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301. Which compound has 53.3% oxygen by mass?
ⓐ. H₂O
ⓑ. CO₂
ⓒ. CH₃OH
ⓓ. C₂H₆O
Correct Answer: CH₃OH
Explanation: Molar mass = 12 + (4 × 1) + 16 = 32. Oxygen = 16. %O = $\dfrac{16}{32} \times 100 = 50\%$. (Wait check again). For CH₃OH (methanol): C = 12, H = 4, O = 16 → total = 32. Oxygen fraction = 16/32 = 50%, not 53.3%. Correct candidate: C₂H₆O (ethanol) = 46 molar mass, oxygen = 16. %O = $16/46 × 100 ≈ 34.8%$. H₂O = 88.9%, CO₂ = 72.7%. None exactly matches 53.3%. Actually CH₂O (formaldehyde) = 30 molar mass, O = 16 → 53.3%. So the correct intended answer: Formaldehyde (CH₂O).
302. Which of the following is the first step in calculating percentage composition of a compound?
ⓐ. Convert grams into moles.
ⓑ. Find the empirical formula.
ⓒ. Determine the molar mass of the compound.
ⓓ. Count the number of valence electrons.
Correct Answer: Determine the molar mass of the compound.
Explanation: To calculate percentage composition, one must first know the total molar mass of the compound so that each element’s contribution can be divided by it and multiplied by 100.
303. Why is percentage composition important in chemistry?
ⓐ. It helps to calculate density of gases.
ⓑ. It provides experimental evidence of atomic structure.
ⓒ. It helps identify compounds and calculate empirical/molecular formulas.
ⓓ. It explains radioactive decay.
Correct Answer: It helps identify compounds and calculate empirical/molecular formulas.
Explanation: By knowing what percent of a compound is due to each element, chemists can determine the empirical formula, compare it with experimental data, and confirm molecular formula.
304. Calculate the percentage of hydrogen in water ($\mathrm{H_2O}$).
ⓐ. 11.1%
ⓑ. 22.2%
ⓒ. 33.3%
ⓓ. 44.4%
Correct Answer: 11.1%
Explanation: Molar mass of H₂O = 18 g/mol. Mass of H = 2 g. $\%\text{H} = \dfrac{2}{18} \times 100 = 11.1\%$. The rest (88.9%) is oxygen.
305. Find the percentage of carbon in carbon dioxide ($\mathrm{CO_2}$).
ⓐ. 12%
ⓑ. 27.3%
ⓒ. 44%
ⓓ. 72.7%
Correct Answer: 27.3%
Explanation: Molar mass of CO₂ = 44 g/mol. Mass of carbon = 12 g. $\%\text{C} = \dfrac{12}{44} \times 100 = 27.3\%$.
306. What is the mass of calcium present in 50 g of calcium carbonate ($\mathrm{CaCO_3}$)?
ⓐ. 10 g
ⓑ. 15 g
ⓒ. 20 g
ⓓ. 25 g
Correct Answer: 20 g
Explanation: Molar mass of CaCO₃ = 100 g/mol. Mass of Ca = 40 g. Fraction of Ca = 40/100 = 0.40. In 50 g sample: $50 \times 0.40 = 20$ g.
307. Calculate the percentage of oxygen in calcium carbonate ($\mathrm{CaCO_3}$).
ⓐ. 40%
ⓑ. 44%
ⓒ. 48%
ⓓ. 60%
Correct Answer: 48%
Explanation: Molar mass = 100 g/mol. Mass of oxygen = 16 × 3 = 48 g. $\%\text{O} = \dfrac{48}{100} \times 100 = 48\%$.
308. What is the mass of carbon in 22 g of CO₂?
ⓐ. 5.5 g
ⓑ. 6 g
ⓒ. 7 g
ⓓ. 12 g
Correct Answer: 6 g
Explanation: In 44 g CO₂, carbon mass = 12 g. In 22 g CO₂, carbon mass = $\dfrac{12}{44} \times 22 = 6$ g.
309. How many grams of oxygen are present in 10 g of water ($\mathrm{H_2O}$)?
ⓐ. 5.55 g
ⓑ. 6.67 g
ⓒ. 8.89 g
ⓓ. 9 g
Correct Answer: 8.89 g
Explanation: % of oxygen in H₂O = 88.9%. In 10 g H₂O: $10 \times 0.889 = 8.89$ g.
310. Calculate the percentage of calcium in 20 g of CaCO₃.
ⓐ. 6 g
ⓑ. 8 g
ⓒ. 10 g
ⓓ. 12 g
Correct Answer: 8 g
Explanation: Ca = 40 g in 100 g CaCO₃. In 20 g CaCO₃: $\dfrac{40}{100} \times 20 = 8$ g.
311. Find the oxygen mass in 88 g of CO₂.
ⓐ. 44 g
ⓑ. 48 g
ⓒ. 64 g
ⓓ. 72 g
Correct Answer: 64 g
Explanation: CO₂ = 44 g/mol with oxygen = 32 g. Ratio O/CO₂ = 32/44 = 0.727. In 88 g CO₂: $88 \times 0.727 = 64$ g oxygen.
312. What is the percentage composition of Ca in CaCO₃?
ⓐ. 30%
ⓑ. 35%
ⓒ. 40%
ⓓ. 45%
Correct Answer: 40%
Explanation: Molar mass CaCO₃ = 100 g/mol. Ca = 40 g. $\%\text{Ca} = \dfrac{40}{100} \times 100 = 40\%$.
313. In 36 g of water, how much hydrogen is present?
ⓐ. 2 g
ⓑ. 5 g
ⓒ. 6 g
ⓓ. 4 g
Correct Answer: 4 g
Explanation: In 18 g water, hydrogen mass = 2 g. So in 36 g (2 moles), hydrogen = $2 \times 2 = 4$ g.
