3.1.2 Amount of Substance PPQs Flashcards
The student rinses a burette with deionised water before filling with sodium hydroxide solution. State and explain the effect, if any, that this rinsing will have on the value of the titre.
Titre value would increase Because the sodium hydroxide solution would be more dilute
how many cm^3 are in 1m^3?
1,000,000cm^3
how do you calculate percentage uncertainty when 2 masses have been found with the same instrument?
multiply the uncertainty by 2 on top of the fraction
13.5 g of Al reacts with an excess of O to form Al2O3; 18.4g was produced. The equation is: 2Al + 1.5O2 ⇒ Al2O3. Calculate the percentage yield.
- Calculate theoretical number of moles (m/Mr): 13.5/27 = 0.5 mol. 1:2 ratio= 0.5/2 = 0.25 mol.
- Calculate theoretical mass (n*Mr): 0.25 * 102 = 25.5g
- Find actual mass: 18.4g
- Use equation: (18.4/25.5) * 100 = 72.2%
Nitration of 1.7g of methyl benzoate (Mr=136) produces methyl 3-nitrobenzoate (Mr=181). The percentage yield is 65%. What mass of methyl 3-nitrobenzoate is produced?
- n(methyl benzoate)= 1.7/136 = 0.0125 mol
- n(methyl 3-nitrobenzoate) = 1:1 ratio (assumed) = 0.0125 mol
- T.M(methyl 3-nitrobenzoate) = 0.0125 * 181 = 2.2625g
- Actual mass / 2.2625g = 0.65
- A.M = 0.65 * 2.2625 = 1.47g
After reaction of some zinc metal with excess sulfuric acid, a student collected 40.8g of ZnSO4 crystals. The yield of crystals was 70%. The equation is: Zn + H2SO4 + 7H2O ⇒ ZnSO4 + H2. What was the original mass of zinc used?
- Actual Mass(ZnSO4) = 40.8g
- 40.8 / Theoretical Mass = 0.7 Theoretical Mass = 40.8/0.7 = 58.29g
- Theoretical Moles(ZnSO4) = 58.29/161.5 = 0.36 mol
- Theoretical Moles(Zn) = 1:1 ratio = 0.36 mol
- Mass(Zn) = 0.36 * 65.4 = 23.54g
Calculate the volume of a balloon containing 0.22 moles of Helium at 25 degrees celsius and 100kPa.
- T(He) = 25 celsius = 298 K
- P(He) = 100kPa = 100 000 Pa
- V(He) = (0.22 mol * 8.32 * 298K) / 100000Pa = 5.45 * 10^-3 m^3
A sample of B2O3 was reacted completely with C and Cl. The 2 gases produced occupied a total volume of 5000cm^3 at a pressure of 100kPa and a temperature of 298K. The equation is: B2O3 + 3C + 3Cl2 ⇒ 2BCl3 + 3CO. Calculate the mass of B2O3 that reacted.
- v(gases) = 5000cm^3 = 5 * 10^-3 m^3
- P(gases) = 100kPa = 100 000 Pa
- n(gases) = (100000Pa * 5*10^-3) / (8.31 * 298K) = 0.2019 mol
- 0.2010 mol is 5 moles in equation
- therefore, 0.0464 mol is 1 mole in equation
- n(B2O3) = 0.0464 mol
- m(B2O3) = 0.0464 mol * 69.6 = 2.81g
An unknown compound consists of 55% C, 9% H, and the remainder is O. The Mr is 88. Calculate the molecular formula.
- C H O
- mass: 55g 9g 36g
- Mr: 12 1 16
- n: 4.583 9 2.25
- ratio: 2 : 4 : 1
- empirical formula = C2H4O
- molecular formula = C4H8O2
10.1g of a white solid contains 4.01g Ca, 1.2g C, and 4.8 g of O. Calculate the empirical formula
- Ca C O
- mass: 4.01g 1.2g 4.8g
- Mr: 40.1 12 16
- n: 0.1 0.1 0.3
- ratio: 1 : 1 : 3
- formula = CaCO3
The heat released when 1.00 g of ethanol (Mr = 46.0) undergoes complete combustion is 29.8 kJ. What is the heat released by each molecule, in joules, when ethanol undergoes complete combustion?
- Calculate the number of moles of ethanol (C₂H₅OH) in 1.00 g: 0.0217
- Calculate the number of ethanol molecules in 0.02174 moles (Avogadro’s tells us that 1 mole of a substance contains approximately 6.022×10^23 molecules.) So, the number of ethanol molecules in 0.02174 moles is: 1.3 * 10^22
- Convert the heat released (29.8 kJ) to joules: 29800J
- Calculate the heat released per molecule: 2.28*10^-18
The equation below represents the complete combustion of butane.
