Topic 5 Formulae, Equations & Amounts of Substances Flashcards
Empirical formulae
The smallest whole number ratio of atoms of each element in a compound.
Calculation of empirical formula from mass or % composition by mass
Divide the masses or %s by the RAM of each element. Divide answers by the smallest answer. Form a whole number ratio. Note: the oxygen mass may need to be obtained by subtraction.
Calculate the empirical formula of a hydrogen-, carbon & oxygen containing compound via combustion analysis.
12/44 x mass of CO2 = mass of carbon
2/18 x mass of water = mass of hydrogen
Mass of oxygen = total mass - mass of carbon - mass of hydrogen
Divide the masses by the RAMs then the smallest answer to obtain the ratio.
Relative molecular mass vs relative formula mass
Relative molecular mass can only be used to refer to molecules —compounds containing covalent bonds— whereas relative formula mass can be used to refer to ionic or covalent compounds.
Molar mass
Mass per mole of a substance (M). Units: g mol^-1. Can be interchanged with Mr in the moles= mass/Mr formula, but note it’s not the same as Mr.
Ideal gas equation: units
pV=nRT
p= pressure in pascals (Pa)
V= volume in m^3
T= temperature in K.
n = amount of substance in moles.
R= the gas constant: 8.31 J mol^-1 K^-1
Mole
The amount of substance that contains the same number of particles as the number of carbon atoms in exactly 12g of the carbon-12 isotope.
Avogadro’s constant (L)
The number of atoms of 12C in exactly 12g of 12C.
6.02 x 10^23
When using moles, what is important to clarify?
Moles of what? Moles of oxygen atoms or oxygen molecules? Moles of ions or moles of electrons? Etc..
Spectator ions
Ions in an organic compound that do not take part in the reaction.
How would you calculate the mass needed to form x g of product?
Calculate the moles of the known species and use the stoichiometric ratio to find the moles of the species required then convert moles to mass.
A 16.7g sample of a Na2CO3.10H2O is heated until a reaction has occurred. A mass of 3.15g of water is obtained. What is the equation for the reaction?
Find the Mr s.
Find the moles.
Find the ratio.
16.7/286.1 = 0.0584
3.15/18 = 0.175
Divide by the smallest to give:
1:3
Na2CO3.10H2O —> Na2CO3.7H2O + 3H2O
Avogadro’s law
States equal volumes of gases under the same conditions of temperature and pressure contain the same number of molecules.
N2(g) + 3H2(g) —> 2NH3(g) What does Avogadro’s law tell us about this reaction?
1 volume of N2 reacts with 3 volumes of H2 to form 2 volumes of ammonia.
2H2S(g) + 3O2(g) —> 2SO2(g) + 2H2O(l)
250 cm^3 of H2S is mixed with 600 cm^3 of oxygen. What is the volume of the gaseous mixture at the end of the reaction?
Solids/liquids can be ignored.
Ratio is 2: 3: 2.
All of the H2S will react, but only 375 cm^3 of oxygen, this will form 250 cm^3 of SO2.
600 - 375 = 225 cm^3 of excess, unreacted oxygen.
Total volume of the resulting mixture = 225 + 250 = 475 cm^3
Molar volume
The volume occupied by 1 mole of any gas.
Molar volume of a gas at RTP
24 = volume in dm^3 / amount in mol Rearrange as required.
Mg(s) + 2HCl(aq) –> MgCl2(aq) + H2(g)
What volume of H2 is formed when 1g of Mg is added to excess HCl?
Calculate the amount in moles from the volume or mass, depending on which is given.
n(Mg)= 1/24.3= 0.0412 mol
Use the reaction ratio to find the amount of the other substance.
1:1 ratio
Convert this amount to a mass or volume, depening on what is required.
24000 x 0.0412 = 988 cm^3
ppm
mg dm^-3
Mass concentration (g dm^-3)
Mass of solute (g) / Volume of solution (dm^3)
Molar concentration (mol dm^-3)
amount in moles/ volume (dm^3)
Primary standards
Substances used to make a standard solution by weighing.
Standard solution
Solution whose concentration is accurately known.
Properties of primary standards
- Solids with high molar masses.
- Available in a high degree of purity.
- Chemically stable: they will neither decompose nor react with substances in the air.
- Water-soluble.
- Won’t absorb water from the air.
- React rapidly and completely with other substances used in titrations.
How can we make a standard solution using sulfamic acid?
- Add ~2.4g of sulfamic acid to the weighing bottle and weigh accurately.
- Transfer the acid to a clean beaker and reweigh the bottle (weighing by difference).
- Add about 100 cm^3 of deionised water to the beaker, and stir until all the acid has dissolved.
- Remove the stirring rod. Wash traces of the solution into the beaker.
- Use a funnel to pour the the solution from the beaker into the flask.
- Rinse the inside of the beaker and transfer the rinsings to the flask.
- Add deionised water to the flask, and make up to the mark.
- Stopper the flask, and invert it several times to make a uniform solution.
