What is the molarity of aqueous NaOH solution containing 0.5 g in 500 cm3 of the solution

11. Molarity, volumes and the concentration of solutions

See also 14.3 dilution of solutions calculations

(a) Explaining the terms solubility, concentration, strength and molarity

• Why are the terms 'concentration', 'strength' and 'molarity' important?

  • Quite a lot of analytical procedures in chemistry involve the use of solutions of accurately known concentration e.g. standard solutions for various analytical purposes including titrations.

  • If you want to analyse an acid solution you need to titrate it with a standard solution of alkali of accurately known concentration e.g. an accurately known molarity (concentration usually expressed in mol/dm3, lots more on this on the rest of this page!).

  • The solubility of a substance is the maximum amount of solute that dissolves in a given volume of solvent.

  • It is important to know the solubility of substance in various liquids, quite often quoted as the maximum solubility of salts in water, but often quoted, not as molarity, but in g salt /100 g of water and plotted in graphs known as solubility curves.

  • This is the maximum concentration possible for a given solute and solvent.

  • 

  • For more on solubility see Important formulae of compounds, salt solubility and water of crystallisation

• Misconceptions

  • There are differences in using the words ‘concentration’ and ‘strength’ in science compared to everyday language

  • In scientific language concentration is a specifically defined term e.g.

    • (i) the mass of solute per unit volume of solvent e.g. g/dm3, g/cm3  (g dm-3, g cm-3), OR,

    • (ii) moles per unit volume of solvent e.g. mol/dm3   (mol dm-3),

      • neither isn't a vague description, they are specific terms and units!

  • 'Strength' in terms of solution concentration is not a scientifically defined term and tends to be used in everyday language to 'crudely' indicate a concentration e.g. 'great/high strength' indicating a very concentrated solution, and conversely, 'low/weak strength' to indicate a low concentration solution.

  • Unfortunately this every use of the term is widespread, so take care, because it does not apply to the science of chemistry!

  • In chemistry, for solutions, the word 'strength' in chemistry is applied to e.g. an acid to indicate how much it ionises in aqueous solution - is it a weak or strong acid or alkali (soluble base).

    • A low strength solution of an acid indicates it is a weak acid and only ionises a few % to give hydrogen ions.

      • e.g. ethanoic acid: CH3COOH(aq)

        
        CH3COO–(aq) + H+(aq)

      • The equilibrium is about 2% to the right, indicating a very weak acid.

    • An acid of high strength indicates that it ionises to a very percent to form hydrogen ions i.e a strong acid.

      • hydrochloric acid: HCl(g) + aq ===> H+(aq) + Cl–(aq)

      • When you dissolve hydrogen chloride in water (aq) you get virtually 100% ionisation into the hydrogen ion and chloride ion indicating a very strong acid.

    • In both cases the term concentration applies to the concentration of the original acid molecules:

      •  e.g. ignoring extent of ionisation, the concentration of CH3COOH or HCl in mol/dm3.

    • For more details see More on acid-base theory and weak and strong acids and their properties

• Revise section 7. moles and mass before proceeding in this section 11 and eventually you may need to be familiar with the use of the apparatus illustrated above, some of which give great accuracy when dealing with solutions and some do not.

• It is very useful to be know exactly how much of a dissolved substance is present in a solution of particular concentration or volume of a solution.
  • So we need a standard way of comparing the concentrations of solutions in some standard units.
  • The more you dissolve in a given volume of solvent, or the smaller the volume you dissolve a given amount of solute in, the more concentrated the solution.
  • Note: A standard solution is one whose precise concentration is known
• Reminders: The dissolved substance is called the solute and the liquid dissolving it is the solvent and the result is a solution.
  • The more of the substance you dissolve in the same volume of liquid, the more concentrated the solution, the solute particles on average are closer together.
  • 
    
     
  • The diagrams represent two substances dissolved in the solvent, the right-hand diagram represents a more concentrated solution e.g. a mixture of two salts in water.
  • The pictures do not mean the right hand one is a stronger solution!
  • Unfortunately, in everyday language, it would be described as such, but this is science and the correct use of scientific language is essential!

