Some Basic Concepts Of Chemistry – JEE Main ChemistryPractice Questions & PYQs

The molecular formula of a compound is a direct multiple of its empirical formula, which represents the simplest ratio of the atoms present in the substance. Calculating this requires converting the elemental mass percentages into moles by dividing each percentage value by the corresponding atomic mass based on a 100-gram sample. For carbon, this yields 4.52 moles; for hydrogen, 9.2 moles; and for oxygen, 2.28 moles.

Dividing each of these quantities by the smallest mole value, 2.28, produces a simple atomic ratio of 2:4:1, corresponding to an empirical formula of $\mathrm{C}_2\mathrm{H}_4\mathrm{O}$. The mass of this empirical unit is calculated as $(2 \times 12) + (4 \times 1) + 16 = 44$ g mol^-1. Dividing the given molar mass of 132 g mol^-1 by this empirical mass using the equation $132 / 44 = 3$ confirms that the molecule is three times the size of the empirical formula, leading to a molecular formula of $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_3$.

The core of this problem lies in identifying the limiting reagent, which restricts the extent of the reaction and determines the maximum quantity of product that can be formed. The balanced chemical equation for the reduction of the iron ore is given by:

$\mathrm{Fe}_3\mathrm{O}_4(\text{s}) + 4\mathrm{CO}(\text{g}) \rightarrow 3\mathrm{Fe}(\text{l}) + 4\mathrm{CO}_2(\text{g})$

Begin by converting the given mass of each reactant into kmol to determine their molar ratio. Dividing 2.32 x 10^3 kg of $\mathrm{Fe}_3\mathrm{O}_4$ by its molar mass of 232 kg/kmol gives 10 kmol of ore. Similarly, dividing 2.8 x 10^2 kg of $\mathrm{CO}$ by its molar mass of 28 kg/kmol results in 10 kmol of gas. According to the reaction stoichiometry, every 1 kmol of $\mathrm{Fe}_3\mathrm{O}_4$ reacts with 4 kmol of $\mathrm{CO}$. Because only 10 kmol of $\mathrm{CO}$ are available but 40 kmol would be required for the complete reaction of all the ore, $\mathrm{CO}$ is identified as the limiting reagent. This means the amount of iron formed is dictated by the 10 kmol of $\mathrm{CO}$ available. Since 4 kmol of $\mathrm{CO}$ produces 3 kmol of $\mathrm{Fe}$, the 10 kmol of $\mathrm{CO}$ will produce 7.5 kmol of $\mathrm{Fe}$. Multiplying this by the molar mass of 56 kg/kmol results in a total of 420 kg of iron produced.

In the Dumas method, we determine the nitrogen content of an organic compound by converting its nitrogen atoms into elemental nitrogen gas. Because this gas is collected over water, the total pressure measured includes both the pressure of the nitrogen gas and the vapor pressure of water. To find the true pressure of the nitrogen, we subtract the aqueous tension from the total pressure, using the relation $P_{N_2} = 715 - 15 = 700$ mm Hg.

Using the ideal gas equation $n = \frac{PV}{RT}$, we convert the given values into consistent units where the pressure is $P = \frac{700}{760}$ atm and the volume is $V = \frac{60}{1000}$ L. Substituting these values, along with the gas constant of 0.0821 L atm K^-1 mol^-1 and the temperature of 300 K, we calculate the moles of nitrogen as $n = 2.24 × 10^{-3}$ mol. Finally, the percentage of nitrogen is determined by the formula $\%N = \frac{n \times 28}{0.4} \times 100$, which yields 15.71%.

The Carius method for sulfur estimation relies on the principle of conservation of mass, where the sulfur in an organic compound is converted quantitatively into barium sulphate. Calculating the molar mass of $\text{BaSO}_4$ yields $137 + 32 + 4 \times 16 = 233 \text{ g mol}^{-1}$, and the number of millimoles produced from the precipitate is $\frac{466 \text{ mg}}{233 \text{ g mol}^{-1}} = 2 \text{ mmol}$. Since each mole of barium sulphate contains exactly one mole of sulfur, the original sample contains $2 \text{ mmol}$ of sulfur, which corresponds to a mass of $2 \text{ mmol} \times 32 \text{ mg mmol}^{-1} = 64 \text{ mg}$. Finally, the percentage of sulfur in the organic compound is calculated by dividing the mass of sulfur by the total mass of the compound and multiplying by one hundred, giving the final result of $\frac{64}{160} \times 100 = 40 \%$.

