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Applying General Mathematics in Chemistry (HL IB Chemistry)

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Applying General Mathematics in Chemistry

Using arithmetic and algebraic calculations in chemistry

  • Chemistry often requires the use of calculations, which can include:
    • Decimals
      • Most chemical calculations use decimals, e.g. the concentration of a chemical
    • Fractions
      • These are most commonly used in uncertainty calculations
      • Most scientific calculators will initially give answers as fractions
        • Make sure you know where the S⇔D button is so that you convert the fraction into a decimal 
    • Percentages
      • There are many percentage calculations, including percentage yield, percentage atom economy, percentage change, percentage difference, percentage error and percentage uncertainty
    • Ratios
      • These are commonly used in moles calculations where the stoichiometry of the balanced chemical equation is not 1 : 1, e.g.

H2SO4 + 2NaOH → Na2SO4 + 2H2O

    • Reciprocals
      • These are most obvious in gas laws, using 1 / V, and concentration versus rate graphs, using 1 / T
    • Logarithmic functions
      • These are most obvious in pH and Arrhenius calculations
    • Exponential functions (Additional Higher level)
      • This is most obvious in Arrhenius calculations
  • Section 1 of the Data Booklet contains a list of the mathematical equations that you are expected to be able to manipulate and work with
    • Careful: This is not an exhaustive list - there are other equations that you will be expected to know that are not given in Section 1 of the Data Booklet, e.g. percentage yield

Examiner Tip

  • Make sure your final answers are written as proper numbers, not left as fractions
    • Leaving them as fractions will typically lose you a mark

What is the mean average?

  • The mean average is often just called the “average”
  • It is the total of all the values divided by the number of values, i.e. add all the numbers together and divide by how many there are
  • For example, two common isotopes of chlorine are chlorine-35 and chlorine-37, which exist in a 3 : 1 ratio
    • The information shows that there are three chlorine-35 atoms for every one chlorine-37 atom
    • Overall, there is a total of four atoms to be considered in the calculation
    • So, the mean average = begin mathsize 14px style fraction numerator 35 plus 35 plus 35 plus 37 over denominator 4 end fraction end style = 35.5
  • Problems with the mean average occur when there are anomalous results 
    • These should not be included in a mean average calculation

What mean average calculations are in chemistry?

  • Two main calculations in chemistry that involve mean average calculations are:
    • Calculating relative atomic mass from isotopic abundance data
      • For more information, see our revision note on Isotopes
    • Calculating average titres as part of a titration calculation
      • These calculations will typically have values that should not be considered because they are not concordant

Worked example

Calculate the average titre for the following experimental results.

Rough titre
/ cm3
Titre 1 
/ cm3
Titre 2
/ cm3
Titre 3
/ cm3
Titre 4
/ cm3
15.50 14.90 15.15 14.95 14.95

 

Answer:

  • The three values that are used for the mean average calculation are:
    • 14.90
    • 14.95
    • 14.95
  • The rough titre should never be used in the mean average calculation
  • Any results that are not concordant are considered to be outlying / anomalous results and should not be included in the mean average calculation
  • So, the average titre = fraction numerator 14.90 plus 14.95 plus 14.95 over denominator 3 end fraction = 14.93 cm3 

Range

  • The range can only be applied to numerical data
  • It is a measure of how spread out the data is, which means that it is the difference between the highest and lowest values
    • You can remember this as "Hi - Lo"
  • It can be expressed as:
    • A range of values
      • e.g. 9.2 - 8.4 
    • A single value
      • e.g. 9.2 - 8.4 = 0.8
  • The range can be affected when the highest and / or lowest data are anomalous results themselves

What is scientific notation?