TOP OF PAGE and sub-index

(b) Measures of concentration and simple calculations of molarity

(b)(i) Concentration in terms of mass of solute per unit volume of solution

A summary of how to do basic concentration calculations and rearrangement of the solution concentration formula

• We will look at moles in (b)(ii)
• The simplest measure of concentration is mass of solute per unit volume of solvent e.g.
  • concentration = mass of solute / volume of solvent
  • Take 5.0 g of salt dissolved in 500 cm3 of water.
  • The concentration can be expressed in several ways.
    • concentration = 5.0/500 = 0.01 g/cm3
    • 1 dm3 = 1000 cm3, so 500 cm3 = 500/1000 = 0.50 dm3
    • concentration = 5.0/0.50 = 10.0 g/dm3
    • For interconversion: g/cm3 x 1000 = g/dm3  AND  g/dm3/1000 = g/dm3
    • -
  • Sometimes the general formula c = m/v is used
    • c = concentration, m = mass, v = volume
    • rearrangements: m = c x v  and  v = m/c
    • -
• Example questions (not using moles)
  • Q1 What is the concentration in g/dm3 if 6.0 g of salt is dissolved in 150 cm3 of water?
    • 150 / 1000 = 0.15 dm3
    • concentration =  mass / volume = 6.0 / 0.15 = 40.0 g/dm3
    • -
  • Q2 Given a salt solution of concentration 16 g/dm3, what mass of salt is in 40 cm3 of the solution?
    • 1 dm3 = 1000 cm3
    • c = m / v = 16 / 1000 = 0.16 g/cm3  (note this a way of converting g/dm3 to g/cm3)
    • Therefore mass of salt: m = c x v = 0.16 x 40 = 6.4 g of salt
    • -
  • Q3 Given 5.0 g of a salt, what volume of water in cm3, should it be dissolved in to give a solution of concentration of 12.5 g/dm3?
    • c = m / v, rearranging gives v = m / c
    • v = 5.0 / 12.5 = 0.40 dm3
    • volume of water needed = 1000 x 0.40 = 400 cm3
    • -
• Its also good to be able to do dilution' calculations in section 14.3 dilution of solutions

TOP OF PAGE and sub-index

(b)(ii) Concentration in terms of moles of solute per unit volume of solution

A summary of how to do basic molarity calculations and rearrangement of the molarity formula

• 
  For most analytical and calculation purposes the concentration of an aqueous solution is usually expressed in terms of moles of dissolved substance per cubic decimetre of solution (reminder mole formula triangle on the right). 1 cubic decimetre (dm3) = 1 litre (l) in old money!
  • concentration = molarity = moles of solute / volume of solvent in dm3  (litres)
  • Make sure you know how to calculate moles, see the triangle on the right!
  • Using concentration units of mol dm-3 (or mol/dm3), the concentration is called molarity, sometimes denoted in shorthand as M (old money again, take care!) and the word molar is used too.
  • Note: 1dm3 = 1 litre = 1000ml = 1000 cm3, so dividing cm3/1000 gives dm3, which is handy to know since most volumetric laboratory apparatus is calibrated in cm3 (or ml), but solution concentrations are usually quoted in molarity, that is mol/dm3 (mol/litre).
  • Concentration is also expressed in a 'non-molar' format of mass per volume e.g. g/dm3
  • You need to know all about moles to proceed further on this page and get into 'molarity' ...
  • ... so read section 7. on moles and mass - essential pre-reading for section 11 ...
  • AND, if you can't understand molarity, you cannot do titration calculations either!
• Equal volumes of solution of the same molar concentration contain the same number of moles of solute i.e. the same number of particles as given by the chemical formula you use in defining a specific molarity.
  • A note on solutions of ionic compounds e.g.
  • A 1.0 molar solution of magnesium chloride MgCl2(aq), contains 1.0 mol/dm3 of magnesium ions (Mg2+), BUT 2.0 mol/dm3 in terms of the chloride ion (Cl-) concentration.
• You need to be able to calculate
  • the number of moles or mass of substance in an aqueous solution of given volume and concentration
  • the concentration of an aqueous solution given the amount of substance and volume of water, for this you use the equation ....  (reminder molarity formula triangle on the right), so, for a substance Z ...
  • 
    (1a) molarity (concentration) of Z = moles of Z / volume in dm3
    • This is sometimes referred to as the molar concentration (mole-concentration),
    • and you need to be able to rearrange this equation ... therefore ...
    • (1b) moles = molarity (concentration) x volume in dm3 and ...
    • (1c)  volume in dm3 = moles / molarity (concentration)
    • You can use the triangle on the right to help you rearrange the equation for the basic definition of molarity, BUT it is much better to know how to rearrange the equation:
      • molarity = moles ÷ volume(dm3)
  • You may also need to know that ...
    • (2) molarity x formula mass of solute = solute concentration in g/dm3
      • This is sometimes referred to as the mass-concentration,
      • and dividing this by 1000 gives the concentration in g/cm3, and
    • 
      (3) concentration in g/dm3 / formula mass = molarity in mol/dm3
      • both equations (2) and (3) result from equations (1) and (4), work it out for yourself.
  • and to sum up, by now you should know:
    • (4) moles Z = mass Z / formula mass of Z
    • (5) 1 mole = formula mass in grams
    • (6) molarity = moles/dm3