Molarity is defined by the ratio of moles of solute to the volume of solution in liters, represented by the formula $M = \frac{n}{V}$. Mixing 10 mL of 2 M NaOH with 20 mL of 1 M HCl results in complete neutralization, as both reactants provide 20 mmol, creating 20 mmol of NaCl in a total volume of 30 mL. Transferring a 10 mL aliquot of this neutralized mixture into the volumetric flask introduces a negligible amount of salt compared to the 2 moles of HCl already present. Since the volume is made up to 100 mL, or 0.1 L, the final molarity of the HCl is determined by dividing 2 moles by 0.1 L, which equals 20 M.

Determining an empirical formula relies on finding the simplest whole-number ratio of atoms by dividing the mass percentage of each element by its respective atomic mass. First, identify the percentage of oxygen by subtracting the sum of the provided components from 100%, which results in 19.24%. Dividing the percentages by their atomic weights yields relative molar amounts of 1.21 for carbon ($14.5/12$), 1.8 for hydrogen ($1.8/1$), 1.81 for chlorine ($64.46/35.5$), and 1.2 for oxygen ($19.24/16$).

Dividing these values by the smallest result, 1.2, provides stoichiometric ratios of approximately 1 for carbon, 1.5 for hydrogen, 1.5 for chlorine, and 1 for oxygen. Multiplying these by 2 eliminates the fractions to produce the empirical formula $C_2H_3Cl_3O_2$. Summing the individual atomic contributions within this formula ($2 \times 12 + 3 \times 1 + 3 \times 35.5 + 2 \times 16$) gives a total formula mass of 165.5, which is expressed as $1655 \times 10^{-1}$.

Chemical reactions are governed by the limiting reagent, the reactant that is fully consumed first and thereby determines the total yield. Calculating the available amounts gives 3.0 mol of aluminium from 81.0 g, and 4.0 mol of oxygen gas from 128.0 g, as oxygen is diatomic with a molar mass of 32.0 g/mol.

The balanced reaction
$4\mathrm{Al} + 3\mathrm{O}_2 \rightarrow 2\mathrm{Al}_2\mathrm{O}_3$
indicates that 4 moles of aluminium require 3 moles of oxygen for completion. Since 3.0 mol of aluminium only needs 2.25 mol of oxygen, aluminium is the limiting reagent and oxygen remains in excess. Because 4 moles of aluminium generate 2 moles of $\mathrm{Al}_2\mathrm{O}_3$, the 3.0 mol of aluminium will produce 1.5 mol of product. Multiplying this amount by the molar mass of $\mathrm{Al}_2\mathrm{O}_3$ at 102 g/mol yields 153 grams.

Molarity defines the concentration of a solution as the number of moles of solute contained within one liter of the total solution volume. By assuming we have exactly $100\text{ g}$ of the solution, the $70\%$ mass/mass concentration indicates that there are $70\text{ g}$ of the acid present. Dividing this mass by the given molar mass of $70\text{ g mol}^{-1}$ confirms that the solution contains exactly $1\text{ mole}$ of solute.

To determine the volume of this solution, divide the total mass of $100\text{ g}$ by the provided density of $1.25\text{ g mL}^{-1}$. This calculation yields a volume of $80\text{ mL}$, which is equivalent to $0.08\text{ L}$. Applying the molarity formula by dividing the moles of solute by the volume in liters results in the final concentration:

$M = \frac{1\text{ mol}}{0.08\text{ L}} = 12.5\text{ M}$

Expressing this value in the required format of $\times 10^{-1}$ leads to $125 \times 10^{-1}\text{ M}$.

Molality defines concentration based on the mass of the solvent, whereas molarity relies on the total volume of the solution. To determine the molality, we must calculate the mass of the solvent present in a given volume of solution, which involves subtracting the mass of the solute from the total mass of that solution.

If we consider 1 liter of the solution, the molarity of 3 M tells us there are 3 moles of NaCl present. Given a density of 1.25 g/mL, the mass of 1000 mL of the solution is 1250 g. Since the molar mass of NaCl is 58.5 g/mol, the mass of the 3 moles of solute is 175.5 g. Subtracting this mass from the total solution mass gives us 1074.5 g, or 1.0745 kg, of solvent. The molality is calculated by dividing the moles of solute by the mass of the solvent in kilograms:

$\frac{3}{1.0745} = 2.79 \text{ m}$

The mass percentage of an element in a compound is calculated by dividing the total atomic mass of the element by the molar mass of the compound and multiplying by 100.
$\text{Mass Percentage} = \frac{\text{Total mass of element}}{\text{Molar mass of compound}} \times 100$
The molecular formula of thyroxine is $C_{15}H_{11}I_4NO_4$. The total molar mass is calculated as $15 \times 12 + 11 \times 1 + 4 \times 127 + 14 + 4 \times 16 = 777$ g/mol. The total mass contributed by the four iodine atoms is $4 \times 127 = 508$ g/mol. Dividing 508 by 777 and multiplying by 100 yields approximately 65.38%, which corresponds to 65 when rounded to the nearest integer.

Mixing solutions of the same solute is governed by the conservation of moles, where the total quantity of solute remains the sum of the amounts provided by each individual portion. The final concentration is found by dividing this total amount of solute by the total final volume of the combined mixture.

The first component contributes $20 \times 2 = 40 \text{ mmol}$ of solute, and the second contributes $400 \times 0.5 = 200 \text{ mmol}$, creating an aggregate of $240 \text{ mmol}$. With the total volume reaching $420 \text{ mL}$, the resulting concentration is defined by the ratio $\frac{240}{420} \text{ M}$. Evaluating this fraction yields $0.57 \text{ M}$, which is expressed as $57 \times 10^{-2} \text{ M}$.

The initial amount of a substance is determined by the sum of the quantity that remains and the quantity that was removed. Begin by finding the moles removed by dividing the 10^21 molecules by Avogadro's constant, 6.02 × 10^23, which results in $10^{21} / 6.02 \times 10^{23} = 1.66 \times 10^{-3}$ moles. Adding this removed amount to the remaining 2.8 × 10^-3 moles gives a total initial quantity of $2.8 \times 10^{-3} + 1.66 \times 10^{-3} = 4.46 \times 10^{-3}$ moles. Multiplying this total molar quantity by the molar mass of carbon dioxide, which is 44 g/mol, provides the initial mass. Performing this final calculation $4.46 \times 10^{-3} \times 44 = 196.24 \times 10^{-3}$ g reveals that the mass taken initially was 196.24 mg.

Combustion analysis relies on the principle that during burning, all carbon in an organic sample is oxidized to carbon dioxide, and all hydrogen is oxidized to water, allowing the original elemental composition to be derived from these combustion products. To find the hydrogen content, note that two hydrogen atoms contribute to the molecular mass of water; therefore, the mass of hydrogen present in 0.127 g of water is calculated as $\frac{0.127 \times 2}{18}$, which is 0.014 g. When measured against the initial 0.210 g of the compound, this gives a hydrogen percentage of 6.72%.

Because the compound contains only carbon, hydrogen, and oxygen, determining the final oxygen percentage requires subtracting the percentages of carbon and hydrogen from the total mass. The carbon mass is derived from 0.307 g of carbon dioxide using the formula $\frac{0.307 \times 12}{44}$, which equals 0.084 g, corresponding to a carbon percentage of 39.87%. Subtracting both the hydrogen and carbon percentages from 100% identifies the oxygen content as 53.41%.

The mass of iron required for 150 kg of wheat at a concentration of 12 ppm is calculated by the expression $\frac{12}{10^6} \times 150 \times 10^3 \text{ g} = 1.8 \text{ g}$. The molar mass of $\mathrm{FeSO}_4 \cdot 7 \mathrm{H}_2\mathrm{O}$ is determined by summing the atomic masses: $56 + 32 + (4 \times 16) + 7 \times (2 \times 1 + 16) = 278 \text{ g mol}^{-1}$. Given that one mole of this compound contains one mole of iron ($56 \text{ g}$), the mass of the salt required to supply 1.8 g of iron is found using the ratio $\frac{278}{56} \times 1.8 \text{ g} \approx 8.9357 \text{ g}$. Rounding this result to the nearest integer yields 9 g.

