in a] ions in a

chemical reaction.chemical reaction.

aacbbx

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x = [H3O+]

The following quadratic equation has been given Solve for x.

2.4x2 + 1.5x – 3.6 = 0

)4.2(2)6.3)(4.2(4)5.1(5.1 2

x

8.456.3425.25.1

x

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8.481.365.1

x

8.4067.65.1

x

8.4067.65.1

xTherefore or8.4

067.65.1 x

x = 0.95 or x = -1.58

Because we are dealing with concentrations, a negative value will not make sense. Therefore, we report the positive x value, 0.95, as our answer. Never round up this number!

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Important formulae so far…Important formulae so far…y = mx + cy = mx + c

xy

xxyym

12

12

aaxx22 + b + bxx + c = 0 + c = 0

aacbbx

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Graphs…..

Quadratic equations…..

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Calculations: The Calculations: The MoleMole Stoichiometry is the study of quantitative aspects of chemicalStoichiometry is the study of quantitative aspects of chemical

formulas and reactionsformulas and reactions Mole: SI unit of the amount of a substanceMole: SI unit of the amount of a substance

DefinitionDefinition::A mole is the number of atoms in exactly 12g of theA mole is the number of atoms in exactly 12g of the

carbon-12 isotopecarbon-12 isotope

This number is called Avogadro’s number and is given by 6.022 This number is called Avogadro’s number and is given by 6.022 ×10×102323

The mole is The mole is NOTNOT just a counting unit, like the dozen, which specifies only the just a counting unit, like the dozen, which specifies only the number of objects. The definition of a mole specifies the number of objects number of objects. The definition of a mole specifies the number of objects in a in a fixed massfixed mass of substance. of substance.

Mass spectrometry tells us that the mass of a carbon-12 atom isMass spectrometry tells us that the mass of a carbon-12 atom is1.99261.9926×10×10-23-23g.g.

No. of carbon-12 atoms = atomic mass (g) mass of one atom (g)

= 12g _

1.9926×10-23g= 6.022 ×1023 atoms

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Other definitions of the MoleOther definitions of the Mole One mole contains Avogadro’s Number One mole contains Avogadro’s Number

(6.022 x 10(6.022 x 102323)) A mole is the amount of a substance of a A mole is the amount of a substance of a

system which contains as many elementary system which contains as many elementary entities as there are atoms in 0.012kg entities as there are atoms in 0.012kg (or12g) of Carbon-12(or12g) of Carbon-12

A mole is that quantity of a substance whose A mole is that quantity of a substance whose mass in grams is the same as its formula mass in grams is the same as its formula weightweight

E.g. Fe55.85E.g. Fe55.85 Iron has an atomic mass or 55.85Iron has an atomic mass or 55.85g molg mol-1-1, so , so

one moleone mole of iron has a mass or 55.85 of iron has a mass or 55.85gg

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One mole of any object One mole of any object alwaysalways means 6.022 means 6.022 × 10× 102323 units units of those of those objects.objects.For example,For example, 1 mol of H1 mol of H22OO contains 6.022 × 10contains 6.022 × 102323 molecules molecules

1 mol of NaCl contains 6.022 × 101 mol of NaCl contains 6.022 × 102323 formula units formula units

Avogadro’s number is used to convert between the number ofAvogadro’s number is used to convert between the number ofmoles and the number of atoms, ions or molecules.moles and the number of atoms, ions or molecules.

Example

0.450mol of iron contains how many atoms?

Number of atoms = number of moles × Avogadro’s number (NA)

Therefore No. of atoms = (0.450mol) × (6.022 × 1023)

= 2.7 × 1023 atoms

Calculating the number of particles

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ExampleExampleHow many molecules are there in 4 moles of hydrogen peroxide (HHow many molecules are there in 4 moles of hydrogen peroxide (H22OO22)?)?

No. of molecules = no. of moles × Avogadro’s number (NA)

= 4mol × (6.022 × 1023 mol-1)

= 24 ×1023 molecules

= 2.4 × 1024 molecules

QuestionsQuestions

How many atoms are there in 7.2 moles of gold (Au)?How many atoms are there in 7.2 moles of gold (Au)?Answer: 4.3 × 1024 atoms

The visible universe is estimated to contain 10The visible universe is estimated to contain 102222 stars. How many moles stars. How many moles of stars are there?of stars are there?

Answer: 1022 stars = 1022 = 0.17 mol.

6.022×1023

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Calculating the mass of one Calculating the mass of one moleculemolecule

Example:Example: What is the mass of one molecule of water? What is the mass of one molecule of water?

Step 1: Calculate the molar mass of water HO

H

waterMolar mass of water = (2 × atomic mass H) + (1 × atomic mass O)

Molar mass H2O = (2 × 1.008g mol-1) + (1 ×16.000g mol-1)

= 18.00 g mol-1

Step 2: Employ Avogadro’s number

Mass of one molecule = Molar mass Avogadro’s no.