• Molarity calculation Example 11.1

  • If 5.00g of sodium chloride is dissolved in exactly 250 cm3 of water in a calibrated volumetric flask,

  • (a) what is the concentration in g/dm3?

    • volume = 250/1000 = 0.25 dm3

    • concentration = mass / volume = 5/0.25 = 20 g/dm3

    • -

  • (b) What is the molarity of the solution?

    • Ar(Na) = 23, Ar(Cl) = 35.5, so Mr(NaCl) = 23 + 35.5 = 58.5

    • mole NaCl = 5.0/58.5 = 0.08547

    • volume = 250/1000 = 0.25 dm3

    • molarity = mol of solute / volume of solvent

    • Molarity = 0.08547/0.25 = 0.342 mol/dm3

  • -

• Molarity calculation Example 11.2

  • 5.95g of potassium bromide was dissolved in 400cm3 of water.

  • 
    (a) Calculate its molarity. [Ar's: K = 39, Br = 80]

    • moles = mass / formula mass, (KBr = 39 + 80 = 119)

    • mol KBr = 5.95/119 = 0.050 mol

    • 400 cm3 = 400/1000 = 0.400 dm3

    • molarity = moles of solute / volume of solution

    • molarity of KBr solution = 0.050/0.400 = 0.125 mol/dm3

    • -

  • (b) What is the concentration in grams per dm3?

    • concentration = mass / volume, the volume = 400 / 1000 = 0.4 dm3

    • concentration = 5.95 / 0.4 = 14.9 g/dm3

    • -

• Molarity calculation Example 11.3

  • 
    What mass of sodium hydroxide (NaOH) is needed to make up 500 cm3 (0.500 dm3) of a 0.500 mol dm-3 (0.5M) solution? [Ar's: Na = 23, O = 16, H = 1]

  • 1 mole of NaOH = 23 + 16 + 1 = 40g

  • molarity = moles / volume,

  • so mol needed = molarity x volume in dm3

  • 500 cm3 = 500/1000 = 0.50 dm3

  • mol NaOH needed = 0.500 x 0.500 = 0.250 mol NaOH

  • therefore mass = mol x formula mass

  • = 0.25 x 40 = 10g NaOH required

  • -

• Molarity calculation Example 11.4

  • (a) How many moles of H2SO4 are there in 250 cm3 of a 0.800 mol dm-3 (0.8M) sulphuric acid solution?

  • (b) What mass of acid is in this solution? [Ar's: H = 1, S = 32, O = 16]

    • 
      (a) molarity = moles / volume in dm3, rearranging equation for the sulfuric acid

      • mol H2SO4 = molarity H2SO4 x volume of H2SO4 in dm3

      • mol H2SO4 = 0.800 x 250/1000 = 0.200 mol H2SO4

    • (b) mass = moles x formula mass

      • formula mass of H2SO4 = 2 + 32 + (4x16) = 98

      • 0.2 mol H2SO4 x 98 = 19.6g of H2SO4 

      • -

• Molarity calculation Example 11.5 This involves calculating concentration in other ways e.g. mass/volume units

  • 
    What is the concentration of sodium chloride (NaCl) in g/dm3 and g/cm3 in a 1.50 molar solution?

  • At. masses: Na = 23, Cl = 35.5, formula mass NaCl = 23 + 35.5 = 58.5

  • since mass = mol x formula mass, for 1 dm3

  • concentration = 1.5 x 58.5 = 87.8 g/dm3, and

  • concentration = 87.75 / 1000 = 0.0878 g/cm3

  • -

• Molarity calculation Example 11.6

  • A solution of calcium sulphate (CaSO4) contained 0.500g dissolved in 2.00 dm3 of water.

  • 
    Calculate the concentration in (a) g/dm3, (b) g/cm3 and (c) mol/dm3.