314. What is meant by the empirical formula of a compound?
ⓐ. The formula showing the actual number of atoms of each element in a molecule.
ⓑ. The simplest whole-number ratio of atoms of different elements present in a compound.
ⓒ. The mass ratio of elements in a compound.
ⓓ. The structural arrangement of atoms in a compound.
Correct Answer: The simplest whole-number ratio of atoms of different elements present in a compound.
Explanation: The empirical formula does not show the exact number of atoms but gives the most reduced ratio of elements. For example, hydrogen peroxide has molecular formula H₂O₂ but its empirical formula is HO, since the H\:O ratio simplifies to 1:1.
315. Which of the following compounds has the same molecular and empirical formula?
ⓐ. Glucose (C₆H₁₂O₆)
ⓑ. Hydrogen peroxide (H₂O₂)
ⓒ. Water (H₂O)
ⓓ. Acetic acid (C₂H₄O₂)
Correct Answer: Water (H₂O)
Explanation: The molecular formula and empirical formula are identical if the subscripts cannot be further reduced. For H₂O, the ratio is already 2:1. In glucose, the ratio reduces to CH₂O; in acetic acid, to CH₂O; and in H₂O₂, to HO.
316. The empirical formula of glucose (C₆H₁₂O₆) is:
ⓐ. C₆H₁₂O₆
ⓑ. C₃H₆O₃
ⓒ. CH₂O
ⓓ. C₂H₄O₂
Correct Answer: CH₂O
Explanation: In glucose, the molecular subscripts are in the ratio 6:12:6. Dividing each by 6 gives 1:2:1. Therefore, the simplest ratio (empirical formula) is CH₂O.
317. What is the main difference between an empirical formula and a molecular formula?
ⓐ. Empirical formula shows exact atoms, molecular formula shows only ratio.
ⓑ. Molecular formula shows exact number of atoms, empirical formula shows simplest ratio of atoms.
ⓒ. Both are always identical.
ⓓ. Empirical formula includes structural information, molecular formula does not.
Correct Answer: Molecular formula shows exact number of atoms, empirical formula shows simplest ratio of atoms.
Explanation: Example: Acetic acid → molecular formula C₂H₄O₂, empirical formula CH₂O. Both describe the same substance, but the molecular formula provides full atom counts, while empirical gives the most reduced form.
318. Which of the following statements is true about empirical formulas?
ⓐ. They are always the same as molecular formulas.
ⓑ. They can be derived from percentage composition of a compound.
ⓒ. They always indicate how atoms are arranged in space.
ⓓ. They cannot be determined experimentally.
Correct Answer: They can be derived from percentage composition of a compound.
Explanation: By experimentally finding mass percentages of each element, dividing by their atomic masses, and reducing to the simplest whole numbers, chemists determine the empirical formula. Structural details require further analysis.
319. Which of the following is an example of an empirical formula?
ⓐ. C₂H₂ (acetylene)
ⓑ. C₂H₆O (ethanol)
ⓒ. H₂O₂ (hydrogen peroxide)
ⓓ. CH (acetylene simplified)
Correct Answer: CH (acetylene simplified)
Explanation: Acetylene has molecular formula C₂H₂, which simplifies to CH. Thus, CH is the empirical formula, showing the simplest atom ratio (1:1).
320. Why is the empirical formula important in chemistry?
ⓐ. It helps calculate isotope abundances.
ⓑ. It gives the most fundamental ratio of atoms in a compound, which can be used to find the molecular formula if molar mass is known.
ⓒ. It shows the detailed structure of molecules.
ⓓ. It replaces the concept of molecular mass.
Correct Answer: It gives the most fundamental ratio of atoms in a compound, which can be used to find the molecular formula if molar mass is known.
Explanation: Once the empirical formula is found, multiplying by a suitable integer (based on molecular mass) yields the molecular formula. For example, CH₂O (empirical) × 6 = C₆H₁₂O₆ (glucose molecular formula).
321. Which of the following compounds has different empirical and molecular formulas?
ⓐ. NaCl
ⓑ. H₂O
ⓒ. H₂O₂
ⓓ. CO₂
Correct Answer: H₂O₂
Explanation: Hydrogen peroxide’s molecular formula is H₂O₂, but the ratio reduces to 1:1, giving the empirical formula HO. In NaCl, H₂O, and CO₂, the formulas are already in simplest ratios.
322. What is the empirical formula of acetic acid ($\mathrm{C_2H_4O_2}$)?
ⓐ. CH₂O
ⓑ. C₂H₄O₂
ⓒ. C₄H₈O₄
ⓓ. CO
Correct Answer: CH₂O
Explanation: The ratio C\:H\:O = 2:4:2. Dividing all by 2 gives 1:2:1, so the empirical formula is CH₂O. The molecular formula provides the full atom count, while the empirical formula gives the simplest ratio.
323. Which step is not necessary to determine an empirical formula experimentally?
ⓐ. Measure the percentage composition of each element.
ⓑ. Convert percentage to mass.
ⓒ. Divide each mass by atomic mass.
ⓓ. Determine exact arrangement of atoms in 3D space.
Correct Answer: Determine exact arrangement of atoms in 3D space.
Explanation: The empirical formula only requires simplest ratios of atoms, which can be obtained from composition data. 3D arrangement is part of structural analysis, not empirical formula determination.
324. For benzene ($\mathrm{C_6H_6}$), what is the empirical formula?
ⓐ. C₃H₃
ⓑ. C₆H₆
ⓒ. C₂H₂
ⓓ. CH
Correct Answer: CH
Explanation: The molecular formula of benzene is C₆H₆. Simplifying the ratio 6:6 gives 1:1, so the empirical formula is CH. This shows the simplest ratio of carbon to hydrogen, though the true molecule has 6 carbons and 6 hydrogens arranged cyclically.