= 18.00g mol-1 6.022×1023mol-1

= 2.992×10-23g

Note: Always check the units you have in your answer to ensure you are correct

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ExampleExampleCalculate the mass of one molecule of ammonium carbonate [(NHCalculate the mass of one molecule of ammonium carbonate [(NH44))22COCO33]]Step 1: Calculate the molar mass

2 Nitrogen atoms8 Hydrogen atoms1 Carbon atom3 Oxygen atoms

2 × 14.01gmol-1

8 × 1.008 gmol-1

1 ×12.01gmol-1

3 × 16.00 gmol-1

= 28.02 gmol-1

= 8.064 gmol-1

= 12.01 gmol-1

= 48.00 gmol-1

Total = 96.09 gmol-1Step 2: Employ Avogadro’s Number, NA

Mass of one molecule = 96.09 gmol-1 .

6.022×1023mol-1

QuestionsQuestionsCalculate the mass of one molecule of:Calculate the mass of one molecule of:

= 1.59 × 10-22g

(a) Ethanoic acid (CH3COOH)

(b) Methane (CH4)

(c) Potassium dichromate (K2Cr2O7)

9.96 × 10-23 g

2.66 × 10-23 g

4.89 × 10-22 g

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Converting between mass and Converting between mass and molesmolesIn the lab, we measure the mass of our reactants in grams using aIn the lab, we measure the mass of our reactants in grams using a

balance. However, when these react they do so in a ratio of moles.balance. However, when these react they do so in a ratio of moles.Therefore, we need to convert between the mass we measure and theTherefore, we need to convert between the mass we measure and thenumber of moles we require.number of moles we require.

The expression relating mass and number of moles is:Mass of sample (g) = no. of moles (mol) × molar mass (gmol-1)

ExampleCalculate the mass in grams in 0.75mol of sodium hydroxide, NaOHStep 1: Find the molar mass of the compound

Na: 22.99 gmol-1

O: 16.00 gmol-1

H: 1.008 gmol-1Mr: 40.00 gmol-1

Step 2: Substitute into the above expression

Mass of sample = 0.75mol × 40.00 gmol-1 = 30g

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QuestionsQuestions

Calculate the mass in grams present in:Calculate the mass in grams present in:(a) 0.57mol of potassium permanganate (KMnO4)

Answer: Molar mass KMnO4 = 158.03 gmol-1

Mass in grams = 0.57mol × 158.03 gmol-1 = 90.07 g

(b) 1.16mol of oxalic acid (H2C2O4)

Answer: Molar mass H2C2O4 = 90.04 gmol-1Mass in grams = 1.16mol × 90.04 gmol-1

= 104.44 g

(c) 2.36mol of calcium hydroxide (Ca(OH)2)Answer: Molar mass Ca(OH)2 = 74.1 gmol-1

Mass in grams = 2.36mol × 74.1 gmol-1 = 174.87 g

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Converting between moles and Converting between moles and massmass

Number of moles = Number of moles = mass of sample (g)mass of sample (g) molar mass (gmolmolar mass (gmol-1-1))

ExampleConvert 25.0g of KMnO4 to moles

Step 1: Calculate the molar mass

K

Mn

O

1 × 39.10 gmol-1

1 × 54.93 gmol-1

4 × 16.00 gmol-1

39.10 gmol-1

54.93 gmol-1

64.00 gmol-1Mr = 158.03 gmol-1

Step 2: Substitute into above expression

No. of moles = 25.0g .

158.03gmol-1= 0.158 mol

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QuestionsQuestions

Calculate the number of moles in:Calculate the number of moles in:

(a) 1.00g of water (H2O)Answer: Molar mass water = 18.02 gmol-1

1.00g H2O = 0.055mol

(b) 3.0g of carbon dioxide (CO2)Answer: Molar mass carbon dioxide = 44 gmol-1

3.0g CO2 = 0.068mol

(c) 500g of sucrose (C12H22O11)Answer: Molar mass sucrose = 342.30 gmol-1

500g C12H22O11 = 1.46mol

(d) 2.00g of silver chloride (AgCl)Answer: Molar mass silver chloride = 143.38 gmol-1

2.00g AgCl = 0.014mol

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Important formulae so far….Important formulae so far….No. of carbon-12 atoms = atomic mass (g)

mass of one atom (g)

No. of atoms = No. of moles × Avogadro’s number (NA)

No. of molecules = No. of moles × Avogadro’s number (NA)

Mass of one molecule = Molar mass Avogadro’s no.