    • (a) concentration = 0.500/2.00 = 0.250 g/dm3, then since 1dm3 = 1000 cm3

    • (b) concentration = 0.250/1000 = 0.00025 g/cm3   (or from 0.500/2000)

    • (c) At. masses: Ca = 40, S = 32, O = 64, formula mass CaSO4 = 40 + 32 + (4 x 16) = 136

      • moles CaSO4 = 0.5 / 136 = 0.00368 mol in 2.00 dm3 of water

      • concentration CaSO4 = 0.00368 / 2 = 0.00184 mol/dm3

      • -

• Molarity calculation Example 11.7

• Its also good to be able to do dilution' calculations in section 14.3 dilution of solutions

There are more questions involving molarity in section 12. on titrations

and section 14.3 on dilution calculations and

TOP OF PAGE and sub-index

(c) APPENDIX 1 on SOLUBILITY and concentration calculations

How do you find out how soluble a substance is in water?

Reminder: solute + solvent ==> solution

i.e. the solute is what dissolves, the solvent is what dissolves it and the resulting homogeneous mixture is the solution.

The solubility of a substance is the maximum amount of it that will dissolve in a given volume of solvent e.g. water.

The resulting solution is known as a saturated solution, because no more solute will dissolve in the solvent.

Solubility can be measured and expressed in with different concentration units e.g. g/100cm3, g/dm3 and molarity (mol/dm3).

Solubility can also be expressed as mass of solute per mass of water e.g. g/100g of water.

You can determine solubility by titration if the solute reacts with a suitable reagent e.g. acid - alkali titration and it is especially suitable for substances of quite low solubility in water e.g. calcium hydroxide solution (alkaline limewater) can be titrated with standard hydrochloric acid solution.

However, many substances like salts are very soluble in water and a simple evaporation method will do which is described below e.g. for a thermally stable salt like sodium chloride.

(1) A saturated solution is prepared by mixing the salt with 25cm3 of water until no more dissolves at room temperature.

(2) The solution is filtered to make sure no undissolved salt crystals contaminate the saturated solution.

(3) Next, an evaporating dish (basin) is accurately weighed. Then, accurately pipette 10 cm3 of the saturated salt solution into the basin and reweigh the dish and contents.

By using a pipette, its possible to express the solubility in two different units.

(4) The basin and solution are carefully heated to evaporate the water.

(5) When you seem to have dry salt crystals, you let the basin cool and reweigh it.

(6) The basin is then gently heated again and then cooled and weighed again.

This is repeated until the weight of the dish and salt is constant, proving that all the water is evaporated

By subtracting the original weight of the dish from the final weight you get the mass of salt dissolved in the volume or mass of saturated salt solution you started with.

You can repeat the experiment to obtain a more accurate and reliable result.

(7) Calculations

By using a pipette it is possible to calculate the solubility in two ways, expressed as two quite different units.

Suppose the dish weighed 95.6g.

With the 10.0 cm3 of salt solution in weighed 107.7g

After evaporation of the water the dish weighed 96.5g

Mass of 10.0 cm3 salt solution = 107.7 - 95.6 = 12.1g

Mass of salt in 10 cm3 of salt solution = 96.5 - 95.6 = 0.9g

Mass of water evaporated = 107.7 - 96.5 = 11.2g

(a) Expressing the solubility in grams salt per 100 g of water

From the mass data above 0.9g of salt dissolved in 11.2g of water

Therefore X g of salt dissolves in 100g of water, X = 100 x 0.9 / 11.2 = 8.0

Therefore the solubility of the salt = 8.0g/100g water

You can scale this up to 80.0g/1000g H2O, or calculate how much salt would dissolve in any given mass of water.

You can also express the solubility as g salt/100g of solution.

0.9g salt is dissolved in 12.1g of solution, X g in 100g of solution

Therefore X = 100 x 0.9 / 12.1 = 7.4, so solubility = 7.4g/100g solution

These calculations do not require the original salt solution to be pipetted. You can just measure out approximately 10cm3 of the salt solution with 10cm3 measuring cylinder, and do the experiment and these calculations in the exactly the same way.

(b) However, if you know the exact volume of salt solution and the mass dissolved in it, then you can calculate the concentration in g/dm3, and if you know the formula mass of the salt, you can calculate the molarity of the solution.

From part (a) we have 0.9g of salt in 10.0 cm3

Therefore X g will dissolve in 1000cm3 solution, X = 1000 x 0.9 / 10 = 90g/1000 cm3

Solubility of salt = 90g/dm3

Suppose the formula mass of the salt was 200, calculate the molarity of the saturated solution.

moles salt = mass / formula mass = 90/200 = 0.45 moles

Therefore solubility of saturated salt solution in terms of molarity = 0.45 mol/dm3

NOTE Solubility varies with temperature, see Gas and salt solubility in water and solubility curves, and it usually (but not always) increases with increase in temperature. So, in the experiment described above, the temperature of the saturated solution should be noted, or perhaps controlled to be saturated at 20oC or 25oC.

TOP OF PAGE and sub-index

(d) APPENDIX 2 - How to make up a standard solution - a solution of precisely known concentration

The method and procedure of how to make up a standard solution of a soluble solid e.g. a salt, is fully described.

Procedure for making up a standard solution of known molarity

The method and procedure of how to make up a standard solution of a soluble solid e.g. a salt, is fully described

Example 1.

Suppose you want to make up 250 cm3 of a salt solution of concentration 20g/dm3 (20g/litre, 20g/1000cm3, 20g/1000ml).

c = m / v, m = c x v, m = 20 x 250 / 1000 = 5g

so 5g of the salt is needed to be made up into an aqueous solution of exactly 250.0 cm3.

The procedure

to

is described in detail after example 2. below.

Example 2.

To prepare a solution of known molarity, you need to work backwards from the volume required and the molarity to see how much solid you need.

Suppose you want to make up 250cm3 of a sodium chloride solution of concentration 0.20 moldm-3

Preliminary calculation:

From molarity formula (on the right): moles = molarity (mol/dm3) x volume (dm3)

and volume in cm3 / 1000 = dm3

moles NaCl needed = 0.20 x 250/1000 = 0.20 x 0.25 = 0.05 mol NaCl

Atomic masses: Na =23 and Cl = 35.5, so molar mass of NaCl = 23 + 35.5 = 58.5

From basic mole formula: mass of NaCl needed = mol NaCl x formula mass NaCl

mass of NaCl needed = 0.05 x 58.5 = 2.925 g (which is ok if you have a 3 decimal place balance!), so

2.295g of pure NaCl salt is needed to made up 250.0 cm3 of solution with a precise concentration of 0.20 mol/dm3.

Procedure to make the standard solution i.e. one of known concentration of solid (in this case)

The amount of water you add to the beaker should be much less than 250cm3 to allow for the transfer and rinsing of the solution into the standard volumetric flask using a 'squeezy' wash bottle!

You can use the wash bottle to rinse down any crystals on the side of the beaker, but watch the volume you use..

Then, the whole of the solution in the beaker is transferred into the flask with the help of a funnel to avoid the risk of spillage.

To make sure every drop of the salt solution ends up in the flask, a wash bottle of pure water is used to rinse out the beaker several times, AND rinse the stirring rod and the funnel too.

This is to ensure nothing is lost in the transfer fro beaker to flask.

The stopper is placed on and the flask carefully shaken quite a few times to ensure the salt solution is completely mixed up.

Finally, check the meniscus lies on the calibration mark, in case another few drops are needed.

Either way, the last drops of water should be added most carefully with a teat pipette.

Job done!

Note on standard solutions of acids and alkalis

You can purchase standard solutions ready for use.

OR, a phial of concentrated acid or alkali, which you dilute into a specified volume to give a specific molarity.

Apart from weighing out a solid, the procedure is the same as 

and

, ensuring every drop from the phial is rinsed down the funnel into the calibrated volumetric flask.

See dilution' calculations in section 14.3 dilution of solutions

TOP OF PAGE and sub-index

(e) Self-assessment Quizzes on molarity calculations:

type in answer

multiple choice

QUIZ on molarity

type in titration answer

multiple choice titration 

QUIZ

(good revision for A level students)

See also Advanced level GCE-AS-A2 acid-alkali titration calculation questions

Above is typical periodic table used in GCSE science-chemistry specifications in doing molarity calculations, and I've 'usually' used these values in my exemplar calculations to cover most syllabuses

WHAT NEXT? e.g. OTHER CALCULATION PAGES

1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

2. Calculating relative formula/molecular mass of a compound or element molecule

3. Law of Conservation of Mass and simple reacting mass calculations

4. Composition by percentage mass of elements in a compound

5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

9. Moles and the molar volume of a gas, Avogadro's Law

10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants-products)