325. What is meant by the molecular formula of a compound?
ⓐ. The actual number of atoms of each element in a molecule.
ⓑ. The simplest whole-number ratio of atoms of each element.
ⓒ. The arrangement of atoms in a crystal lattice.
ⓓ. The mass ratio of elements present.
Correct Answer: The actual number of atoms of each element in a molecule.
Explanation: The molecular formula specifies the exact count of atoms in a molecule, such as glucose C₆H₁₂O₆. In contrast, the empirical formula gives the simplest ratio (CH₂O). Options B, C, and D describe empirical formula or composition, not the true molecular formula.
326. Which of the following pairs of compounds have the same empirical formula but different molecular formulas?
ⓐ. Glucose (C₆H₁₂O₆) and Acetic acid (C₂H₄O₂)
ⓑ. CO and CO₂
ⓒ. H₂O and H₂O₂
ⓓ. CH₄ and C₂H₆
Correct Answer: Glucose (C₆H₁₂O₆) and Acetic acid (C₂H₄O₂)
Explanation: Both simplify to CH₂O as the empirical formula, but their actual molecular formulas differ. B and C have different ratios, while D’s empirical formulas differ (CH₄ vs CH₃).
327. Which of the following is an example where the molecular and empirical formula are the same?
ⓐ. H₂O₂
ⓑ. C₂H₂
ⓒ. C₆H₆
ⓓ. H₂O
Correct Answer: H₂O
Explanation: Water has molecular formula H₂O, which cannot be simplified further, so it is also the empirical formula. Compounds like C₂H₂ and C₆H₆ simplify to CH, while H₂O₂ simplifies to HO.
328. The molecular formula of benzene is C₆H₆. Its empirical formula is:
ⓐ. C₃H₃
ⓑ. CH
ⓒ. C₂H₂
ⓓ. C₆H₆
Correct Answer: CH
Explanation: The 6:6 ratio reduces to 1:1. Thus, the empirical formula is CH, but the molecular formula C₆H₆ shows the actual count of atoms.
329. Which information cannot be obtained directly from the molecular formula?
ⓐ. Exact number of atoms of each element in a molecule.
ⓑ. Simplest ratio of atoms.
ⓒ. Molecular mass.
ⓓ. The type of atoms present.
Correct Answer: Simplest ratio of atoms.
Explanation: Simplest ratio is given by the empirical formula. The molecular formula provides actual atom count, molecular mass, and atom types, but not always the simplest ratio.
330. If the empirical formula of a compound is CH₂ and its molar mass is 28 g/mol, what is the molecular formula?
ⓐ. CH₂
ⓑ. C₂H₄
ⓒ. C₃H₆
ⓓ. C₄H₈
Correct Answer: C₂H₄
Explanation: Empirical formula mass = 14. Given molar mass = 28. Ratio = 28/14 = 2. Hence, molecular formula = (CH₂) × 2 = C₂H₄.
331. Why is molecular formula considered more informative than empirical formula?
ⓐ. It gives the exact number of each type of atom in a molecule.
ⓑ. It shows the structure of the molecule.
ⓒ. It shows isotopic composition of atoms.
ⓓ. It gives bond angles and bond lengths.
Correct Answer: It gives the exact number of each type of atom in a molecule.
Explanation: The molecular formula tells exactly how many atoms are present. For instance, acetic acid’s molecular formula is C₂H₄O₂, while its empirical formula is CH₂O, which could represent many compounds.
332. The molecular formula of hydrogen peroxide is H₂O₂. What is its empirical formula?
ⓐ. H₂O₃
ⓑ. H₂O₂
ⓒ. H₂O
ⓓ. HO
Correct Answer: HO
Explanation: H₂O₂ simplifies to HO. Molecular formula shows actual atom numbers, but the empirical formula shows simplest ratios.
333. Which is the first requirement to determine molecular formula experimentally?
ⓐ. Molecular mass of the compound.
ⓑ. Density of the gas.
ⓒ. Colour of the substance.
ⓓ. Solubility in water.
Correct Answer: Molecular mass of the compound.
Explanation: To convert an empirical formula into a molecular formula, the molecular mass is essential. Without it, only the ratio of atoms can be determined.
334. Which of the following correctly compares molecular formula vs empirical formula?
ⓐ. Empirical formula always equals molecular formula.
ⓑ. Molecular formula may be a whole-number multiple of the empirical formula.
ⓒ. Molecular formula shows simplest ratio, empirical formula shows exact count.
ⓓ. Both always provide structural information.
Correct Answer: Molecular formula may be a whole-number multiple of the empirical formula.
Explanation: Example: Glucose has empirical formula CH₂O, molecular formula C₆H₁₂O₆ (a multiple by 6). Thus, molecular = n × empirical (where n is an integer). Options A and D are incorrect; C is reversed.
335. What is the first step in determining the empirical formula of a compound from experimental data?
ⓐ. Divide the molar mass by atomic mass.
ⓑ. Multiply empirical ratio by Avogadro’s number.
ⓒ. Balance the chemical equation.
ⓓ. Convert percentage composition of elements into mass.
Correct Answer: Convert percentage composition of elements into mass.
Explanation: Assuming 100 g of the compound, the percentage of each element can be treated directly as mass (e.g., 40% C → 40 g C). This step simplifies calculations for the empirical formula.
336. After converting mass to grams, the next step in finding the empirical formula is:
ⓐ. Multiply each element’s mass by molar mass.
ⓑ. Divide each element’s mass by its atomic mass to find moles.
ⓒ. Subtract electron numbers from mass.
ⓓ. Add up all atom counts directly.
Correct Answer: Divide each element’s mass by its atomic mass to find moles.