C4H10 + 6.5O2 -> 4CO2 +5H2O
20 cm3 of butane are completely burned in 0.20 dm3 of oxygen. Which statement is correct?
A 40 cm3 of carbon dioxide are formed
B 0.065 dm3 of oxygen react
C 70 cm3 of oxygen remain
D 0.50 dm3 of steam are formed
Let’s go through each of the statements step by step.
- A. 40 cm³ of carbon dioxide are formed
- From the balanced equation for the combustion of butane we know that 2 moles of butane produce 8 moles of carbon dioxide. So the ratio of butane to carbon dioxide is 1:4. Therefore, 20 cm³ of butane will produce 80cm3 ofcarbondioxide. So, this statement is incorrect.
- B. 0.065 dm³ of oxygen react
- We already found that 20 cm³ of butane would need 130 cm³ (0.13 dm³) of oxygen to completely combust. So, this statement is also incorrect.
- C. 70 cm³ of oxygen remain
- Initially, we have 200 cm³ of oxygen. After the reaction, the remaining oxygen would be 70cm3 ofoxygen. So, this statement is correct.
- D. 0.50 dm³ of steam are formed
- From the balanced equation, 2 moles of butane produce 10 moles of water (steam). So the ratio of butane to water is 1:5. Therefore, 20 cm³ of butane will produce 100cm3 ofwater(steam) Which is equivalent to 0.10 dm³. So, this statement is also incorrect.
- Therefore, the correct statement is C. 70 cm³ of oxygen remain.
A student does an experiment to determine the percentage by mass of sodium chlorate(I), NaClO, in a sample of bleach solution.
Method:
* Dilute a 10.0 cm3 sample of bleach solution to 100 cm3 with distilled water.
* Transfer 25.0 cm3 of the diluted bleach solution to a conical flask and acidify using sulfuric acid.
* Add excess potassium iodide to the conical flask to form a brown solution containing l2(aq).
* Add 0.100 mol dm–3 sodium thiosulfate solution (Na2S2O3) to the conical flask from a burette until the brown solution containing l2(aq) becomes a colourless solution containing l–(aq).
The student uses 33.50 cm3 of sodium thiosulfate solution.
The density of the original bleach solution is 1.20 g cm–3
The equations for the reactions in this experiment are:
ClO–(aq) + 2 H+(aq) + 2 l–(aq) → Cl–(aq) + H2O(l) + l2(aq)
2 S2O32–(aq) + l2(aq) → 2 l–(aq) + S4O62–(aq)
Use all the information given to calculate the percentage by mass of NaClO in the original bleach solution
- n(S2O32-) = 33.50 x 0.100 ÷1000 = 0.00335
- n(I2) = 0.00335 ÷ 2 = 0.001675 (from eqn 2)
- n(ClO–) in 25 cm3 pipette = 0.001675 (from eqn 1)
- n(ClO–) in 100 cm3 flask = 0.001675 x 4 = 0.00670 = n(NaClO) in original 10 cm3 sample
- mass (NaClO) = 0.00670 x 74.5 = 0.499 g
- mass (bleach) = 10.0 x 1.20 = 12 g
- % by mass of NaClO = = 4.16 %
A 30 cm3 sample of nitrogen was reacted with a 60 cm3 sample of fluorine according to the equation: 0.5N2 + 1.5F2 -> NF3.
What is the volume of the gas mixture after the reaction, at constant temperature and pressure?
- 1 volume of N₂ reacts with 3 volumes of F₂ to produce 2 volumes of NF₃.
- We can determine the limiting reagent by calculating the required volume of F₂ for the given N₂: 30cm × 3 = 90cm
- Since we only have 60 cm³ of F₂, fluorine is the limiting reagent. Therefore, all of the F₂ will be used up in the reaction, and we need to calculate how much N₂ reacts with it: 60cm / 3 = 20cm
- Next, we can calculate the volume of NF₃ produced: (2 x 60cm) / 3 = 40cm
- determine the total volume of the gas mixture after the reaction: Initial volume of nitrogen = 30 cm³, Volume of nitrogen remaining after reaction = 30 cm³ - 20 cm³ = 10 cm³,
Volume of fluorine remaining after reaction = 0 cm³ (all fluorine reacted) - Volume of nitrogen trifluoride produced = 40 cm³ - Total volume of the gas mixture after the reaction: 10cm
(remainingN₂) + 40cm (producedNF₃) = 50cm
