JEE Chemistry Purification And Characterisation Of Organic Compounds MCQs Set 01

Answer: (d) \( \ce{C3H4} \)

Explanation:
1. Moles: C = 90/12 = 7.5, H = 10/1 = 10
2. Ratio C : H = 7.5 : 10 = 3 : 4

Teacher's Note:
1. Convert every percentage or mass to moles first and divide by the smallest value to get the simplest ratio.
2. Check the answer by recomputing the formula mass and comparing it with the given molecular mass.

Answer: (a) \( \ce{C4H8N4} \)

Explanation:
1. Molar mass = 22400/200 = 112 g/mol
2. Moles: C = 48/12 = 4, H = 8, N = 56/14 = 4, so the ratio is CH2N
3. Mass of CH2N = 12 + 2 + 14 = 28, and 112/28 = 4, giving \( \ce{C4H8N4} \)

Teacher's Note:
1. Use the standard factor method: the fraction of the element in the weighed product times its mass gives the mass of the element.
2. Divide by the mass of the sample and multiply by 100 for the percentage.

Answer: (d) 940

Explanation:
1. Each molecule must contain at least one S atom (32 u)
2. Minimum molar mass = 32 x 100/3.4 = 941, i.e. about 940

Teacher's Note:
1. Convert every percentage or mass to moles first and divide by the smallest value to get the simplest ratio.
2. Check the answer by recomputing the formula mass and comparing it with the given molecular mass.

Answer: (b) \( \ce{CH2} \)

Explanation:
1. H in 1.8 g of water = 1.8 x 2/18 = 0.2 g
2. C = 1.4 - 0.2 = 1.2 g, so moles C = 0.1 and H = 0.2
3. Ratio 1 : 2 gives \( \ce{CH2} \)

Teacher's Note:
1. Write the working as mass of element, then moles, then ratio. This avoids slips with atomic masses.
2. Round only when a ratio is within about 0.1 of a whole number.

Answer: (c) \( \ce{H2SO4} \)

Explanation:
1. Fuming nitric acid oxidises the sulphur of the compound to sulphuric acid
2. Barium chloride then precipitates it as \( \ce{BaSO4} \) for weighing

Teacher's Note:
1. Match the technique to the property that differs between the components: boiling point, solubility, volatility or adsorption.
2. Questions often ask why one method is chosen over another.

Answer: (b) 25 and 73.3

Explanation:
1. C in 0.22 g \( \ce{CO2} \) = 12/44 x 0.22 = 0.06 g, so % C = 0.06/0.24 x 100 = 25
2. % O = 100 - 25 - 1.66 = 73.3 (the compound contains only C, H and O)

Teacher's Note:
1. Convert every percentage or mass to moles first and divide by the smallest value to get the simplest ratio.
2. Check the answer by recomputing the formula mass and comparing it with the given molecular mass.

Answer: (c) \( \ce{C5H7N} \)

Explanation:
1. Moles: C = 74/12 = 6.17, H = 8.65, N = 17.3/14 = 1.24
2. Dividing by 1.24 gives 5 : 7 : 1, so \( \ce{C5H7N} \)

Teacher's Note:
1. Write the working as mass of element, then moles, then ratio. This avoids slips with atomic masses.
2. Round only when a ratio is within about 0.1 of a whole number.

Answer: (a) Equivalent weight x basicity

Explanation:
1. Equivalent weight = molecular weight / basicity
2. Hence molecular weight = equivalent weight x basicity

Teacher's Note:
1. Use the standard factor method: the fraction of the element in the weighed product times its mass gives the mass of the element.
2. Divide by the mass of the sample and multiply by 100 for the percentage.

Answer: (b) Different molecular weight

Explanation:
1. Same empirical formula means the same percentage composition
2. Molecular formula = (empirical formula) x n with different n, so the molecular weights differ (and so do the vapour densities)

Teacher's Note:
1. Convert every percentage or mass to moles first and divide by the smallest value to get the simplest ratio.
2. Check the answer by recomputing the formula mass and comparing it with the given molecular mass.

