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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. JF Tutorial: Mole Calculations. Shane Plunkett [email protected] Recommended reading

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jf tutorial mole calculations

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

JF Tutorial: Mole Calculations

Shane Plunkett

[email protected]

  • 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

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

Carrying out Calculations

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

In chemistry, must deal with several mathematical functions.

  • Calculators:
  • Sharp & Casio Type in 6.022
  • Press the exponential function [EXP]
  • Key in 23
questions
Questions

How would you write the following:

7.84 × 108

2.3 × 10-4

9.22 × 106

1.5 × 10-8

(a) 784000000

(b) 0.00023

(c) 9220000

(d) 0.000000015

Calculate the following:

1.13 × 1011

(a) (1.38 × 104) × (8.21 × 106)

2.05 × 10-3

(b) (8.56 × 10-8) × (2.39 × 104)

2 logarithms
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
Calculators
  • Sharp: Press the [LOG] function

Type the number

Hit answer

  • Casio: Key in the number

Press the [LOG] function

Questions

Calculate the common logarithms of the following:

  • 10
  • 1,000,000
  • 0.001
  • 853

log 10

log 1000000

log 0.001

log 853

1

6

-3

2.931

natural logarithms
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
questions1
Questions

What is the natural log of:

  • 50
  • 1.25 × 105
  • 2.36 × 10-3
  • 8.98 × 1013

ln 50

ln 1.25x105

ln 2.36x10-3

ln 8.98x1013

3.91

11.74

-6.05

32.13

3 graphs
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

Slope

Intercept

x-axis

slide10
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
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:
  • this expression finds the roots or the solution for x of the quadratic equation
find the roots of the equation x 2 6x 8 0

Example:

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

ax2 + bx + c = 0

a = 1

b = -6

c = 8

x =

x =

x =

=

= 2

or

= 4

Therefore, x =

question
Question

x = [H3O+]

The following quadratic equation has been given Solve for x.

2.4x2 + 1.5x – 3.6 = 0

You have been asked to calculate the concentration of [H3O+] ions in a

chemical reaction.

slide14

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
Important formulae so far…

y = mx + c

Graphs…..

ax2 + bx + c = 0

Quadratic

equations…..

calculations the mole
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
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
slide18

One mole of any object always means 6.022 × 1023units of those objects.For example, 1 mol of H2Ocontains 6.022 × 1023 molecules 1 mol of NaCl contains 6.022 × 1023 formula units

Calculating the number of particles

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

example
Example

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

= 4mol × (6.022 × 1023 mol-1)

How many molecules are there in 4 moles of hydrogen peroxide (H2O2)?

= 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
Calculating the mass of one molecule

Step 1: Calculate the molar mass of water

Example: What is the mass of one molecule 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

example1
Example

Step 1: Calculate the molar mass

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

2 Nitrogen atoms

8 Hydrogen atoms

1 Carbon atom

3 Oxygen atoms

Calculate the mass of one molecule of ammonium carbonate [(NH4)2CO3]

Total = 96.09 gmol-1

Step 2: Employ Avogadro’s Number, NA

Mass of one molecule =

96.09 gmol-1 .

6.022×1023mol-1

= 1.59 × 10-22g

Questions

Calculate the mass of one molecule of:

  • 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
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

Calculate the mass in grams in 0.75mol of sodium hydroxide, NaOH

Step 1: Find the molar mass of the compound

Na: 22.99 gmol-1

Mr: 40.00 gmol-1

O: 16.00 gmol-1

H: 1.008 gmol-1

Step 2: Substitute into the above expression

Mass of sample =

0.75mol × 40.00 gmol-1

= 30g

questions2
Questions

(a) 0.57mol of potassium permanganate (KMnO4)

Answer: Molar mass KMnO4 = 158.03 gmol-1

Calculate the mass in grams present in:

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
Converting between moles and mass

Example

Convert 25.0g of KMnO4 to moles

Number of moles = mass of sample (g)

molar mass (gmol-1)

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

25.0g .

158.03gmol-1

No. of moles =

= 0.158 mol

questions3
Questions

(a) 1.00g of water (H2O)

Answer: Molar mass water = 18.02 gmol-1

Calculate the number of moles in:

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 far1
Important formulae so far….

Defining the mole:

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

mass of one atom (g)

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

Number of moles = mass of sample (g)

molar mass (gmol-1)

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

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

Calculating the mass of an individual molecule:

Mass of one molecule = Molar mass

Avogadro’s no.

Most important equation:

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

calculating mass percentage from a chemical formula
Calculating mass percentage from a chemical formula

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

Many of the elements in the periodic table of the elements occur in

combination with other elements to form compounds.

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

Two N atoms: 28.016 gmol-1

Four H atoms: 4.032 gmol-1

Three O atoms: 48.00 gmol-1

Molar mass NH4NO3 = 80.05 gmol-1

step 2 determine the mass composition for each element
Step 2: Determine the mass % composition for each element

Mass fraction of N = 28.016g

80.05g

Nitrogen: 28.016g N in one mol of ammonium nitrate

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 nh 4 no 3 is
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
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
Step 2: Determine the number of moles of each element present

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

Since we are dealing with percentages, we can express the mass % as

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

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

question1
Question

Answer: No. of mol Carbon = 2.27mol

No. of mol Oxygen = 4.54mol

Ratio 1:2 Empirical formula CO2

Determine the empirical formula of a compound that contains 27.3

mass% Carbon and 72.7 mass% Oxygen.

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
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
Molarity

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

Some chemical reactions involve aqueous solutions of reactants

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 c 2 h 5 oh solution containing 1 77g of ethanol

Example

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

Molar mass of ethanol, C2H5OH:

Step 1: Determine the number of moles of ethanol

2 × carbon atoms

2 × 12.01 gmol-1

24.02 gmol-1

1 × oxygen atom

1 × 16.00 gmol-1

16.00 gmol-1

6 × hydrogen atoms

6 × 1.008 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
Step 2: Convert to molarity

1 L = 1000mL

 Have 0.085 L of ethanol

Have 85.0mL 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

example2
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 caco 3 used in the reaction
Step 2: Determine no. of moles of CaCO3 used in the reaction

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.

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

Molar mass of CaCO3:

40.08 gmol-1

1 × calcium atom

1 × 40.08 gmol-1

12.01 gmol-1

1 × carbon atom

1 × 12.01 gmol-1

3 × oxygen atoms

3 × 16.00 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

questions4
Questions

Answer: 4.62 × 10-2 mol NaCl

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

(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
Important formulae…

Calculating the number of moles:

No. of moles = mass in g

molar mass

Calculating the molarity or concentration:

Molarity = no. of moles

volume in L

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