Methyl orange in acid
Red
Methyl orange in alkali
Yellow
When is methyl orange used?
For strong acid-weak base and strong acid-strong base titrations.
Phenolphthalein in acid
Colourless
Phenolphthalein in alkali
Pink
When is phenolphthalein used?
For weak acid-strong base and strong acid-strong base titrations.
Use a white tile
So that the colour change can be seen more clearly.
Add about 3 drops of indicator
Many indicators are weak acids, so would affect the endpoint. Hence, the same volume of indicator must be used in repeated titrations.
Fill the burette so that the space between the tap and the tip is filled with solution.
If not, then as the liquid in the burette goes down, some of the liquid will fill the space and not enter the conical flask.
Set up the burette with the tip inside the neck of the conical flask.
Minimises the risk of spilling the solution out of the flask.
Record the burette reading to the nearest 0.05 cm^3 using a light background to see the bottom of the meniscus.
This increases the accuracy of the reading of the meniscus.
Swirl the mixture when adding from the burette.
Ensures continuous mixing of the two solutions.
Add the solution from the burette steadily initially then more slowly then dropwise close to the endpoint.
This decreases the chance of overshooting the endpoint.
Stop adding the solution just as the indicator changes colour.
To increase the accuracy of the titre. Adding more solution decreases the accuracy.
Repeat the titration to obtain concordant results.
Concordant titres are within 0.20 cm^3 of each other.
Rinse the pipette and burette with both deionised water and the solution to be used. Rinse the conical flask only with deionised water.
If the conical flask were mixed with the solution, there would e an unknown extra amount of substance being titrated.
Equivalence point
The point at which there are exactly the right amounts of substances to complete the reaction.
Endpoint
The point at which the indicator changes colour. Ideally, this should coincide wit the equivalence point.
Meniscus
The curving of the upper surface of a liquid in a container. The lowest/horizontal part of the meniscus should be read.
Titre
Volume added from the burette during a titration.
Error
The difference between an experimental value and the accepted/correct value.
Accuracy
A measure of how close values are to the accepted/correct value.
Precision
A measure of how close values are to each other.
Random errors
Errors caused by unpredictable variation in conditions, or by a difference in recording that is difficult to get exactly right.
Systematic errors
Errors that are constant or predictable, usually due to the apparatus being used.
Measurement uncertainty
The potential error involved when using a piece of apparatus to make a measurement.
% Uncertainty
Measurement uncertainty / Value recorded x 100
Theoretical yield
Maximum mass of a product, assuming complete reaction and no losses.
Actual yield
The actual mass obtained in a reaction.
Percentage Yield
Actual yield/theoretical yield x 100
Why might the mass of the product be less than the maximum possible?
- The reaction is reversible, so may not be complete.
- There are side-reactions that lead to unwanted products.
- The product may need to be purified, which may result in a loss of product.
1 tonne
1 x 10^6 g
Atom economy
The molar mass of the desired product divided by the sum of the molar masses of all the products, expressed as a %.
Besides % yield, what other factors should be considered when assessing the suitability of an industrial process?
The scarcity of non-renewable resources.
The cost of raw materials.
How much energy is required.
Multistep reactions
Have lower atom economies.
Elimination & substitution reactions
Have lower atom economies.
Addition reactions
100% atom economy
What is the difference between % yield & atom economy?
% yield indicates how efficient a reaction is at converting reactants to products. Atom economy indicates the % of atoms from the starting materials that end up in the desired product.
Displacement reaction
A reaction in which one element replaces another element in a compound. A redox reaction, so don’t say one element is more reactive, say it is a better oxidising or reducing agent.
Precipitation reaction
Reactions in which an insoluble solid is one of the products.
Limewater equation
Ca(OH)2(aq) + CO2 (g) –> CaCO3(s) + H2O(l)
Forms a white precipitate!
How can we use a precipitation reaction to work out equations?
Place the same volume of one reactant into a test tube. Add different volumes of the other reactant to each tube. Centrifuge each tube for the same length of time. Use the same concentrations of reactants. The depth of the precipitate is proportional to the mass. The first test tube after which the depth of the precipitate doesn’t change shows the ratio in which the reaction occurs.
Metal + acid –>
Salt + H2 Redox & neutralisation.
Metal oxide + acid –>
Salt + H2O Neutralisation, but not redox. The reactivity of the metal is not relevant as the metal is present as ions.
Metal hydroxide + acid –>
Salt + H2O Neutralisation, but not redox. The reactivity of the metal is not relevant as the metal is present as ions.
Metal carbonate + acid –>
Salt + water + CO2 Neutralisation, but not redox.
How can we test for a carbonate of hydrogen carbonate ion?
Add aqueous acid and test the gas given off with limewater.
Calculate the volume of C3H8 in dm^3 needed to produce 14.7g of 1-bromopropane. Percentage yield is 31%. Ratio is 1:1.
Moles of product x 100/31 x 24 = volume of reactant required
14.7/122.9 = 0.1196
0.1196 x 100/31 x 24 = 9.26 dm^3