11. Molarity, volumes and solution concentrations (and diagrams of apparatus) (this page)

12. How to do acid-alkali titration calculations, diagrams of apparatus, details of procedures

13. Electrolysis products calculations (negative cathode and positive anode products)

14. Other calculations e.g. % purity, % percentage & theoretical yield, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

16. Gas calculations involving PVT relationships, Boyle's and Charles Laws

17. Radioactivity & half-life calculations including dating materials

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CALCULATIONS IN CHEMISTRY and quantitative chemical analysis

Doc Brown's Chemistry KS4 science GCSE 9-1, IGCSE, O Level and GCE AS A2 Advanced A Level Revision Notes

Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB.

F/H (foundation/higher) represent easier/harder UK KS4 GCSE/IGCSE/KS4 science-chemistry courses * There are 16 linked sections listed below * EMAIL query? comment

e.g. spotted a silly error or request for type of GCSE calculation I don't seem to have covered

In sections 1-16 you can study definitions of the terms used and examples of how to do chemical calculations.

Sections 2 to 6 illustrate the basic 'calculation' requirements for most UK GCSE Science students.

Sections 1, and 7 to 15 cover most extra material for higher GCSE student in Science, triple award or IGCSE Chemistry.

These revision notes and practice questions on how to do chemical calculations and worked examples should prove useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry science courses.

Page 3

Answers to the more advanced level mole concept calculations

Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) Online Chemical Calculations

ANSWERS to Part 7 - the more advanced mole Q's 

The ORIGINAL mole Questions

Index of all my online chemical calculation quizzes

All my GCSE/IGCSE/US grade 8-10 Chemistry Revision notes

Advanced UK A/AS level, IB and US grade 11-12 pre-university chemistry

email doc brown

Use your mobile phone or ipad etc. in 'landscape' mode

This is a BIG website, you need to take time to explore it [

Quantitative chemistry calculations This page has the answers to the more advanced mole based questions from calculations section 7. for more advanced chemistry students. Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB. These revision notes and practice questions on how to do chemical calculations and worked examples should prove useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry science courses.

Spotted any careless error? EMAIL query ? comment or request a type of GCSE calculation not covered?

Mole calculations introduction * Molar gas volume * Advanced Redox titration Q's * Non-redox titration Q's

(g) e.g. Mg + 2HCl ==> MgCl2 + H2

1 mole magnesium gives 1 mole hydrogen, mole Mg = 2 ÷ 24.3 = 0.0823

so mole H2 = 0.0823, so volume H2 = 0.0823 x 24 = 1.975 dm3

(h) both 1 mole of Na2CO3 or NaHCO3 will give 1 mole of CO2

(1) VCO2 = mol Na2CO3 x 24000 = 0.76 ÷ 106 x 24000 = 172cm3

(2) VCO2 = mol NaHCO3 x 24000 = 0.76÷ 84 x 24000 = 217cm3

(i) Zn + 2HCl ==> ZnCl2 + H2 , mole H2 = mole HCl ÷ 2

mol HCl = 50 ÷ 1000 x 0.2 = 0.01 mol

so mole H2 = 0.005, VH2 = 0.005 x 24000 = 120cm3

(j) CaCO3 + 2HCl ==> CaCl2 + H2O + CO2

mole CO2 = mole HCl ÷ 2, mol HCl = 75 ÷ 1000 x 0.05 = 0.00375 mol

so mole CO2 = 0.001875, VCO2 = 0.001875 x 24000 = 45cm3

Above is typical periodic table used in GCSE science-chemistry specifications in doing chemical calculations, and I've 'usually' used these values in my exemplar calculations to cover most syllabuses

OTHER CALCULATION PAGES

1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

2. Calculating relative formula/molecular mass of a compound or element molecule

3. Law of Conservation of Mass and simple reacting mass calculations

4. Composition by percentage mass of elements in a compound

5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

9. Moles and the molar volume of a gas, Avogadro's Law

10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants-products)

11. Molarity, volumes and solution concentrations (and diagrams of apparatus)

12. How to do acid-alkali titration calculations, diagrams of apparatus, details of procedures

13. Electrolysis products calculations (negative cathode and positive anode products)

14. Other calculations e.g. % purity, % percentage & theoretical yield, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

16. Gas calculations involving PVT relationships, Boyle's and Charles Laws

17. Radioactivity & half-life calculations including dating materials

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