The thermal decomposition of limestone proceeds according to the balanced chemical equation $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$, which establishes a stoichiometric ratio where one mole of calcium carbonate produces one mole of calcium oxide. First, determine the mass of pure $\text{CaCO}_3$ present in the sample by calculating $75\%$ of $150 \text{ kg}$, which equals $112.5 \text{ kg}$. The molar masses of $\text{CaCO}_3$ and $\text{CaO}$ are $100 \text{ g/mol}$ and $56 \text{ g/mol}$ respectively. Applying these values, the mass of $\text{CaO}$ produced is calculated as $\frac{56}{100} \times 112.5 \text{ kg}$, resulting in exactly $63 \text{ kg}$.

The reaction between sodium iodide and silver nitrate proceeds according to the stoichiometric equation $NaI + AgNO_3 \rightarrow AgI + NaNO_3$, indicating that one mole of silver iodide is produced for every mole of sodium iodide consumed. The molar mass of silver iodide is calculated by summing the atomic weights of silver and iodine, yielding $108 + 127 = 235$ g/mol. By dividing the 4.74 g of precipitate by this molar mass, we find that the reaction produced $\frac{4.74}{235} = 0.02$ moles of silver iodide. Since the stoichiometry is 1:1, there must have been 0.02 moles of sodium iodide in the original solution. Finally, convert the volume of the solution from 20 mL to 0.02 liters and divide the moles of sodium iodide by this volume, resulting in a molarity of $\frac{0.02}{0.02} = 1$ M.

In chemical reactions involving multiple reactants, the quantity of the final product is dictated by the limiting reagent, which is the substance completely consumed first, thereby restricting the total possible yield. To identify this reagent, we first determine the amount of hydrochloric acid by multiplying the volume by the concentration, yielding $0.250 \times 0.76 = 0.19$ moles of $\mathrm{HCl}$. Comparing this to the stoichiometry of the reaction, which requires 2 moles of $\mathrm{HCl}$ for every mole of $\mathrm{CaCO}_3$, it is clear that the 1000 g of $\mathrm{CaCO}_3$ is present in excess, leaving $\mathrm{HCl}$ as the limiting reagent.

Once the limiting reagent is established, the stoichiometry dictates that 2 moles of $\mathrm{HCl}$ produce 1 mole of $\mathrm{CaCl}_2$. Consequently, $0.19$ moles of $\mathrm{HCl}$ will yield $0.095$ moles of $\mathrm{CaCl}_2$. To find the mass, we multiply this molar amount by the molar mass of $\mathrm{CaCl}_2$, which is calculated as $40 + 2 \times 35.5 = 111\,\mathrm{g}\,\mathrm{mol}^{-1}$. Therefore, the total mass of the product formed is $0.095 \times 111 = 10.545\,\mathrm{g}$.

The mass of carbon in an organic compound can be determined from the combustion product $\mathrm{CO}_2$ by using the ratio of the molar mass of carbon to that of $\mathrm{CO}_2$. Since the molar mass of $\mathrm{CO}_2$ is 44 g/mol and contains 12 g of carbon, the mass of carbon in 220 mg of the product is:

$\frac{12}{44} \times 220 = 60$

This represents 60 mg of carbon. The percentage composition in the 500 mg compound is calculated as:

$\frac{60}{500} \times 100 = 12$

The percentage of carbon in the compound is 12 %.

Since the nitration of nitrobenzene proceeds with a 1:1 molar ratio, the number of moles of the starting reactant must equal the moles of the product obtained. First, determine the molar mass for nitrobenzene ($C_6H_5NO_2$) as 123 g/mol and for m-dinitrobenzene ($C_6H_4(NO_2)_2$) as 168 g/mol.

Using the equivalence of moles, set up the proportion $\frac{X}{123} = \frac{4.2}{168}$ to solve for the unknown mass $X$. This calculation yields $X = 3.075$ g, which rounds to 3 g when expressed as the nearest integer.

The balanced chemical equation
$\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$
establishes that one mole of magnesium is required to produce one mole of hydrogen gas. Because one mole of any gas at standard temperature and pressure occupies 22400 mL, the number of moles of hydrogen produced from 220 mL is determined by the fraction $\frac{220}{22400}$. Multiplying this molar value by the molar mass of magnesium, 24 g/mol, gives a mass of 0.236 g, which is equivalent to 236 mg.