# Jf tutorial: Mole Calculations

 Date 19.11.2017 Size 33.39 Kb.

## JF Tutorial: Mole Calculations

• Shane Plunkett
• plunkes@tcd.ie
• Some Mathematical Functions
• What is a mole?
• Converting between moles and mass
• Calculating mass % from a chemical formula
• Determining empirical and molecular formulae from mass
• 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

## Carrying out Calculations

• In chemistry, must deal with several mathematical functions.
• 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 number
• 602 213 670 000 000 000 000 000
• It 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

## Questions

• (a) 784000000
• (b) 0.00023
• (c) 9220000
• (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

## Common Decimal Prefixes

 Prefix Symbol Number Word Exponential Notation tera T 1,000,000,000,000 trillion 1012 giga G 1,000,000,000 billion 109 Mega M 1,000,000 million 106 kilo k 1,000 thousand 103 hecto h 100 hundred 102 deca da 10 ten 101 deci d 0.1 tenth 10-1 centi c 0.01 hundredth 10-2 milli m 0.001 thousandth 10-3 micro  0.000001 millionth 10-6 nano n 0.000000001 billionth 10-9 pico p 0.000000000001 trillionth 10-12 femto f 0.000000000000001 quadrillionth 10-15

## 2. Logarithms

• Makes dealing with a wide range of numbers more convenient, especially pH
• Two 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

## Calculators

• Sharp: Press the [LOG] function
• Type the number
• Casio: Key in the number
• Press the [LOG] function
• Questions
• 10
• 1,000,000
• 0.001
• 853
• 1
• 6
• -3
• 2.931
• Calculate the common logarithms of the following:
• log 10
• log 1000000
• log 0.001
• log 853

## Natural logarithms

• Natural log of x is denoted ln x
• the difference here is, instead of base 10, we have base e (where e = 2.71828)
• Gives the power to which e must be raised to equal x
• lnx or logex = n or en = 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

## Questions

• What is the natural log of:
• 50
• 1.25 × 105
• 2.36 × 10-3
• 8.98 × 1013
• 3.91
• 11.74
• -6.05
• 32.13
• ln 50
• ln 1.25x105
• ln 2.36x10-3
• ln 8.98x1013

## 3. Graphs

• Experimental data often represented in graph form, especially in straight lines
• Equation of straight line given by
• y = mx + c
• where x and y are the axes values
• m is the slope of the graph
• c is the intercept of the plot
• y- axis
• x-axis
• Slope
• Intercept

## Sign of slope tells you the direction of the line

• Sign of slope tells you the direction of the line
• Magnitude of slope tells you steepness of line
• Slope found by taking two x values and the two corresponding y values and substituting these into the following relation:
• 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 = 1.67
• 5 – 2 3

• May be encountered when dealing with concentrations
• Involve x2 (x-squared terms)
• Take the form ax2 + bx + c = 0
• Can be solved by:

## Find the roots of the equation x2 – 6x + 8 = 0

• Example:
• a = 1
• b = -6
• c = 8
• x =
• x =
• x =
• =
• Therefore, x =
• or
• = 4
• = 2
• ax2 + bx + c = 0

## Question

• You have been asked to calculate the concentration of [H3O+] ions in a
• chemical reaction.
• x = [H3O+]
• The following quadratic equation has been given Solve for x.
• 2.4x2 + 1.5x – 3.6 = 0
• Therefore
• or
• 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!

## Important formulae so far…

• y = mx + c
• ax2 + bx + c = 0
• Graphs…..
• equations…..

## Calculations: The Mole

• Stoichiometry is the study of quantitative aspects of chemical
• formulas and reactions
• Mole: SI unit of the amount of a substance
• Definition: A mole is the number of atoms in exactly 12g of the
• carbon-12 isotope
• This number is called Avogadro’s number and is given by 6.022 ×1023
• The mole is NOT just a counting unit, like the dozen, which specifies only the number of objects. The definition of a mole specifies the number of objects in a fixed mass of substance.
• Mass spectrometry tells us that the mass of a carbon-12 atom is
• 1.9926×10-23g.
• No. of carbon-12 atoms = atomic mass (g)
• mass of one atom (g)
• = 12g _
• 1.9926×10-23g
• = 6.022 ×1023 atoms

## Other definitions of the Mole

• One mole contains Avogadro’s Number (6.022 x 1023)
• A mole is the amount of a substance of a system which contains as many elementary entities as there are atoms in 0.012kg (or12g) of Carbon-12
• A mole is that quantity of a substance whose mass in grams is the same as its formula weight
• E.g. Fe55.85
• Iron has an atomic mass or 55.85g mol-1, so one mole of iron has a mass or 55.85g

## One mole of any object always means 6.022 × 1023 units of those objects. For example, 1 mol of H2O contains 6.022 × 1023 molecules 1 mol of NaCl contains 6.022 × 1023 formula units