  • Scientific notation is also known as standard form
  • It is a system of writing and working with very large or very small numbers
    • For example, Avogadro's number 602 000 000 000 000 000 000 000 is 6.02 x 1023 in scientific notation
  • Numbers in scientific notation are written as:

a × 10n

  • They follow these rules:
    • a is a number above 1 and below 10
    • For large numbers, n is an integer that is greater than 0 
      • i.e It shows how many times a is multiplied by 10
    • For small numbers, n is an integer that is less than 0 
      • i.e It shows how many times a is divided by 10
    • n < 0 for small numbers i.e how many times a is divided by 10
  • For example:

Applying scientific notation to numbers

Diagram showing how to convert large and small numbers into scientific notation

The scientific notation for numbers greater than 1 has the x 10 raised to a positive power while the scientific notation for numbers less than 1 has the x 10 raised to a negative power 

  • When rounding a number in standard form to a certain number of significant figures, only the value of a is rounded (the × 10n value will not be significant)
    • For example, 4.37 × 106 to 2 significant figures is 4.4 × 106

Orders of magnitude

  • When a number is expressed to an order of 10, this is an order of magnitude
    • Example: If a number is described as 3 × 108 then that number is actually 3 × 100 000 000
    • The order of magnitude of 3 × 108 is just 108
  • Orders of magnitude follow rules for rounding
    • The order of magnitude of 6 × 108 is 109 as the magnitude is rounded up
  • A quantity is one order of magnitude larger than another quantity if it is about ten times larger
    • Similarly, two orders of magnitude would be 100 times larger, or 102
  • In chemistry, orders of magnitude can be very large or very small

Approximation and estimation

  • Approximation and estimation are both methods used to obtain values that are close to the true or accurate values
    • While they share some similarities, they have distinct characteristics and are used in different contexts

Approximation

  • Approximation involves finding a value that is close to the actual value of a quantity
    • It may not necessarily be very precise or accurate
  • It is often used when an exact calculation is challenging or time-consuming and a reasonably close value is sufficient 
  • Approximations are typically quick and easy to calculate
  • For example, the pH of a strong acid is commonly accepted as being between pH 1.0 and pH 3.0
    • A reasonable approximation would be to say that the pH of a strong acid is pH 1.0 
    • The approximation is not exact, but it is reasonable as well as being aligned with a value that most people accept as a strong acid

Estimation

  • Estimation involves making an educated guess or assessment based on available information or data
  • It is used when the true value of a quantity is unknown or cannot be directly measured
  • For example, estimating the percentage yield of an industrial reaction involves:
    • Taking a known amount of reactant
    • Calculating the theoretical mass of the product that should be made
    • Completing the reaction and measuring the mass of the product that is made
    • Performing the percentage yield calculation 
    • The percentage yield calculation is correct for that amount of reactant under the specific reaction conditions used but can be applied to the industrial reaction to give an estimated percentage yield 

Appreciate when some effects can be ignored and why this is useful

  • During calculations using acid and base dissociation constants the assumption made is that the value of [H+] = [HA] 
    • Or the concentration of hydrogen ions is the same as the concentration of the acid
  • This is because
    • The weak acid has a low degree of dissociation so you can assume the concentration of the acid is the same value at equilibrium and use it in the expression for Ka
      • CH3COOH rightwards harpoon over leftwards harpoon CH3COO + H+
      • A point to remember is that Ka values are different depending on the acid, so the degree of dissociation also varies. Therefore the error using this assumption for some acids will be larger than others 
    • All the H+ ions are assumed to come from the acid as water will contribute a very small number of H+ ions
  • So the Ka expression:

Kafraction numerator stretchy left square bracket H to the power of plus stretchy right square bracket stretchy left square bracket A to the power of minus stretchy right square bracket over denominator stretchy left square bracket H A stretchy right square bracket end fraction

can be simplified to 

Ka = fraction numerator stretchy left square bracket H to the power of plus stretchy right square bracket to the power of bold 2 over denominator stretchy left square bracket HA stretchy right square bracket end fraction

    • Remember to give the full expression in an exam when asked

Percentage change and percentage difference

  • Percentage change and percentage difference are commonly used to express the relative change between two values
    • They are useful for comparing experimental results, determining reaction yields and analysing other chemical data

Percentage change

  • Percentage change is used to express the relative change between an initial value and a final value
  • It is calculated using the following formula:

Percentage Change = fraction numerator bold left parenthesis bold Final bold space bold Value bold minus bold Initial bold space bold Value bold right parenthesis over denominator bold Initial bold space bold value end fraction bold cross times bold 100

Worked example

During the course of a chemical reaction, the initial of chemical species A increases from 0.05 mol dm–3 to 0.08 mol dm–3.

Calculate the percentage change in concentration.

 

Answer:

  • Percentage change = fraction numerator left parenthesis Final space Value minus Initial space Value right parenthesis over denominator Initial space value end fraction cross times 100

  • Percentage change = fraction numerator open parentheses 0.08 minus 0.05 close parentheses over denominator 0.05 end fraction cross times 100 = 60%

  • So, the concentration of the chemical species increased by 60% during the reaction

Percentage difference

  • Percentage difference is used to compare two values to determine how much they differ from each other as a percentage
  • It is calculated using the following formula:

Percentage Difference = fraction numerator stretchy left parenthesis value space 1 space minus space value 2 stretchy right parenthesis over denominator stretchy left parenthesis average space of space value 1 space and space value space 2 stretchy right parenthesis end fraction bold cross times bold 100

Worked example

The melting points of different samples B and C are measured:

  • B = 75°C
  • C = 81°C

Calculate the percentage difference in melting points.

 

Answer:

  • Percentage difference = fraction numerator stretchy left parenthesis value space 1 minus value 2 stretchy right parenthesis over denominator stretchy left parenthesis average space of space value 1 space and space value space 2 stretchy right parenthesis end fraction cross times 100

  • Percentage difference = fraction numerator open parentheses 75 minus 81 close parentheses over denominator open parentheses bevelled fraction numerator open parentheses 75 plus 81 close parentheses over denominator 2 end fraction close parentheses end fraction cross times 100

  • Percentage difference = fraction numerator negative 6 over denominator 78 end fraction cross times 100 = –7.69%

    • When calculating percentage difference, you can ignore the minus sign in front of the calculation
  • So, the melting points of B and C differ by 7.69%

Percentage Error

  • Percentage error is used to express the difference between a final calculated answer and an accepted or literature value
  • It is calculated using the following formula

Percentage error = fraction numerator bold accepted bold space bold value bold minus bold experimental bold space bold value over denominator bold accepted bold space bold value end fraction bold cross times bold 100

  • You should be able to comment on any differences between the experimental and literature values

Worked example

Experimental results showed the enthalpy of combustion of propan-1-ol to be –1.5 x 103 kJ mol–1.

The literature value for this enthalpy change is -2021 kJ mol-1. Calculate the percentage error.

 

Answer:

  • Percentage error = fraction numerator accepted space value minus experimental space value over denominator accepted space value end fraction cross times 100

  • Percentage error = fraction numerator 2021 minus 1500 over denominator 2021 end fraction cross times 100 = 25%

Percentage uncertainty

  • Percentage uncertainties are a way to compare the significance of an absolute uncertainty on a measurement
    • This is not to be confused with percentage error, which is a comparison of a result to a literature value
  • It is calculated using the following formula

Percentage uncertainty = fraction numerator bold absolute bold space bold uncertainty over denominator bold measured bold space bold value end fraction bold cross times bold 100 

Diagram with examples of percentage uncertainty calculations for common laboratory apparatus

Diagram showing how to calculate percentage uncertainty for analogue and digital measurements

The absolute uncertainty for analogue measurements is ± half a division and for digital measurements is ± the last significant division

Examiner Tip

  • Percentage uncertainty can be reduced by:
    • Using equipment with a smaller uncertainty
    • Increasing the measured value, e.g. using a sample with greater mass or volume

Mathematical skills linked to graphs and tables

  • There are several specification points linked to graphs and tables
    • For more information about working specifically with graphs, see our revision note on Graphing in Chemistry
  • These include, but are not limited to, being able to:
    • Distinguish between qualitative and quantitative data, incorporating continuous and discrete variables
    • Understand direct and inverse proportionality, as well as positive and negative correlations between variables
    • Determine rates of change from tabulated data

