All Free Chemistry MCQs with Answers

51. The total number of atoms present in one mole of sulphuric acid, H2SO4, is

• A. 6.02 x 10^23
• B. 4.21 x 10^24
• C. 4.214 x 10^24
• D. 98

Explanation: One molecule contains 2 + 1 + 4, that is 7 atoms, so one mole contains 7 x 6.02 x 10^23, which is 4.214 x 10^24 atoms. The value 6.02 x 10^23 is the number of molecules, not of atoms, and is the trap in this question. The number 98 is the molar mass in grams per mole.

52. A compound has the empirical formula CH2O and a relative molecular mass of 180. Its molecular formula is

• A. CH2O
• B. C2H4O2
• C. C3H6O3
• D. C6H12O6

Explanation: The empirical formula mass is 12 + 2 + 16, that is 30, and 180 divided by 30 gives 6, so every subscript is multiplied by six to give C6H12O6, which is glucose. The empirical formula gives only the simplest whole number ratio of atoms, while the molecular formula gives the actual number in one molecule. The two coincide only when that multiplier happens to be one.

53. The percentage by mass of carbon in carbon dioxide, CO2, is about

• A. 27.3 per cent
• B. 12 per cent
• C. 44 per cent
• D. 72.7 per cent

Explanation: The molar mass is 12 + 32, that is 44, so the carbon fraction is 12 divided by 44 multiplied by 100, giving 27.3 per cent. The figure 72.7 per cent is the oxygen content, which is why it appears as a distractor, and the two must add to 100. Percentage composition is the first step in working out an empirical formula from experimental data.

54. The limiting reactant in a chemical reaction is the one that

• A. is present in the largest amount by mass
• B. is completely used up first and so determines the amount of product formed
• C. has the highest molar mass
• D. remains unreacted at the end

Explanation: The reaction stops when one reactant runs out, so that reactant fixes the maximum yield no matter how much of the others is present. It must be identified by comparing moles against the balanced equation, not by comparing masses, because a small mass of a light reactant can be many moles. The reactant left over at the end is the excess reactant.

55. In the reaction N2 + 3H2 gives 2NH3, if 1 mole of nitrogen is mixed with 2 moles of hydrogen, the limiting reactant is

• A. nitrogen, because it has the larger molar mass
• B. neither, because both are used up exactly
• C. hydrogen, because 3 moles would be needed to react with 1 mole of nitrogen
• D. nitrogen, because fewer moles of it are present

Explanation: The equation demands three moles of hydrogen for every mole of nitrogen, so 1 mole of nitrogen would need 3 moles of hydrogen and only 2 are available. Hydrogen therefore runs out first and limits the yield to two thirds of a mole of ammonia. Choosing the reactant present in the smaller number of moles without checking the ratio is the standard mistake.

56. In the reaction 2H2 + O2 gives 2H2O, 4 moles of hydrogen are mixed with 3 moles of oxygen. The amount of the excess reactant left over is

• A. 2 moles of hydrogen
• B. 3 moles of oxygen
• C. nothing, both are exactly consumed
• D. 1 mole of oxygen

Explanation: Four moles of hydrogen require only two moles of oxygen, so hydrogen is limiting and 3 minus 2, that is 1 mole of oxygen, remains unreacted. Four moles of water are produced. The point of the calculation is that the excess is found by subtracting what was actually used from what was supplied, not by comparing the starting amounts directly.

57. The theoretical yield of a reaction is

• A. the maximum mass of product calculated from the balanced equation and the limiting reactant
• B. the mass of product actually obtained in the laboratory
• C. always equal to the mass of the limiting reactant
• D. the mass of the excess reactant left over

Explanation: Theoretical yield is a calculation, assuming the reaction goes to completion with no losses and no side reactions, and it is the standard against which real performance is judged. What is actually collected is the actual yield and is nearly always smaller. Comparing the two as a percentage tells a chemist how efficient the process was.

58. A reaction has a theoretical yield of 25 g but only 20 g of product is obtained. The percentage yield is

• A. 125 per cent
• B. 80 per cent
• C. 5 per cent
• D. 20 per cent

Explanation: Percentage yield is actual divided by theoretical multiplied by 100, so 20 divided by 25 multiplied by 100 gives 80 per cent. Dividing the wrong way round gives 125 per cent, which should be recognised as impossible because a reaction cannot produce more than the equation allows. A figure above 100 per cent in real work means the product is still wet or impure.

59. The actual yield of a reaction is usually less than the theoretical yield for all of the following reasons EXCEPT

• A. some product is lost during filtration and transfer
• B. side reactions produce unwanted products
• C. the balanced equation was written incorrectly
• D. the reaction is reversible and does not go to completion

Explanation: Physical losses, competing side reactions and reversibility are all genuine reasons why a real yield falls short. A wrongly balanced equation is not a reason for a low yield at all; it simply makes the calculated theoretical yield wrong, so the comparison itself becomes meaningless. Recognising which factors are chemical and which are arithmetic is the point of the question.

60. How many grams of calcium oxide are produced when 1 mole of calcium carbonate decomposes completely, given CaCO3 gives CaO + CO2?

• A. 100 g
• B. 44 g
• C. 40 g
• D. 56 g

Explanation: The equation shows a one to one ratio, so 1 mole of CaCO3 gives 1 mole of CaO, whose molar mass is 40 + 16, that is 56 g. The value 100 g is the molar mass of the calcium carbonate that decomposed and 44 g is the mass of carbon dioxide released, and note that 56 plus 44 equals 100, which confirms mass is conserved.