Quantitative Chemistry (review)

1. Sulfur dioxide is an important chemical used in the manufacture of sulfuric acid by a process called the Contact process. Industrially, there are two methods for producing sulfur dioxide.

Method 1: burning sulfur in air S(l) + O2(g)  →  SO2(g)

Method 2: roasting iron sulfide in air 4FeS2(s) + 11O2(g)  →  2Fe2O3(s) + 8SO2(g)


a) Explain why the reaction for method 1 has a percentage atom economy of 100%.

Answer: There is only one product / no undesired product, so none of the reactants are wasted


b) Calculate the percentage atom economy for producing sulfur dioxide by method 2.
Relative formula masses (Mr): FeS2 = 120 ; O2 = 32 ; Fe2O3 = 160 ; SO2 = 64

Answer: Atom economy = 100 × (sum of the Mr of desired products ÷ sum of the Mr of all the reactants) = 100 × (8 × 64) ÷ [(4 × 120) + (11 × 32)] = 61.5 %


c) 100 kg of iron sulfide (FeS2 ) were roasted in an excess of air. Calculate the maximum theoretical yield of sulfur dioxide that this mass of FeS2 can produce. Relative formula masses (Mr): FeS2 = 120 ; SO2 = 64

Answer: mol FeS2 = mass ÷ Mr = 100 000 ÷ 120 = 833.3 mol
mol SO2 = 2 × mol FeS2 (from the 4 : 8 ratio in the balanced equation) =1666.7 mol SO2
mass SO2 = mol × Mr = 1666.7 × 64 = 106666 g = 106.7 kg


d) The actual yield of sulfur dioxide in method 2 is 95.0 kg. Calculate the percentage yield for this reaction.

Answer: % yield = 100 × (actual yield max. theoretical yield) = 100 × (95.0 ÷ 107) = 88.8 %


2. A student carried out a titration in order to determine the concentration of sulfuric acid in a solution. They titrated 25.0 cm3 portions of the sulfuric acid solution with a 0.850 mol/dm3 solution of potassium hydroxide. The equation for the reaction is: H2SO4 + 2KOH  →  K2SO4 + 2H2O. The student’s results are shown in the table below.


a) Calculate the mean value for the volume of potassium hydroxide solution added. Give your answer to two decimal places. Write down the uncertainty in these three titration volumes.

Answer: mean volume = (23.55 + 23.40 + 23.45) ÷ 3 = 23.47 cm3 ± 0.08 cm3


b) Use the answer and the information in the table above to calculate the concentration of the sulfuric acid solution in mol/dm3.

Answer: mol KOH = concentration × volume = 0.850 mol/dm3 × (23.47 ÷ 1000) = 0.01995 mol ;
mol H2SO4 = 12 × mol KOH (from 1 : 2 ratio in balanced equation) = 0.009975 mol ;
concentration of H2SO4 = mol ÷ volume = 0.009975 ÷ (25.0 ÷ 1000) = 0.399 mol/dm3


c) Use the answer above to calculate the mass of sulfuric acid dissolved in 25.0 cm3 of the sulfuric acid solution. Relative formula mass (Mr) H2SO4 = 98

Answer: mol H2SO4 in 25.0 cm3 = 0.009975 mol ;
mass H2SO4 = mol × Mr = 0.009975 × 98 = 0.978 g


3. Urea has the formula CH4N2O, and is an important chemical used in agriculture, the chemical industry and in the manufacture of explosives. Urea can be manufactured by reacting ammonia with carbon dioxide: 2NH3(g) + CO2(g)  →  CH4N2O(g) + H2O(g)

a) Give the number of moles of urea that will form when 0.46 mol of ammonia gas are reacted with an excess of carbon dioxide gas.

Answer: Volume ratio of NH3 : CH4N2O = 2 : 1 ; mol CH4N2O = 12 × mol NH3 = 0.23 mol


b) In one reaction, 5.00 g of ammonia reacts with 5.00 g of carbon dioxide. Show that the carbon dioxide is the limiting reactant in this reaction. Relative formula masses (Mr): NH3 = 17 ; CO2 = 44

Answer: mol NH3 = mass ÷ Mr = 5.00 g ÷ 17 = 0.294 mol ;
mol CO2 = mass ÷ Mr = 5.00 g ÷ 44 = 0.114 mol ;
so, 0.294 mol of NH3 will react with 0.147 mol of CO2 . There are fewer moles of CO2 , so CO2 will run out before the NH3 . CO2 is the limiting reactant.


c) Urea is highly soluble in water. A solution of urea was prepared by dissolving 0.75 g of urea into 400 cm3 of water. Calculate the concentration of this urea solution in g/dm3.

Answer: concentration = mass ÷ volume = 0.75 g ÷ (400 ÷ 1000) = 1.88 g/dm3


d) Convert the concentration of urea in the answer above from g/dm3 into mol/dm3. Relative formula mass (Mr) CH4N2O = 70

Answer: mol/dm3 = g/dm3 ÷ Mr = 1.88 ÷ 70 = 0.0269 mol/dm3


4. When heated, ammonium dichromate [(NH4)2Cr2O7 ] decomposes to form chromium oxide (Cr2O3 ), nitrogen gas (N2 ) and water. A student heated 7.56 g of ammonium dichromate until it had completely decomposed. 4.56 g of chromium oxide, 0.84 g of nitrogen gas, and 2.16 g of water were formed.

a) Calculate the relative formula mass of (NH4)2Cr2O7 . Relative atomic masses (Ar): N = 14 ; H = 1 ; Cr = 52 ; O = 6

Answer: Mr (NH4)2Cr2O7 = (2 × 14) + (8 × 1) + (2 × 52) + (7 × 16) = 252


b) Use the answer and the information above, to determine the balanced equation for this reaction. Relative formula masses (Mr): Cr2O3 = 152 ; N2 = 28 ; H2O = 18

Answer: mol (NH4)2Cr2O7 = mass ÷ Mr = 7.56 ÷ 252 = 0.03 mol ;
mol Cr2O3 = 4.56 ÷ 152 = 0.03 mol ;
mol N2 = 0.84 ÷ 28 = 0.03 mol ;
mol H2O = 2.16 ÷ 18 = 0.12 mol
Dividing each number of moles by 0.03 mol to give the whole number ratio. Ratio (NH4)2Cr2O7 : Cr2O3 : N2 : H2O = 1 : 1 : 1 : 4
Correct symbol equation: (NH4)2Cr2O7  →  Cr2O3 + N2 + 4H2O