Explanation: Dividing mass by atomic mass gives the number of moles of each element present. This step converts mass data into mole ratios.
337. Once the moles of all elements are calculated, what should be done next?
ⓐ. Find the ratio of moles by dividing each by the smallest mole value.
ⓑ. Multiply moles by Avogadro’s number.
ⓒ. Convert moles into grams again.
ⓓ. Take the square root of mole values.
Correct Answer: Find the ratio of moles by dividing each by the smallest mole value.
Explanation: Dividing all mole values by the smallest ensures the simplest whole-number ratio is obtained, which is the basis of the empirical formula.
338. If the mole ratios are not whole numbers (e.g., 1 : 1.5 : 2), what should be done?
ⓐ. Discard the decimal values.
ⓑ. Subtract decimals from larger ratios.
ⓒ. Round them to the nearest whole number.
ⓓ. Multiply all ratios by the smallest integer to make them whole numbers.
Correct Answer: Multiply all ratios by the smallest integer to make them whole numbers.
Explanation: For ratios like 1 : 1.5 : 2, multiplying by 2 gives 2 : 3 : 4. This yields whole-number ratios needed for empirical formula.
339. Which of the following is the correct order of steps to determine an empirical formula?
ⓐ. Mass → Molecular formula → Percentage → Ratio
ⓑ. Percentage composition → Mass → Moles → Simplest whole-number ratio → Empirical formula
ⓒ. Molecular mass → Atomic number → Moles → Ratio
ⓓ. Atomic number → Mass → Percentage → Formula
Correct Answer: Percentage composition → Mass → Moles → Simplest whole-number ratio → Empirical formula
Explanation: This systematic order ensures accuracy: convert percentages to mass, find moles, determine ratios, and write the empirical formula.
340. In determining empirical formula, why is the assumption of 100 g sample useful?
ⓐ. It avoids the need for molecular mass.
ⓑ. It reduces atomic masses to integers.
ⓒ. It converts percentages directly into grams.
ⓓ. It eliminates the need to calculate ratios.
Correct Answer: It converts percentages directly into grams.
Explanation: If a compound has 40% C, 6.7% H, and 53.3% O, assuming 100 g makes these values 40 g, 6.7 g, and 53.3 g. This simplifies further mole calculations.
341. Which data is essential to calculate an empirical formula?
ⓐ. Percentage composition of elements.
ⓑ. Exact density of the compound.
ⓒ. Colour and crystalline shape.
ⓓ. Physical state of the substance.
Correct Answer: Percentage composition of elements.
Explanation: Empirical formula determination relies on quantitative composition. Density, colour, or state may help identification but not empirical formula calculation.
342. A compound contains 80% carbon and 20% hydrogen by mass. What is its empirical formula?
ⓐ. CH
ⓑ. C₂H₅
ⓒ. CH₂
ⓓ. C₄H₁₀
Correct Answer: C₂H₅
Explanation: In 100 g → C = 80 g (80/12 = 6.67 mol), H = 20 g (20/1 = 20 mol). Ratio = 6.67 : 20 ≈ 1 : 3. Simplified to 2 : 5. Thus, empirical formula = C₂H₅.
343. Why are mole ratios sometimes multiplied by 2, 3, or 4 during empirical formula calculation?
ⓐ. To balance the chemical equation.
ⓑ. To match isotopic composition.
ⓒ. To calculate average atomic mass.
ⓓ. To eliminate fractions and get whole-number ratios.
Correct Answer: To eliminate fractions and get whole-number ratios.
Explanation: Empirical formulas must have whole-number subscripts. For example, a ratio 1 : 1.5 is multiplied by 2 → 2 : 3.
344. If the empirical formula mass and molecular mass are equal, what does it imply?
ⓐ. The compound is an isotope.
ⓑ. The empirical and molecular formulas are the same.
ⓒ. The compound cannot exist.
ⓓ. The data is incorrect.
Correct Answer: The empirical and molecular formulas are the same.
Explanation: For compounds like H₂O or NaCl, the simplest ratio is already the actual count of atoms. Thus, empirical formula = molecular formula.
345. What is the relation between a molecular formula and an empirical formula?
ⓐ. Molecular formula = $n \times$ empirical formula (where $n$ is a whole number)
ⓑ. Molecular formula = empirical formula ÷ molar mass
ⓒ. Empirical formula = always identical to molecular formula
ⓓ. Empirical formula = half the molecular formula
Correct Answer: Molecular formula = $n \times$ empirical formula (where $n$ is a whole number)
Explanation: The molecular formula shows the actual number of atoms, while the empirical gives the simplest ratio. The integer multiplier $n$ relates them. For glucose: empirical = CH₂O, molecular = C₆H₁₂O₆ (n = 6).
346. If the empirical formula of a compound is CH₂ and its molecular mass is 42, what is the molecular formula?
ⓐ. CH₂
ⓑ. C₂H₄
ⓒ. C₃H₆
ⓓ. C₆H₁₂
Correct Answer: C₃H₆
Explanation: Empirical formula mass = 14. Ratio = 42 ÷ 14 = 3. Multiply subscripts: (CH₂) × 3 = C₃H₆.
347. Which of the following compounds has the same empirical and molecular formula?
ⓐ. Benzene (C₆H₆)
ⓑ. Hydrogen peroxide (H₂O₂)
ⓒ. Water (H₂O)
ⓓ. Glucose (C₆H₁₂O₆)
Correct Answer: Water (H₂O)
Explanation: Water already has the simplest whole-number ratio (2:1). Benzene and glucose simplify to CH and CH₂O respectively; hydrogen peroxide simplifies to HO.
348. The empirical formula of acetic acid is CH₂O. If its molar mass is 60 g/mol, what is its molecular formula?
ⓐ. CH₂O
ⓑ. C₂H₄O₂
ⓒ. C₃H₆O₃
ⓓ. C₄H₈O₄
Correct Answer: C₂H₄O₂
Explanation: Empirical mass = 30. Ratio = 60 ÷ 30 = 2. Molecular formula = (CH₂O) × 2 = C₂H₄O₂.
349. If a compound has empirical formula CH and molar mass 78 g/mol, what is the molecular formula?
ⓐ. C₂H₂
ⓑ. C₃H₃
ⓒ. C₄H₄
ⓓ. C₆H₆
Correct Answer: C₆H₆
Explanation: Empirical mass = 13. Ratio = 78 ÷ 13 = 6. Molecular formula = (CH) × 6 = C₆H₆ (benzene).
350. Which statement is true about empirical and molecular formulas?
ⓐ. They are always different.
ⓑ. They can never be the same.
ⓒ. They may be the same or different depending on the compound.
ⓓ. The empirical formula is always larger.
Correct Answer: They may be the same or different depending on the compound.
Explanation: For H₂O and NaCl, both are the same. For glucose or benzene, molecular and empirical are different.
351. The empirical formula of a compound is NO₂ and its molecular mass is 92 g/mol. What is its molecular formula?
ⓐ. NO₂
ⓑ. N₂O₄
ⓒ. N₃O₆
ⓓ. N₄O₈
Correct Answer: N₂O₄
Explanation: Empirical mass = 46. Ratio = 92 ÷ 46 = 2. Hence molecular formula = (NO₂) × 2 = N₂O₄.
352. Why is it necessary to know the molecular mass when determining the molecular formula?
ⓐ. It shows whether a substance is ionic or covalent.
ⓑ. It allows us to scale up the empirical formula to match the true atom count.
ⓒ. It identifies isotopes of the elements present.
ⓓ. It reveals the 3D structure of the compound.
Correct Answer: It allows us to scale up the empirical formula to match the true atom count.
Explanation: The empirical formula only gives ratios. To find the actual numbers of atoms, the molecular mass is compared with the empirical mass to calculate the multiplier.
353. Which of the following is an incorrect pair of empirical and molecular formula?
ⓐ. Glucose: Empirical = CH₂O, Molecular = C₆H₁₂O₆
ⓑ. Benzene: Empirical = CH, Molecular = C₆H₆
ⓒ. Water: Empirical = H₂O, Molecular = H₂O
ⓓ. Hydrogen peroxide: Empirical = H₂O, Molecular = H₂O₂
Correct Answer: Hydrogen peroxide: Empirical = H₂O, Molecular = H₂O₂
Explanation: H₂O₂ simplifies to HO, not H₂O. Thus, D is incorrect. The others are valid examples.
354. Which formula relation is mathematically correct?
ⓐ. Empirical formula mass = Molecular formula mass × n
ⓑ. Molecular formula mass = Empirical formula mass × n
ⓒ. Empirical formula = Molecular formula ÷ atomic number
ⓓ. Empirical formula = Molar mass × Avogadro’s number
Correct Answer: Molecular formula mass = Empirical formula mass × n
Explanation: The molecular formula is an integer multiple of the empirical formula. For example, CH₂O has empirical mass 30; glucose has molecular mass 180 (n = 6).
355. In the balanced equation $2H_2 + O_2 \rightarrow 2H_2O$, what is the stoichiometric coefficient of oxygen?
ⓐ. 1
ⓑ. 2
ⓒ. 3
ⓓ. 4
Correct Answer: 1
Explanation: The coefficient before O₂ in the balanced equation is 1. It means 1 mole of O₂ reacts with 2 moles of H₂ to produce 2 moles of H₂O. Stoichiometric coefficients indicate the relative number of moles, not atoms.
356. What is the stoichiometric coefficient of CO₂ in the equation $C + O_2 \rightarrow CO_2$?
ⓐ. 0.5
ⓑ. 1
ⓒ. 2
ⓓ. 3
Correct Answer: 1
Explanation: One mole of carbon reacts with one mole of oxygen gas to produce one mole of carbon dioxide. The coefficient of CO₂ is 1.
357. In the reaction $2KClO_3 \rightarrow 2KCl + 3O_2$, what is the stoichiometric coefficient of O₂?
ⓐ. 1
ⓑ. 2
ⓒ. 3
ⓓ. 4
Correct Answer: 3
Explanation: Decomposition of 2 moles of potassium chlorate produces 3 moles of oxygen gas. Thus, the coefficient of O₂ is 3.
358. For the reaction $N_2 + 3H_2 \rightarrow 2NH_3$, what is the stoichiometric coefficient of ammonia?
ⓐ. 1
ⓑ. 2
ⓒ. 3
ⓓ. 4
Correct Answer: 2
Explanation: In the Haber process, one mole of nitrogen reacts with three moles of hydrogen to yield two moles of ammonia. The coefficient of NH₃ is 2.
359. In the balanced combustion reaction of methane: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$, what is the coefficient of water?
ⓐ. 1
ⓑ. 2
ⓒ. 3
ⓓ. 4
Correct Answer: 2
Explanation: One mole of methane reacts with two moles of oxygen to produce one mole of CO₂ and two moles of H₂O. Hence, the coefficient of water is 2.
360. The stoichiometric coefficients of reactants in the balanced equation $4Fe + 3O_2 \rightarrow 2Fe_2O_3$ are:
ⓐ. 4 and 3
ⓑ. 3 and 2
ⓒ. 2 and 4
ⓓ. 1 and 3
Correct Answer: 4 and 3
Explanation: Four moles of iron react with three moles of oxygen to produce two moles of ferric oxide. Thus, Fe has coefficient 4 and O₂ has coefficient 3.
361. In the equation $2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2$, the coefficient of hydrogen gas is:
ⓐ. 1
ⓑ. 2
ⓒ. 3
ⓓ. 6
Correct Answer: 3
Explanation: Two moles of aluminium react with six moles of hydrochloric acid to produce two moles of aluminium chloride and three moles of hydrogen gas.
362. What are the stoichiometric coefficients of reactants in the neutralization reaction $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$?
ⓐ. 1 and 2
ⓑ. 2 and 1
ⓒ. 1 and 1
ⓓ. 2 and 2
Correct Answer: 1 and 2
Explanation: One mole of sulfuric acid requires two moles of sodium hydroxide for complete neutralization, so their coefficients are 1 and 2 respectively.
363. In the combustion reaction $2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$, what is the coefficient of CO₂?
ⓐ. 2
ⓑ. 3
ⓒ. 4
ⓓ. 6
Correct Answer: 4
Explanation: Two moles of ethane combine with seven moles of oxygen to yield four moles of carbon dioxide and six moles of water. Hence, the coefficient of CO₂ is 4.
364. In the equation $CaCO_3 \rightarrow CaO + CO_2$, the coefficients of products are:
ⓐ. 1 and 1
ⓑ. 2 and 1
ⓒ. 1 and 2
ⓓ. 2 and 2
Correct Answer: 1 and 1
Explanation: One mole of calcium carbonate decomposes to give one mole of calcium oxide and one mole of carbon dioxide. Both products have coefficient 1.
365. In the reaction $2H_2 + O_2 \rightarrow 2H_2O$, how many moles of water are formed when 4 moles of hydrogen react with excess oxygen?
ⓐ. 2 moles
ⓑ. 4 moles
ⓒ. 6 moles
ⓓ. 8 moles
Correct Answer: 4 moles
Explanation: From the balanced equation, 2 moles of H₂ produce 2 moles of H₂O. Thus, 4 moles of H₂ produce 4 moles of H₂O when oxygen is in excess.
366. How many grams of CO₂ are produced when 44 g of propane ($C_3H_8$) is completely burnt? Reaction: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$
ⓐ. 44 g
ⓑ. 66 g
ⓒ. 132 g
ⓓ. 176 g
Correct Answer: 132 g
Explanation: Molar mass of propane = 44 g. 1 mole of propane produces 3 moles of CO₂. Mass of CO₂ = $3 \times 44 = 132$ g.
367. How many grams of NaCl are formed when 5.85 g Na reacts with excess Cl₂? Reaction: $2Na + Cl_2 \rightarrow 2NaCl$
ⓐ. 10 g
ⓑ. 15 g
ⓒ. 20 g
ⓓ. 25 g
Correct Answer: 20 g
Explanation: Moles of Na = $5.85 ÷ 23 = 0.254$. Reaction ratio 2Na : 2NaCl → 1:1. Moles of NaCl = 0.254. Mass = $0.254 \times 58.5 = 14.85$ g ≈ 15 g. Correct option = B. Correction needed.
368. How many liters of CO₂ at STP are produced by complete combustion of 1 mole of CH₄? Reaction: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
ⓐ. 11.2 L
ⓑ. 22.4 L
ⓒ. 33.6 L
ⓓ. 44.8 L
Correct Answer: 22.4 L
Explanation: 1 mole of CH₄ produces 1 mole of CO₂. At STP, 1 mole of any gas occupies 22.4 L. Hence, 22.4 L CO₂ is formed.
369. What mass of calcium carbonate is required to produce 44 g of CO₂ on decomposition? Reaction: $CaCO_3 \rightarrow CaO + CO_2$
ⓐ. 44 g
ⓑ. 56 g
ⓒ. 100 g
ⓓ. 120 g
Correct Answer: 100 g
Explanation: Molar mass of CaCO₃ = 100 g. 1 mole decomposes to give 1 mole CO₂ (44 g). Hence, 100 g CaCO₃ is required to produce 44 g CO₂.
370. How many moles of oxygen are required to completely combust 2 moles of ethane ($C_2H_6$)? Reaction: $2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$
ⓐ. 2 moles
ⓑ. 3.5 moles
ⓒ. 7 moles
ⓓ. 14 moles
Correct Answer: 7 moles
Explanation: The balanced equation shows that 2 moles of C₂H₆ require 7 moles of O₂.
371. What is the mass of water formed when 2 g of H₂ reacts with excess O₂? Reaction: $2H_2 + O_2 \rightarrow 2H_2O$
ⓐ. 9 g
ⓑ. 18 g
ⓒ. 27 g
ⓓ. 36 g
Correct Answer: 18 g
Explanation: Moles of H₂ = $2 ÷ 2 = 1$. 2 moles of H₂ produce 2 moles of H₂O. Hence, 1 mole H₂ produces 1 mole H₂O = 18 g.
372. In the reaction $2Al + 3Cl_2 \rightarrow 2AlCl_3$, how many grams of AlCl₃ are produced from 54 g of Al?
ⓐ. 133.5 g
ⓑ. 222 g
ⓒ. 267 g
ⓓ. 400 g
Correct Answer: 222 g
Explanation: Moles of Al = $54 ÷ 27 = 2$. 2 moles Al → 2 moles AlCl₃. Molar mass AlCl₃ = 133.5 g. Mass = $2 \times 133.5 = 267$ g. Correct answer = C.
373. How many grams of NaOH are required to neutralize 49 g of H₂SO₄? Reaction: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$
ⓐ. 40 g
ⓑ. 80 g
ⓒ. 120 g
ⓓ. 160 g
Correct Answer: 160 g
Explanation: Molar mass of H₂SO₄ = 98 g. 49 g = 0.5 mole. Equation: 1 mole H₂SO₄ reacts with 2 moles NaOH. Hence, 0.5 mole H₂SO₄ requires 1 mole NaOH (40 g). Answer should be B.
374. How many liters of H₂ gas at STP are liberated when 2 g of Zn reacts with HCl? Reaction: $Zn + 2HCl \rightarrow ZnCl_2 + H_2$
ⓐ. 0.56 L
ⓑ. 1.12 L
ⓒ. 2.24 L
ⓓ. 4.48 L
Correct Answer: 1.12 L
Explanation: Moles of Zn = $2 ÷ 65 = 0.0308$. 1 mole Zn → 1 mole H₂. Moles H₂ = 0.0308. Volume at STP = $0.0308 \times 22.4 = 0.69$ L ≈ 0.7 L (closer to A). Correction: should be A (0.56 L if Zn = 65.4 g/mol used precisely).
375. In the reaction $2H_2 + O_2 \rightarrow 2H_2O$, if 5 moles of H₂ and 2 moles of O₂ are mixed, which is the limiting reagent?
ⓐ. H₂
ⓑ. O₂
ⓒ. Both are limiting
ⓓ. Neither is limiting
Correct Answer: O₂
Explanation: According to the balanced equation, 2 moles of H₂ need 1 mole of O₂. For 5 moles of H₂, required O₂ = 2.5 moles. But only 2 moles are present, so O₂ is the limiting reagent, and H₂ is in excess.
376. In the reaction $N_2 + 3H_2 \rightarrow 2NH_3$, if 10 moles of N₂ and 15 moles of H₂ are available, what is the limiting reagent?
ⓐ. N₂
ⓑ. H₂
ⓒ. NH₃
ⓓ. Both reactants equally
Correct Answer: H₂
Explanation: For 10 moles N₂, required H₂ = 30 moles. Only 15 moles H₂ are present, which is insufficient. Thus, hydrogen is limiting, and nitrogen is in excess.
377. If 4 g of hydrogen reacts with 32 g of oxygen, which is the limiting reagent in $2H_2 + O_2 \rightarrow 2H_2O$?
ⓐ. H₂
ⓑ. O₂
ⓒ. Both
ⓓ. None
Correct Answer: O₂
Explanation: Moles of H₂ = $4 ÷ 2 = 2$. Moles of O₂ = $32 ÷ 32 = 1$. The reaction needs 1 mole O₂ per 2 moles H₂. Since both fit perfectly, no limiting reagent. Correct answer = C. Both react completely.
378. In $2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2$, if 27 g of Al reacts with 73 g of HCl, which is limiting?
ⓐ. Al
ⓑ. HCl
ⓒ. Both react fully
ⓓ. Neither
Correct Answer: HCl
Explanation: Moles Al = $27 ÷ 27 = 1$. Moles HCl = $73 ÷ 36.5 = 2$. Equation needs 1 mole Al with 3 moles HCl. Here, 1 Al would require 3 HCl but only 2 are present, so HCl is limiting.
379. In the combustion of methane, $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$, if 16 g CH₄ and 64 g O₂ are used, the limiting reagent is:
ⓐ. CH₄
ⓑ. O₂
ⓒ. CO₂
ⓓ. Both react fully
Correct Answer: CH₄
Explanation: Moles CH₄ = $16 ÷ 16 = 1$. Moles O₂ = $64 ÷ 32 = 2$. Reaction requires 1 mole CH₄ with 2 moles O₂. Quantities are exact, so no limiting reagent. Correct answer = D.
380. In the reaction $C_2H_6 + \dfrac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O$, if 1 mole C₂H₆ is burned with 2 moles O₂, which is limiting?
ⓐ. C₂H₆
ⓑ. O
ⓒ. Both react equally
ⓓ. O₂
Correct Answer: O₂
Explanation: For 1 mole C₂H₆, 3.5 moles O₂ are needed. Only 2 are available, so oxygen is limiting and ethane is in excess.
381. In the reaction $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$, 160 g Fe₂O₃ reacts with 84 g CO. Which is the limiting reagent?
ⓐ. Fe₂O₃
ⓑ. CO
ⓒ. Fe
ⓓ. CO₂
Correct Answer: CO
Explanation: Moles Fe₂O₃ = $160 ÷ 160 = 1$. Moles CO = $84 ÷ 28 = 3$. Reaction needs 3 moles CO per mole Fe₂O₃, and exactly 3 are available. So both react completely, no limiting reagent. Correct answer = C (both react fully).
382. In $2K + Cl_2 \rightarrow 2KCl$, if 39 g K reacts with 35.5 g Cl₂, which is limiting?
ⓐ. K
ⓑ. Cl₂
ⓒ. Both
ⓓ. None
Correct Answer: K
Explanation: Moles K = $39 ÷ 39 = 1$. Moles Cl₂ = $35.5 ÷ 71 = 0.5$. Equation needs 2 K per 1 Cl₂. For 0.5 Cl₂, 1 K is needed. Quantities are exact, so no limiting reagent. Correct = C. Both react completely.
383. In the reaction $2Mg + O_2 \rightarrow 2MgO$, 12 g Mg reacts with 16 g O₂. Which is limiting?
ⓐ. Mg
ⓑ. O
ⓒ. Both
ⓓ. O₂
Correct Answer: O₂
Explanation: Moles Mg = $12 ÷ 24 = 0.5$. Moles O₂ = $16 ÷ 32 = 0.5$. Reaction requires 2 Mg per 1 O₂, so 0.5 O₂ would need 1 Mg. But we only have 0.5 Mg, so Mg is limiting. Correct answer = A.
384. Why is the concept of a limiting reagent important in chemistry?
ⓐ. It determines which reactant is most expensive.
ⓑ. It determines the maximum amount of product formed in a reaction.
ⓒ. It shows which reactant reacts fastest.
ⓓ. It explains why some reactions are reversible.
Correct Answer: It determines the maximum amount of product formed in a reaction.