Qualitative and quantitative data

  • Qualitative data usually describes something in words, not numbers 
    • For example:
      • Copper sulfate solution is blue
      • A more dilute copper sulfate solution is pale blue, while a more concentrated copper sulfate solution is a darker blue

  • Quantitative data uses numbers to count / measure something
    • For example:
      • The neutralisation of 25.0 cm3 sodium hydroxide by 25.0 cm3 hydrochloric acid increases the temperature of the system by 2.5 oC

  • Discrete data is quantitative
    • It consists of separate, distinct and countable values
    • For example:
      • The stoichiometric coefficients representing the relative number of molecules or atoms involved in the reaction are discrete values and must be integers
      • Electrons can only occupy certain discrete energy levels, e.g. 1s, 2s, etc

  • Continuous data is also quantitative
    • It is based on measurements and can include decimal numbers or fractions
    • This allows for an infinite number of values 
    • For example:
      • The temperature of an exothermic reaction as time progresses
      • The volume of gas produced during the thermal decomposition of calcium carbonate

Direct and inverse proportionality

  • There are a number of terms that are commonly applied to trends, particularly in graphs

  • Directly proportional
    • This applies to a trend that has a clearly linear relationship
    • Mathematically, this can be described as y = kx, where k can be positive or negative
    • In most situations, it is clear that k is positive
    • This means that the relationship can be described as "when one variable increases, the other increases" or "if x doubles, then y doubles"
    • A directly proportional relationship is always a straight line through the origin with a fixed gradient 

  • Inversely proportional
    •  Mathematically, this can be described as y = straight k over straight x, where k can be positive or negative
    • This means that the relationship can be described as "when one variable increases, the other decreases" or "if x doubles, then y halves" 
    • When plotted, inverse proportionality is not a straight line and does not pass through the origin

  • Positive correlation
    • This term is best applied to the gradient of a graph
    • The gradient of the graph is positive / slopes or curves upwards
    • It describes a relationship where as x increases, y also increases

  • Negative correlation
    • This term is, also, best applied to the gradient of a graph
    • The gradient of the graph is negative / slopes or curves downwards
    • It describes a relationship where as x increases, y decreases

Examiner Tip

  • Careful: A common mistake made by students is to describe any graph with a straight line going diagonally upwards as directly proportional
    • This is not correct because direct proportionality must go through the origin
    • A graph that does not go through the origin can correctly be described as proportional, but it is not directly proportional 

Determine rates of change from tabulated data

  • To determine rates of change from tabulated data, you can use the average rate of change or gradient, if the data has been plotted as a graph
  • The average rate of change between two points on a graph or in a table is:

Rate of change = fraction numerator bold the bold space bold change bold space bold in bold space bold the bold space bold dependent bold space bold variable bold space bold left parenthesis bold y bold minus bold axis bold right parenthesis over denominator bold the bold space bold change bold space bold in bold space bold the bold space bold independent bold space bold variable bold space bold left parenthesis bold x bold minus bold axis bold right parenthesis end fraction

Worked example

An experiment is run to measure the amount of chemical Y produced as time progresses:

Time / seconds  10 20 30 40 50
Amount of Y / cm3  3.0 7.0 10.0 14.0 18.0

 

Calculate the rate of change for this reaction between 10 and 30 seconds.

 

Answer:

  • Rate of change = fraction numerator the space change space in space the space dependent space variable space left parenthesis straight y minus axis right parenthesis over denominator the space change space in space the space independent space variable space left parenthesis straight x minus axis right parenthesis end fraction

  • Rate of change = begin mathsize 14px style fraction numerator open parentheses 10.0 minus 3.0 close parentheses over denominator open parentheses 30 minus 10 close parentheses end fraction end style = 0.35 cm3 s–1 

  • So, on average, the amount of Y increases by 3.5 cm3 every 10 seconds over the interval from t = 10 to t = 30 seconds

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Richard

Author: Richard

Expertise: Chemistry

Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.