• Avogadro’s number is used to convert between the number of
• 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

## Example

• How many molecules are there in 4 moles of hydrogen peroxide (H2O2)?
• No. of molecules = no. of moles × Avogadro’s number (NA)
• = 4mol × (6.022 × 1023 mol-1)
• = 24 ×1023 molecules
• = 2.4 × 1024 molecules
• Questions
• How many atoms are there in 7.2 moles of gold (Au)?
• Answer: 4.3 × 1024 atoms
• The visible universe is estimated to contain 1022 stars. How many moles of stars are there?
• Answer: 1022 stars = 1022 = 0.17 mol.
• 6.022×1023

## Calculating the mass of one molecule

• Example: What is the mass of one molecule of water?
• Step 1: Calculate the molar mass of water
• Molar 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
• = 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

## Example

• Calculate the mass of one molecule of ammonium carbonate [(NH4)2CO3]
• 2 Nitrogen atoms
• 8 Hydrogen atoms
• 1 Carbon atom
• 3 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-1
• Step 2: Employ Avogadro’s Number, NA
• Mass of one molecule =
• 96.09 gmol-1 .
• 6.022×1023mol-1
• Questions
• Calculate the mass of one molecule of:
• = 1.59 × 10-22g
• Ethanoic acid (CH3COOH)
• Methane (CH4)
• Potassium dichromate (K2Cr2O7)
• 9.96 × 10-23 g
• 2.66 × 10-23 g
• 4.89 × 10-22 g

## Converting between mass and moles

• In 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.
• Therefore, we need to convert between the mass we measure and the
• 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)
• Example
• Step 1: Find the molar mass of the compound
• Na: 22.99 gmol-1
• O: 16.00 gmol-1
• H: 1.008 gmol-1
• Mr: 40.00 gmol-1
• Step 2: Substitute into the above expression
• Mass of sample =
• 0.75mol × 40.00 gmol-1
• = 30g

## Questions

• 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-1
• Mass 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

## Converting between moles and mass

• Number of moles = mass of sample (g)
• molar mass (gmol-1)
• Example
• Convert 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-1
• Mr = 158.03 gmol-1
• Step 2: Substitute into above expression
• No. of moles =
• 25.0g .
• 158.03gmol-1
• = 0.158 mol

## Questions

• 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

## 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
• 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:
• Calculating the mass of an individual molecule:

## Calculating mass percentage from a chemical formula

• Many of the elements in the periodic table of the elements occur in
• combination with other elements to form compounds.
• 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.
• Example
• Ammonium 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

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

• Nitrogen: 28.016g N in one mol of ammonium nitrate
• Mass 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 nitrate
• As above, the mass % composition of O is found to be 60%

## Therefore, the mass % composition of ammonium nitrate (NH4NO3) is:

• % Nitrogen: 35%
• % Hydrogen: 5%
• % Oxygen: 60%
• Question
• What is the mass % composition of C12H22O11?
• % Hydrogen: 6.5%
• % Oxygen: 51.4%

## Determining empirical formula from mass

• The empirical formula of a compound tells you the relative number of
• atoms of each element present in that compound. It gives you the
• 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

## Step 2: Determine the number of moles of each element present

• Since we are dealing with percentages, we can express the mass % as
• 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 moles
• Number 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 3.12mol
• Step 3: Determining the ratios of elements
• Sulfur: 1.56mol
• Oxygen: 3.12mol
• Ratio 1.56 : 3.12
• Ratio must be in whole numbers. Here we must divide across by 1.56
• Therefore, we have a ratio of 1:2 giving an empirical formula of SO2

## Question

• Determine the empirical formula of a compound that contains 27.3
• mass% Carbon and 72.7 mass% Oxygen.
• Answer: No. of mol Carbon = 2.27mol
• No. 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?

## Important 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

## Molarity

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

## What is molarity of an 85.0mL ethanol (C2H5OH) solution containing 1.77g of ethanol?

• Step 1: Determine the number of moles of ethanol
• Example
• Molar mass of ethanol, C2H5OH:
• 2 × carbon atoms
• 1 × oxygen atom
• 6 × 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

## Step 2: Convert to molarity

• Have 85.0mL ethanol
• 1 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
• (b) 1.567mol of silver nitrate (AgNO3) in 250mL solution

## Example

• An antacid tablet is not pure CaCO3; it contains starch, flavouring, etc.
• If it takes 41.3mL of 0.206 M HCl to react with all the CaCO3 in one
• tablet, how many grams of CaCO3 are in the tablet. You are given the
• 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

## Step 2: Determine no. of moles of CaCO3 used in the reaction

• 2HCl(aq) + CaCO3(s) CaCl2(aq) + H2O(l) + CO2(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 atom
• 1 × carbon atom
• 3 × 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-1
• 100 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

## Questions

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

## 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: