Some Basic Concepts of Chemistry — Practice Questions with Answers

Law of conservation of mass: Matter can neither be created nor destroyed in a chemical reaction — the total mass of reactants equals the total mass of products.

Law of definite proportions: A given chemical compound always contains its constituent elements in a fixed, definite proportion by mass, regardless of the source or method of preparation of that compound.

Empirical formula gives the simplest whole-number ratio of atoms of each element in a compound.

Molecular formula gives the actual number of atoms of each element present in one molecule of the compound — always a whole-number multiple of the empirical formula.

Example: Glucose has molecular formula $C_6H_{12}O_6$, but its empirical formula is $CH_2O$ (simplest ratio 1:2:1), since the molecular formula is exactly 6 times the empirical formula.

The limiting reagent is the reactant that is completely consumed first in a chemical reaction, thereby limiting (determining) the maximum possible amount of product that can form.

Identifying the limiting reagent is essential because once it is fully consumed, the reaction cannot proceed further, regardless of how much of the other reactant(s) remain in excess — calculating theoretical yield based on the reactant present in excess (rather than the actual limiting reagent) would give an incorrect, overestimated prediction of product formed.

Molarity = moles of solute / volume of solution (in litres). Molality = moles of solute / mass of solvent (in kg).

Molality is preferred in contexts involving temperature changes because it is based on mass (which doesn't change with temperature), while molarity is based on solution volume, which can expand or contract with temperature changes — making molarity values temperature-dependent, while molality values remain constant regardless of temperature.

Percentage yield is a measure of the actual efficiency of a chemical reaction, comparing the actual amount of product obtained experimentally to the maximum theoretically possible amount (calculated from stoichiometry, assuming the limiting reagent reacts completely).

$\text{Percentage yield} = \dfrac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%$

Percentage yield is typically less than 100% in practice, due to factors like incomplete reactions, side reactions, or loss of product during handling/purification.

Avogadro's number ($6.022\times10^{23}$) provides the essential bridge between the incredibly small scale of individual atoms/molecules and the practical, measurable macroscopic quantities (like grams) used in the laboratory.

By defining a mole as containing exactly this number of particles, chemists can conveniently work with laboratory-scale masses (grams) while still being able to precisely calculate and relate these back to the actual number of individual atoms or molecules involved in a reaction — this connection is fundamental to all stoichiometric calculations in chemistry.

Significant figures indicate the precision/reliability of a measured quantity — reporting a value with an appropriate number of significant figures accurately reflects the actual precision of the measuring instrument or method used, avoiding the false impression of greater precision than genuinely exists.

When performing calculations using measured quantities, following significant figure rules (e.g., a calculated result cannot be more precise than the least precise measurement used) ensures that reported results are not misleadingly over-precise, maintaining scientific accuracy and honesty in reporting experimental or calculated chemical data.

The law of multiple proportions states that when two elements combine to form more than one compound, the different masses of one element that combine with a fixed mass of the other element are always in a ratio of small whole numbers.

Example: Carbon and oxygen form two compounds — carbon monoxide (CO) and carbon dioxide ($CO_2$). For a fixed mass of carbon, the masses of oxygen combining in CO versus $CO_2$ are in the simple whole-number ratio of 1:2, consistent with this law.

Stoichiometry is the quantitative study of the relationships between the amounts of reactants consumed and products formed in a chemical reaction, based on the balanced chemical equation.

In industrial processes, accurate stoichiometric calculations are essential for determining the precise quantities of raw materials needed, predicting expected product yields, minimising wasteful excess reactant use, and ensuring cost-effective, efficient large-scale chemical production — errors in stoichiometric planning can lead to significant financial losses or inefficient resource use at industrial scale.

Mass percentage expresses the concentration of a component as the mass of that component divided by the total mass of the solution, multiplied by 100 — based purely on mass.

Mole fraction expresses concentration as the number of moles of a component divided by the total number of moles of all components present — based on the actual particle/mole count, not mass. Since different substances have different molar masses, the same solution can have quite different mass percentage and mole fraction values for the same component.

Number of moles $= \dfrac{\text{Given mass}}{\text{Molar mass}} = \dfrac{22}{44}$

Number of moles $= \mathbf{0.5\text{ mol}}$

Assume 100 g of compound: C = 40 g, H = 6.7 g, O = 53.3 g

Moles: $n_C = 40/12 = 3.33$; $n_H = 6.7/1 = 6.7$; $n_O = 53.3/16 = 3.33$

Simplest ratio (divide by smallest, 3.33): $C: 3.33/3.33=1$; $H: 6.7/3.33=2.01\approx2$; $O: 3.33/3.33=1$

Empirical formula = $\mathbf{CH_2O}$ (ratio 1:2:1)

Moles of NaOH $= \dfrac{4}{40} = 0.1\text{ mol}$

Volume in litres $= 250\text{ mL} = 0.25\text{ L}$

Molarity $= \dfrac{\text{moles}}{\text{volume in L}} = \dfrac{0.1}{0.25}$

Molarity $= \mathbf{0.4\text{ M}}$

From the balanced equation, 1 mole $N_2$ requires 3 moles $H_2$. For 3 moles $N_2$, required $H_2 = 3\times3 = 9$ moles.

Available $H_2$ = 6 moles, which is less than the required 9 moles — so $H_2$ is the limiting reagent.

From the equation, 3 moles $H_2$ produce 2 moles $NH_3$, so 6 moles $H_2$ produce: $NH_3 = \dfrac{2}{3}\times6 = \mathbf{4\text{ moles}}$

Moles of glucose $= \dfrac{5}{180} = 0.0278\text{ mol}$

Mass of solvent in kg $= 500\text{ g} = 0.5\text{ kg}$

Molality $= \dfrac{\text{moles of solute}}{\text{mass of solvent in kg}} = \dfrac{0.0278}{0.5}$

Molality $\approx \mathbf{0.0556\text{ m} \approx 0.056\text{ mol/kg}}$

Total moles $= 2 + 8 = 10$

Mole fraction of ethanol: $x_{ethanol} = \dfrac{2}{10} = 0.2$

Mole fraction of water: $x_{water} = \dfrac{8}{10} = 0.8$

Verification: $0.2 + 0.8 = 1$ ✓ (mole fractions of all components always sum to 1)

Percentage yield $= \dfrac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%$

Percentage yield $= \dfrac{3.6}{4.0}\times100$

Percentage yield $= \mathbf{90\%}$

Moles of $NH_3 = \dfrac{3.4}{17} = 0.2\text{ mol}$

Number of molecules $= \text{moles}\times N_A = 0.2\times6.022\times10^{23}$

Number of molecules $= \mathbf{1.2044\times10^{23}\text{ molecules}}$

Molar mass of $H_2O = 2(1) + 16 = 18\text{ g/mol}$

Mass % of H: $\dfrac{2}{18}\times100 = 11.11\%$

Mass % of O: $\dfrac{16}{18}\times100 = 88.89\%$

Verification: $11.11\% + 88.89\% = 100\%$ ✓

Balanced equation: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

(Balancing carbon: 3 CO₂ needed; balancing hydrogen: 4 H₂O needed; balancing oxygen: RHS has $3(2)+4(1)=10$ oxygen atoms, requiring 5 $O_2$ molecules on the left)

From the balanced equation, 1 mole $C_3H_8$ requires 5 moles $O_2$. For 4 moles $C_3H_8$:

$O_2$ required $= 4\times5 = \mathbf{20\text{ moles}}$