Jf tutorial: Mole Calculations



Download 33.39 Kb.
Date19.11.2017
Size33.39 Kb.

JF Tutorial: Mole Calculations

  • Shane Plunkett
  • plunkes@tcd.ie
  • Some Mathematical Functions
  • 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

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
  • Example: Avogadro’s 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
  • Hit answer
  • 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

4. Quadratic Equations

  • 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…..
  • Quadratic
  • 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
  • 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

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
  • 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:
  • 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%
  • 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%

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? 
  • Answer: C5H8O4NNa

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
  • Answer: 0.5378 M
  • (b) 1.567mol of silver nitrate (AgNO3) in 250mL solution
  • Answer: 6.268 M
  • Answer: 0.426 M

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?
  • 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

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:


Download 33.39 Kb.

Share with your friends:




The database is protected by copyright ©sckool.org 2020
send message

    Main page