Answer: (d) \( \ce{CH2O} \)

Explanation:
1. Moles: C = 24/12 = 2, H = 4, O = 32/16 = 2
2. Ratio 1 : 2 : 1 gives \( \ce{CH2O} \)

Teacher's Note:
1. Use the standard factor method: the fraction of the element in the weighed product times its mass gives the mass of the element.
2. Divide by the mass of the sample and multiply by 100 for the percentage.

Answer: (a) \( \ce{CH3NH2} \)

Explanation:
1. Moles: C = 3.23, H = 16, N = 3.23
2. Ratio 1 : 5 : 1 gives \( \ce{CH5N} \), which is \( \ce{CH3NH2} \)

Teacher's Note:
1. Convert every percentage or mass to moles first and divide by the smallest value to get the simplest ratio.
2. Check the answer by recomputing the formula mass and comparing it with the given molecular mass.

Answer: (b) \( \ce{C2H4O} \)

Explanation:
1. Moles: C = 4.54, O = 2.28, H = 9.1
2. Dividing by 2.28 gives C : H : O = 2 : 4 : 1

Teacher's Note:
1. Write the working as mass of element, then moles, then ratio. This avoids slips with atomic masses.
2. Round only when a ratio is within about 0.1 of a whole number.

Answer: (a) \( \ce{C6H6} \)

Explanation:
1. Moles: C = 7.69, H = 7.69, so the empirical formula is \( \ce{CH} \) (mass 13)
2. n = 78/13 = 6, so the molecular formula is \( \ce{C6H6} \)

Teacher's Note:
1. Use the standard factor method: the fraction of the element in the weighed product times its mass gives the mass of the element.
2. Divide by the mass of the sample and multiply by 100 for the percentage.

Answer: (c) \( \ce{C2H6} \)

Explanation:
1. Moles: C = 6.67, H = 20, ratio 1 : 3 giving \( \ce{CH3} \)
2. Doubling gives the saturated hydrocarbon \( \ce{C2H6} \) (ethane)

Teacher's Note:
1. Write the working as mass of element, then moles, then ratio. This avoids slips with atomic masses.
2. Round only when a ratio is within about 0.1 of a whole number.

Answer: (c) \( \ce{C12O9} \)

Explanation:
1. Moles: C = 4.17, O = 3.125, ratio 4 : 3 so the empirical formula is \( \ce{C4O3} \) (mass 112)
2. 290/112 = 2.6, nearest whole number 3 gives \( \ce{C12O9} \) (mass 288)

Teacher's Note:
1. Use the standard factor method: the fraction of the element in the weighed product times its mass gives the mass of the element.
2. Divide by the mass of the sample and multiply by 100 for the percentage.

Answer: (d) \( \ce{C3H8O} \)

Explanation:
1. Moles: C = 5, H = 13.3, O = 1.67
2. Dividing by 1.67 gives 3 : 8 : 1

Teacher's Note:
1. Write the working as mass of element, then moles, then ratio. This avoids slips with atomic masses.
2. Round only when a ratio is within about 0.1 of a whole number.

Answer: (c) \( \ce{CH4O} \)

Explanation:
1. O = 64 - 24 - 8 = 32 g
2. Moles C = 2, H = 8, O = 2, so the ratio is 1 : 4 : 1

Teacher's Note:
1. Convert every percentage or mass to moles first and divide by the smallest value to get the simplest ratio.
2. Check the answer by recomputing the formula mass and comparing it with the given molecular mass.

Answer: (a) \( \ce{C2H4O2} \)

Explanation:
1. Moles: C = 6/12 = 0.5, H = 1, O = 8/16 = 0.5, so the empirical formula is \( \ce{CH2O} \) (30)
2. Molecular mass = 2 x 30 = 60, n = 2, giving \( \ce{C2H4O2} \)

Teacher's Note:
1. Write the working as mass of element, then moles, then ratio. This avoids slips with atomic masses.
2. Round only when a ratio is within about 0.1 of a whole number.

Answer: (c) \( \ce{CH3} \)

Explanation:
1. Moles: C = 6.67, H = 20
2. Ratio 1 : 3

Teacher's Note:
1. Convert every percentage or mass to moles first and divide by the smallest value to get the simplest ratio.
2. Check the answer by recomputing the formula mass and comparing it with the given molecular mass.

Answer: (b) \( \ce{CH2O} \)

Explanation:
1. O = 100 - 40 - 6.7 = 53.3%
2. Moles C = 3.33, H = 6.7, O = 3.33, ratio 1 : 2 : 1; the empirical formula is the simplest ratio, \( \ce{CH2O} \)

Teacher's Note:
1. Write the working as mass of element, then moles, then ratio. This avoids slips with atomic masses.
2. Round only when a ratio is within about 0.1 of a whole number.

Answer: (c) \( \ce{CH4N} \)

Explanation:
1. Moles: C = 3.33, H = 13.33, N = 3.33
2. Ratio 1 : 4 : 1

Teacher's Note:
1. Convert every percentage or mass to moles first and divide by the smallest value to get the simplest ratio.
2. Check the answer by recomputing the formula mass and comparing it with the given molecular mass.

Answer: (a) \( \ce{CHClO} \)

Explanation:
1. Moles: C = 1.54, H = 1.55, Cl = 55.04/35.5 = 1.55, O = 1.55
2. The ratio is 1 : 1 : 1 : 1

Teacher's Note:
1. Write the working as mass of element, then moles, then ratio. This avoids slips with atomic masses.
2. Round only when a ratio is within about 0.1 of a whole number.

Answer: (a) 18.52

Explanation:
1. S in 0.35 g \( \ce{BaSO4} \) = 32/233 x 0.35 = 0.0481 g
2. % S = 0.0481/0.2595 x 100 = 18.52

Teacher's Note:
1. Convert every percentage or mass to moles first and divide by the smallest value to get the simplest ratio.
2. Check the answer by recomputing the formula mass and comparing it with the given molecular mass.

Answer: (d) Catalytic agent

Explanation:
1. \( \ce{CuSO4} \) (with \( \ce{K2SO4} \)) is added to the digestion mixture
2. It speeds up the conversion of nitrogen to ammonium sulphate and so acts as a catalyst

Teacher's Note:
1. Learn each test with its observation and the compound responsible for the colour or precipitate.
2. Remember which method suits which kind of sample (volatile, non-volatile, heat-sensitive).

Answer: (b) \( \ce{CHCl3} \)

Explanation:
1. Moles: C = 0.84, H = 0.84, Cl = 2.51, ratio 1 : 1 : 3
2. Empirical mass 119.5; molecular mass = 2 x 60 = 120, so \( \ce{CHCl3} \)

Teacher's Note:
1. Write the working as mass of element, then moles, then ratio. This avoids slips with atomic masses.
2. Round only when a ratio is within about 0.1 of a whole number.

Answer: (c) 46.66

Explanation:
1. Urea is \( \ce{CO(NH2)2} \), molar mass 60
2. % N = 28/60 x 100 = 46.67

Teacher's Note:
1. Use the standard factor method: the fraction of the element in the weighed product times its mass gives the mass of the element.
2. Divide by the mass of the sample and multiply by 100 for the percentage.

Answer: (a) Use of anhydrous copper sulphate

Explanation:
1. White anhydrous \( \ce{CuSO4} \) turns blue on forming the pentahydrate
2. Litmus only indicates acidity or basicity, and taste and smell are unsafe and non-specific

Teacher's Note:
1. Learn each test with its observation and the compound responsible for the colour or precipitate.
2. Remember which method suits which kind of sample (volatile, non-volatile, heat-sensitive).

Answer: (c) Mass spectrometer

Explanation:
1. The three isotopes differ in mass
2. A mass spectrometer separates ions according to the mass-to-charge ratio

Teacher's Note:
1. Match the technique to the property that differs between the components: boiling point, solubility, volatility or adsorption.
2. Questions often ask why one method is chosen over another.

Answer: (a) \( \ce{C3H6O3} \)

Explanation:
1. Empirical mass = 30, n = 90/30 = 3
2. Molecular formula = \( \ce{C3H6O3} \)

Teacher's Note:
1. Write the working as mass of element, then moles, then ratio. This avoids slips with atomic masses.
2. Round only when a ratio is within about 0.1 of a whole number.