Mass of sample (g) = no. of moles (mol) × molar mass (gmol-1)

Number of moles = mass of sample (g) molar mass (gmol-1)

Defining the mole:

Calculating the number of atoms or molecules, given the number of moles:

Most important equation:Most important equation:

Calculating the mass of an individual molecule:

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Calculating mass percentage from a chemical Calculating mass percentage from a chemical formulaformula

Many of the elements in the periodic table of the elements occur inMany of the elements in the periodic table of the elements occur incombination with other elements to form combination with other elements to form compoundscompounds..

A chemical formula of a compound tells you the composition of that compound in terms of the number of atoms of each element present.

The mass percentage composition allows you to determine the fraction of the total mass each element contributes to the compound.

ExampleAmmonium nitrate (NH4NO3) is an important compound in the fertiliser industry. What is the mass % composition of ammonium nitrate?

Step 1: Calculate the molar mass of ammonium nitrate

Molar mass NH4NO3 = 80.05 gmol-1

Two N atoms: 28.016 gmol-1

Four H atoms: 4.032 gmol-1

Three O atoms: 48.00 gmol-1

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Step 2: Determine the mass % composition for each elementStep 2: Determine the mass % composition for each element

Nitrogen: 28.016g N in one mol of ammonium nitrateNitrogen: 28.016g N in one mol of ammonium nitrateMass fraction of N = 28.016g

80.05g

Mass % composition of N = 28.016g × 100% 80.05g

= 34.99% ≈ 35%

Hydrogen: 4.032g H in one mol of ammonium nitrate

Mass fraction of N = 4.032g 80.05g

Mass % composition of H = 4.032g × 100%80.05g

= 5.04% ≈ 5%Oxygen: 48.00g O in one mol of ammonium nitrateAs above, the mass % composition of O is found to be 60%

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Therefore, the mass % composition of ammonium nitrate (NHTherefore, the mass % composition of ammonium nitrate (NH44NONO33) is:) is:

% Nitrogen:% Nitrogen: 35%35%% Hydrogen:% Hydrogen: 5%5%% Oxygen:% Oxygen: 60%60%

To check your answer, make sure it adds up to 100%

Question What is the mass % composition of C12H22O11?

Answer: % Carbon: 42.1%

% Hydrogen: 6.5%

% Oxygen: 51.4%

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Determining empirical formula from Determining empirical formula from massmass

The empirical formula of a compound tells you the relative number ofThe empirical formula of a compound tells you the relative number ofatoms of each element present in that compound. It gives you theatoms of each element present in that compound. It gives you thesimplest ratio of the elements in the compound.simplest ratio of the elements in the compound.

For example, the empirical formula of glucose (C6H12O6) is CH2O, giving the C:H:O ratio of 1:2:1

If you know the mass % composition and the molar mass of elements present in a compound, you can work out the empirical formula

Example

What is the empirical formula of a compound which has a mass % composition of 50.05% S and 49.95% O?

Step 1: Find the atomic masses of the elements present

Sulfur (S) : 32.066 gmol-1 Oxygen (O) : 16.000 gmol-1

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Step 2: Determine the number of moles of each element presentStep 2: Determine the number of moles of each element presentSince we are dealing with percentages, we can express the mass % asSince we are dealing with percentages, we can express the mass % as

grams if we assume we have 100g of the compound.grams if we assume we have 100g of the compound.

Therefore, 100g of our compound contains 50.05g of sulfur and 49.95g of oxygen.

Convert number of grams to number of molesNumber of mol Sulfur = mass of sulfur in sample (g)

atomic mass of sulfur (gmol-1)= 50.05g .

32.066 gmol-1= 1.56 mol

Similarly, the no. of mol of Oxygen is found to be 3.12molStep 3: Determining the ratios of elements

Sulfur: 1.56molOxygen: 3.12mol

Ratio 1.56 : 3.12

Ratio must be in whole numbers. Here we must divide across by 1.56Therefore, we have a ratio of 1:2 giving an empirical formula of SO2

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QuestionQuestionDetermine the empirical formula of a compound that contains 27.3Determine the empirical formula of a compound that contains 27.3mass% Carbon and 72.7 mass% Oxygen.mass% Carbon and 72.7 mass% Oxygen.

Answer: No. of mol Carbon = 2.27molNo. of mol Oxygen = 4.54mol

Ratio 1:2 Empirical formula CO2

Monosodium glutamate (MSG) has the following mass percentage composition: 35.51% C, 4.77 % H, 37.85% O, 8.29% N, and 13.60% Na. What is its molecular formula if its molar mass is 169 gmol-1? 

