2-1 Mole calculations
Propene can be made by the dehydration of propan-2-ol. What is the percentage yield when 30 g of propene (Mr = 42.0) are formed from 50 g of propan-2-ol (Mr = 60.0)?
Percentage yield = Actual yield / Theoretical yield n(Propene) = 0.714285 n(propan-2-ol) = 0.83 PY = (0.714285 / 0.83) x 100 = 86%
What are the uncertainties between readings and measurements?
The uncertainty of a reading (one judgement) is at least ±0.5 of the smallest scale reading. The uncertainty of a measurement (two judgements) is at least ±1 of the smallest scale reading.
In a car airbag, sodium azide (NaN3) decomposes to form sodium metal and nitrogen gas. 2NaN3(s) → 2Na(s) + 3N2(g) The sodium metal then reacts with potassium nitrate to produce more nitrogen gas. 10Na(s) + 2KNO3(s) → N2(g) + 5Na2O(s) + K2O(s) If 2.00 mol of sodium azide react in this way, how many molecules of N2 will be formed? (The Avogadro constant L = 6.022 × 1023 mol-1)
10NaN3(s) --> 10Na(s) + 15N2(g) 10Na(s) + 2KNO3(s) --> K2O(s) + 5Na2O(s) + N2(g) Cancel out to get 10NaN3 + 2KNO3 --> 16N2(g) + K2O(s) + 5Na2O(s) 10 moles of NaN3 produces 16 moles of N2, therefore, 1 mol of NaN3 will produce 1.6 moles of N2 (divide both by 10) and therefore 2 moles will produce 3.2 moles of N2 gas so 3.2 x L = 1.927x10²⁴ = 1.93x10²⁴
You are given the following titre readings: Titre / cm3 10.00 9.50 9.75 9.65 Calculate the mean titre.
9.65 + 9.75 /2 = 9.70 cm3
How do you calculate the maximum percentage apparatus uncertainty in the final result?
Add all the individual equipment uncertainties together.
2 mol of ideal gas X are stored in a flask of fixed volume. Which of the following changes would lead to the greatest increase in pressure inside the flask? A: Increasing the temperature from 20 °C to 200 °C B: Adding another 1 mol of gas X into the flask at fixed temperature C: Adding 0.5 mol of argon gas and increasing the temperature from 20 °C to 150 °C D: Removing 0.5 mol of gas X and increasing the temperature from 20 °C to 300 °C
C
There are 392 mol of pure gold in a bar measuring 10 cm by 10 cm by 40 cm. What is the density of gold in kg dm−3?
Density = Mass / volume Mass = 392 x 196.97 =77212g = 77.2kg Volume = 10x10x40=4000cm³ =40dm³ D = 77.2 /40 = 1.93 kg dm⁻3
What happens and why do we put water into the conical flash during a titration?
Distilled water can be added to the conical flask during a titration to wash the sides of the flask so that all the acid on the side is washed into the reaction mixture to react with the alkali. It does not affect the titration reading as water does not react with the reagents or change the number of moles of acid added.
How can you decrease an apparatus uncertainty?
Either decrease the sensitivity uncertainty by using apparatus with a greater resolution (finer scale divisions ) or you can increase the size of the measurement made.
Relationship between % uncertainty and %difference between the actual value and calculated value/
If the %uncertainty due to the apparatus < percentage difference between the actual value and the calculated value then there is a discrepancy in the result due to other errors. If the %uncertainty due to the apparatus > percentage difference between the actual value and the calculated value then there is no discrepancy and all errors in the results can be explained by the sensitivity of the equipment.
What happens if the jet space in a titration is not filed?
If the jet space is not filled properly prior to commencing the titration it will lead to errors if it then fills during the titration, leading to a larger than expected titre reading.
Calculating the percentage difference between the actual value and the calculated value
If we calculated an Mr of 203 and the real value is 214, then the calculation is as follows: Calculate difference 214-203 = 11 % = 11/214 x100 =5.41%
30 cm3 of xenon are mixed with 20 cm3 of fluorine. The gases react according to the following equation. Xe(g) + F2(g) → XeF2(g) Assume that the temperature and pressure remain constant. What is the final volume of gas after the reaction is complete?
1 mol of any gad will occupy the same volume in the same conditions. So 20cm³ of Xe, F₂, XeF₂ are of the same number of moles. Also 1:1 ration, we can deduce 20cm³ of each reactant reacts to give 20cm³ of XeF₂. We still have 10cm³ of Xe in excess hence 20cm³ of XeF₂ + 10cm³ Xe -> 30cm³ in total
How would you reduce the uncertainty in a burette reading?
It is necessary to make the titre a larger volume. This could be done by: increasing the volume and concentration of the substance in the conical flask or by decreasing the concentration of the substance in the burette.
A student was given a powder made from a mixture of anhydrous barium chloride and anhydrous magnesium chloride. The student dissolved 1.056 g of the powder in water in a conical flask and added an excess of sulfuric acid. A white precipitate formed and was filtered off, washed and dried. The mass of this solid was 0.764 g. Identify the white precipitate and calculate the percentage, by mass, of magnesium chloride in the powder.
Precipitate : BaSO₄ Moles of Barium sulfate = mass/Mr (= 0.764 / 233.4) = 0.003273 moles Mass of Barium chloride = 208.3 × 0.003273 = 0.6818 g Percentage of Magnesium chloride = 1.056-0.6818 / 1.056 = 35.4%
A pure solid is thought to be calcium hydroxide. The solid can be identified from its relative formula mass. The relative formula mass can be determined experimentally by reacting a measured mass of the pure solid with an excess of hydrochloric acid. The equation for this reaction is Ca(OH)2 + 2HCl - > CaCl2 + 2H2O The unreacted acid can then be determined by titration with a standard sodium hydroxide solution. You are provided with 50.0 cm3 of 0.200 mol dm−3 hydrochloric acid. Outline, giving brief practical details, how you would conduct an experiment to calculate accurately the relative formula mass of the solid using this method.
Stage 1: The acid must be in excess and calculation of amount of solid that permits this Moles of acid = 50.0 × 0.200 / 1000 = 1.00 × 10⁻² mol 2 mol of acid react with 1 mol of calcium hydroxide therefore moles of solid weighed out must be less than half the moles of acid = 0.5 × (1.00 × 10⁻²) = 5.00 × 10⁻³ mol Mass of solid must be < (5.00 × 10⁻³) × 74.1 = < 0.371 g Stage 2: Experimental method Measure out 50 cm³ of acid using a pipette and add the weighed amount of solid in a conical flask Titrate against 0.100 (or 0.200) mol dm⁻³ NaOH added from a burette and record the volume (v) when an added indicator changes colour Stage 3: Calculate Mr from the experimental data Moles of calcium hydroxide = 5.00 × 10⁻³ -(v/2 × conc NaOH) / 1000 = z mol Mr = mass of solid / z
Another student is required to make up 250 cm3 of an aqueous solution that contains a known mass of MHCO3. The student is provided with a sample bottle containing the MHCO3. Describe the method, including apparatus and practical details, that the student should use to prepare the solution.
Stage 1: transfers known mass of solid a) Weigh the sample bottle containing the solid on a (2 dp) balance b) Transfer to beaker* and reweigh sample bottle c) Record the difference in mass Stage 2: Dissolves in water a) Add distilled / deionised water b) Stir (with a glass rod) or swirl c) Until all solid has dissolved Stage 3: Transfer, washing and agitation a) Transfer to volumetric / graduated flask. b) With washings c) Make up to 250 cm3 / mark with water d) Shakes/inverts/mixes
The student identified use of the burette as the largest source of uncertainty in the experiment. Using the same apparatus, suggest how the procedure could be improved to reduce the percentage uncertainty in using the burette. Justify your suggested improvement.
Use a larger mass of solid OR use a more concentrated solution of MHCO3 So a larger titre/reading will be needed
How would you reduce uncertainties in measuring mass?
Using a more accurate balance or a larger mass will reduce the uncertainty in weighing a solid Weighing sample before and after addition and then calculating difference will ensure a more accurate measurement of the mass added.
Calculate the amount, in moles, of MHCO3 in 250 cm3 of the solution. Then calculate the experimental value for the Mr of MHCO3. Give your answer to the appropriate number of significant figures.
mol MHCO3 =9.984x10⁻⁴ ×10 (= 9.89 × 10⁻³ ) Mr = 1.464/9.894 x 10⁻³ = 148
Calculate the mass of ammonia in flask Q. Give your answer to the appropriate number of significant figures.
pv=nRT n= (pv)/(RT) = 0.04091 Mass= 0.04091 x 17 =0.696g
When the tap is opened, ammonia passes into flask P. The temperature decreases by 5 °C. The final pressure in both flasks is 75.0 kPa. Calculate the volume, in cm3, of flask P.
v=(nRT) /p = 1.337x10⁻³ volume in p = total volume - volume of Q = 1.337 x 10⁻3 - 1x10⁻³ = 3.37 x 10⁻⁴ m³ =337cm³
What kind of uncertainties do each types of equipments have?
•balance 0.001 g •volumetric flask 0.1 cm3 •25 cm3 pipette 0.1 cm3 •burette (start & end readings and end point ) 0.15 cm3
What is the general method for a titration?
•rinse equipment (burette with acid, pipette with alkali, conical flask with distilled water) •pipette 25 cm3 of alkali into conical flask •touch surface of alkali with pipette ( to ensure correct amount is added) •adds acid solution from burette •make sure the jet space in the burette is filled with acid •add a few drops of indicator and refer to colour change at end point •phenolphthalein [pink (alkali) to colourless (acid): end point pink colour just disappears] [use if NaOH is used] •methyl orange [yellow (alkali) to red (acid): end point orange] [use if HCl is used] •use a white tile underneath the flask to help observe the colour change •add acid to alkali whilst swirling the mixture and add acid dropwise at end point •note burette reading before and after addition of acid •repeats titration until at least 2 concordant results are obtained- two readings within 0.1 of each other