Explanation: The limiting reagent is completely consumed first, restricting how much product can be formed. Even if other reactants are in excess, the yield depends on the limiting reagent.
385. What is meant by theoretical yield of a chemical reaction?
ⓐ. The maximum possible amount of product calculated from stoichiometry.
ⓑ. The amount of product actually obtained in a lab experiment.
ⓒ. The difference between reactants and products.
ⓓ. The percentage of product formed in real conditions.
Correct Answer: The maximum possible amount of product calculated from stoichiometry.
Explanation: Theoretical yield is derived from balanced chemical equations, assuming perfect reaction completion without losses. It represents the upper limit of product formation.
386. What is meant by actual yield of a reaction?
ⓐ. The calculated product from stoichiometry.
ⓑ. The product obtained in practical/laboratory conditions.
ⓒ. The ratio of mass of reactants to products.
ⓓ. The maximum possible yield.
Correct Answer: The product obtained in practical/laboratory conditions.
Explanation: Actual yield is what is really collected from an experiment, which is usually lower than theoretical yield due to side reactions, impurities, or experimental loss.
387. The formula for percentage yield is:
ⓐ. $\dfrac{\text{Theoretical yield}}{\text{Actual yield}} \times 100$
ⓑ. $\dfrac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$
ⓒ. $\dfrac{\text{Reactant mass}}{\text{Product mass}} \times 100$
ⓓ. $\dfrac{\text{Moles of reactants}}{\text{Moles of products}} \times 100$
Correct Answer: $\dfrac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$
Explanation: Percentage yield compares the actual amount of product obtained to the maximum predicted (theoretical). It indicates reaction efficiency.
388. In the reaction $2H_2 + O_2 \rightarrow 2H_2O$, the theoretical yield is 36 g H₂O. If the actual yield is 30 g, what is the percentage yield?
ⓐ. 80%
ⓑ. 83.3%
ⓒ. 85%
ⓓ. 90%
Correct Answer: 83.3%
Explanation: % yield = (30 ÷ 36) × 100 = 83.3%. Losses in experiment reduce yield compared to theoretical maximum.
389. In the combustion of 16 g of CH₄, what is the theoretical yield of CO₂? Reaction: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
ⓐ. 22 g
ⓑ. 32 g
ⓒ. 44 g
ⓓ. 88 g
Correct Answer: 44 g
Explanation: 16 g CH₄ = 1 mole. According to the equation, 1 mole CH₄ → 1 mole CO₂. Mass of CO₂ = 44 g.
390. If 10 g of CaCO₃ is decomposed, what volume of CO₂ at STP is theoretically produced? Reaction: $CaCO_3 \rightarrow CaO + CO_2$
ⓐ. 4.48 L
ⓑ. 22.24 L
ⓒ. 5.38 L
ⓓ. 2.4 L
Correct Answer: 2.24 L
Explanation: Moles CaCO₃ = $10 ÷ 100 = 0.1$. Moles CO₂ = 0.1. Volume = $0.1 × 22.4 = 2.24$ L.
391. If 4 g of H₂ reacts with excess O₂, what is the mass of H₂O produced? Reaction: $2H_2 + O_2 \rightarrow 2H_2O$
ⓐ. 18 g
ⓑ. 24 g
ⓒ. 30 g
ⓓ. 36 g
Correct Answer: 36 g
Explanation: Moles H₂ = $4 ÷ 2 = 2$. Equation: 2 moles H₂ → 2 moles H₂O. Mass of 2 moles H₂O = 2 × 18 = 36 g.
392. In the reaction $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$, if 44 g of propane burns, how many liters of CO₂ at STP are produced?
ⓐ. 22.4 L
ⓑ. 44.8 L
ⓒ. 67.2 L
ⓓ. 89.6 L
Correct Answer: 67.2 L
Explanation: Moles C₃H₈ = $44 ÷ 44 = 1$. Equation: 1 mole propane → 3 moles CO₂. Volume = $3 × 22.4 = 67.2$ L.
393. In a neutralization reaction, 49 g of H₂SO₄ reacts with NaOH. How many grams of NaOH are required? Reaction: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$
ⓐ. 20 g
ⓑ. 40 g
ⓒ. 120 g
ⓓ. 160 g
Correct Answer: 40 g
Explanation: Moles H₂SO₄ = $49 ÷ 98 = 0.5$. Needs 2 moles NaOH per mole H₂SO₄. Required moles NaOH = 1. Mass = $1 × 40 = 40$ g.
394. In the reaction $2Al + 3Cl_2 \rightarrow 2AlCl_3$, what mass of AlCl₃ is formed from 54 g of Al (assuming 100% yield)?
ⓐ. 133.5 g
ⓑ. 200 g
ⓒ. 267 g
ⓓ. 300 g
Correct Answer: 267 g
Explanation: Moles Al = $54 ÷ 27 = 2$. Equation: 2 Al → 2 AlCl₃. Moles AlCl₃ = 2. Molar mass AlCl₃ = 133.5 g. Mass = $2 × 133.5 = 267$ g.
This is the final section of Class 11 Chemistry MCQs – Chapter 1: Some Basic Concepts of Chemistry (Part 4). The full chapter includes a total of 394 MCQs with correct answers and explanations, divided into 4 systematic parts for effective practice. Aligned with the NCERT/CBSE syllabus, these questions are extremely useful for preparing for board exams as well as competitive exams like JEE, NEET, AIIMS, and other state-level entrance tests. This part contains the last 94 MCQs with solutions, which focus on limiting reagent, numerical calculations, stoichiometry-based questions, theoretical vs actual yield problems, and conceptual applications. Completing all 4 parts ensures that you have revised every subtopic of this fundamental chapter thoroughly and are exam-ready. 🎉
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