Answer: C5H8O4NNa

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Important calculationsImportant calculations

Calculating mass percentage from a chemical formula Step 1: Calculate the molar mass Step 2: Determine the mass % composition for each element

Determining empirical formula from mass Step 1: Find the atomic masses of the elements present Step 2: Determine the number of moles of each element present Step 3: Determining the ratios of elements

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Molarity Molarity Some chemical reactions involve aqueous solutions of reactantsSome chemical reactions involve aqueous solutions of reactants

The concentration of a solution is the amount of solute present in a given quantity of solvent or solution

This concentration may be expressed in terms of molarity (M) or molar concentration:

M = Molarity = no. of molesvolume in Litres

Molarity is the number of moles of solute in 1 Litre (L) of solution

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What is molarity of an 85.0mL ethanol (CWhat is molarity of an 85.0mL ethanol (C22HH55OH) solution containing OH) solution containing 1.77g of ethanol?1.77g of ethanol?

Step 1: Determine the number of moles of ethanolStep 1: Determine the number of moles of ethanol

Example

Molar mass of ethanol, C2H5OH:

2 × carbon atoms1 × oxygen atom6 × hydrogen atoms

2 × 12.01 gmol-1

1 × 16.00 gmol-1

6 × 1.008 gmol-1

24.02 gmol-1

16.00 gmol-1

6.048 gmol-1

46.07 gmol-1

No. of moles = mass in g molar mass

No. of moles ethanol = 1.77g .

46.07 gmol-1= 0.038 mol

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Step 2: Convert to molarity Step 2: Convert to molarity

Have 85.0mL ethanolHave 85.0mL ethanol1 L = 1000mL

Have 0.085 L of ethanol

Molarity = no. of moles volume in L

= 0.038 mol 0.085 L

= 0.45 molL-1 ≡ 0.45 M

Questions

Calculate the molarities of each of the following solutions:(a) 2.357g of sodium chloride (NaCl) in 75mL solution

Answer: 0.5378 M(b) 1.567mol of silver nitrate (AgNO3) in 250mL solution

Answer: 6.268 M(c) 10.4g of calcium chloride (CaCl2) in 2.20 × 102 mL of solution

Answer: 0.426 M

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Example Example An antacid tablet is not pure CaCOAn antacid tablet is not pure CaCO33; it contains starch, flavouring, etc. ; it contains starch, flavouring, etc. If it takes 41.3mL of 0.206 M HCl to react with all the CaCOIf it takes 41.3mL of 0.206 M HCl to react with all the CaCO33 in one in onetablet, how many grams of CaCOtablet, how many grams of CaCO33 are in the tablet. You are given the are in the tablet. You are given thefollowing balanced equation:following balanced equation:

2HCl(aq) + CaCO3(s) CaCl2(aq) + H2O(l) + CO2(g)

Step 1: Determine the no. of moles of HCl that react

Have 0.206 M HCl solution have 0.206 mol in one litre

Have 41.3 mL of HCl solution have 0.0413 L of HCl solution

Molarity = no. of moles volume in L

no. of moles = Molarity × volume in L= 0.206 molL-1 × 0.0413L= 0.0085 mol

≡ 8.5 × 10-3 mol HCl

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Step 2: Determine no. of moles of CaCOStep 2: Determine no. of moles of CaCO33 used in the reaction used in the reaction

2HCl2HCl(aq)(aq) + CaCO + CaCO3(s) 3(s) CaCl CaCl2(aq) 2(aq) + H+ H22OO((ll) ) + CO+ CO2(g)2(g)

From the balanced equation, we can see that 2 moles of HCl are required to react with one mole of CaCO3

Therefore, if 8.5 × 10-3 mol of HCl are present in the reaction, we must have 4.25 × 10-3 mol of CaCO3 present.

Molar mass of CaCO3:1 × calcium atom1 × carbon atom3 × oxygen atoms

1 × 40.08 gmol-1

1 × 12.01 gmol-1

3 × 16.00 gmol-1

40.08 gmol-1

12.01 gmol-1

48.00 gmol-1100 gmol-1

No. of mols = mass in g molar mass

Mass in g = no. of mols × molar mass

= (4.25 × 10-3 mol) × (100 gmol-1)= 0.425 g CaCO3 present in tablet

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Questions Questions

(a) How many moles of NaCl are present in 25.00mL of 1.85M NaCl(a) How many moles of NaCl are present in 25.00mL of 1.85M NaCl(aq)(aq)??

Answer: 4.62 × 10-2 mol NaCl

(b) What volume of a 1.25 × 10-3 M solution of C6H12O6(aq) contains 1.44 × 10-6 mol of glucose?

Answer: 1.15 mL

(c) If stomach acid, given as 0.1 M HCl, reacts completely with an antacid tablet containing 500mg of CaCO3, what volume of acid in millilitres will be consumed? The balanced equation is:

CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l)

Answer: 100mL acid

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Important formulae…Important formulae…

No. of moles = mass in g molar mass

Molarity = no. of moles volume in L

Calculating the number of moles:

Calculating the molarity or concentration: