Numericals on Mole Concept Class 11 with Answers

Numericals on mole concept look long, but they use the same few steps again and again, so numericals on mole concept become easy with practice. This page groups 40 solved numericals on mole concept by type, so that you can practise one type at a time. If you need the theory first, read eSaral’s Mole Concept notes for Class 11, JEE and NEET. For the NCERT questions on SI units, significant figures and the laws of chemical combination, use eSaral’s NCERT Solutions for Class 11 Chemistry Chapter 1.

Formulas Used in Numericals on Mole Concept

How to Solve Numericals on Mole Concept: 4 Steps

Use this method for all numericals on mole concept.

1. Write the data with units, and convert the units first (mL to L, mg to g).
2. Convert the given quantity to moles using the mass, the number of particles, the gas volume or the molarity.
3. Use the balanced equation or the formula to find the moles of the quantity that you need.
4. Convert the moles to the answer: mass, volume or number of particles, with the right unit and significant figures.

Numericals on Mole Concept: Moles, Mass and Number of Particles

These numericals on mole concept practise the three basic conversions: mass, number of particles and gas volume to moles.

Q1. Calculate the number of moles in (i) 11 g of $CO_2$, (ii) $3.01\times 10^{22}$ molecules of $CO_2$, (iii) 1.12 L of $CO_2$ at STP.

(i) $M = 44\ \text{g/mol}$, so $n = \dfrac{11}{44} = 0.25$ mol.

(ii) $n = \dfrac{3.01\times 10^{22}}{6.02\times 10^{23}} = 0.05$ mol.

(iii) $n = \dfrac{1.12}{22.4} = 0.05$ mol.

Q2. Calculate the number of atoms in (i) 52 mol of Ar, (ii) 52 u of He, (iii) 52 g of He.

(i) $52 \times 6.022\times 10^{23} = 3.13\times 10^{25}$ atoms.

(ii) One He atom has a mass of 4 u, so the number of atoms is $\dfrac{52}{4} = 13$.

(iii) $n = \dfrac{52}{4} = 13$ mol, so the number of atoms is $13 \times 6.022\times 10^{23} = 7.83\times 10^{24}$.

Q3. In three moles of ethane ($C_2H_6$), calculate (i) the moles of carbon atoms, (ii) the moles of hydrogen atoms, (iii) the number of molecules of ethane.

One mole of $C_2H_6$ has 2 mol of C atoms and 6 mol of H atoms.

(i) $3 \times 2 = 6$ mol of C. (ii) $3 \times 6 = 18$ mol of H. (iii) $3 \times 6.022\times 10^{23} = 1.81\times 10^{24}$ molecules.

Q4. Which one has the largest number of atoms: 1 g Au, 1 g Na, 1 g Li or 1 g $Cl_2$? (Au = 197, Na = 23, Li = 7, Cl = 35.5)

Atoms per gram are proportional to $\dfrac{1}{\text{atomic mass}}$, and for $Cl_2$ to $\dfrac{2}{71}$:

$$\text{Au: } 0.0051, \quad \text{Na: } 0.0435, \quad \text{Li: } 0.1429, \quad Cl_2\text{: } 0.0282 \quad (\times\, 6.022\times 10^{23})$$

Answer: 1 g of Li has the largest number of atoms.

Q5. Chlorophyll contains 2.68% of magnesium by mass. Calculate the number of Mg atoms in 5.00 g of chlorophyll. (Mg = 24)

Mass of Mg $= \dfrac{2.68}{100}\times 5.00 = 0.134$ g, so $n = \dfrac{0.134}{24}$ mol.

$$N = \frac{0.134}{24}\times 6.022\times 10^{23} = 3.36\times 10^{21}\ \text{atoms}$$

Q6. How many oxygen atoms are present in 0.5 mol of $Al_2(SO_4)_3$?

One formula unit has $3 \times 4 = 12$ oxygen atoms. So 0.5 mol has $6$ mol of O atoms.

$$N = 6 \times 6.022\times 10^{23} = 3.61\times 10^{24}\ \text{atoms}$$

Q7. Find the volume at STP of (i) 14 g of $N_2$, (ii) $6.023\times 10^{22}$ molecules of $NH_3$, (iii) 0.1 mol of $SO_2$.

(i) $n = \dfrac{14}{28} = 0.5$ mol, so $V = 0.5 \times 22.4 = 11.2$ L.

(ii) $n = \dfrac{6.023\times 10^{22}}{6.023\times 10^{23}} = 0.1$ mol, so $V = 2.24$ L.

(iii) $V = 0.1 \times 22.4 = 2.24$ L.

Numericals on Mole Concept: Average Atomic Mass

Numericals on mole concept with isotopes use the weighted average of the isotopic masses.

Q8. Calculate the average atomic mass of chlorine from the data: $^{35}Cl$ (34.9689 u, 75.77%) and $^{37}Cl$ (36.9659 u, 24.23%).

$$M_{avg} = \frac{34.9689\times 75.77 + 36.9659\times 24.23}{100} = 35.45\ \text{u}$$

Q9. Find the molar mass of naturally occurring argon from the data: $^{36}Ar$ (35.96755 u, 0.337%), $^{38}Ar$ (37.96272 u, 0.063%) and $^{40}Ar$ (39.9624 u, 99.600%).

$$M_{avg} = \frac{35.96755\times 0.337 + 37.96272\times 0.063 + 39.9624\times 99.600}{100} \approx 39.95\ \text{g/mol}$$

Numericals on Mole Concept: Percentage Composition

In these numericals on mole concept, the mass of the element in one mole is divided by the molar mass.

Q10. Calculate the mass percentage of iron, sulphur and oxygen in ferric sulphate, $Fe_2(SO_4)_3$. (Fe = 56, S = 32, O = 16)

Formula mass $= 2\times 56 + 3(32 + 64) = 400$ u.

$$\%\,Fe = \frac{112}{400}\times 100 = 28\%, \qquad \%\,S = \frac{96}{400}\times 100 = 24\%, \qquad \%\,O = \frac{192}{400}\times 100 = 48\%$$

Q11. Calculate the mass percent of each element in sodium sulphate, $Na_2SO_4$.

Molar mass $= 2\times 23 + 32 + 4\times 16 = 142$ g/mol.

$$\%\,Na = \frac{46}{142}\times 100 = 32.39\%, \quad \%\,S = \frac{32}{142}\times 100 = 22.54\%, \quad \%\,O = \frac{64}{142}\times 100 = 45.07\%$$

Q12. How much copper can be obtained from 100 g of copper sulphate, $CuSO_4$? (Cu = 63.5)

Molar mass of $CuSO_4 = 63.5 + 32 + 64 = 159.5$ g/mol, and it has 63.5 g of Cu.

$$\text{Cu} = \frac{63.5}{159.5}\times 100 = 39.81\ \text{g}$$

Numericals on Mole Concept: Empirical and Molecular Formula

The numericals on mole concept below follow one method: percentages, moles, the simplest ratio, and then the molecular formula.

Q13. An oxide of iron has 69.9% iron and 30.1% oxygen by mass, and its molar mass is 159.8 g/mol. Find its molecular formula. (Fe = 55.85, O = 16)

The ratio Fe : O $= 1 : 1.5 = 2 : 3$, so the empirical formula is $Fe_2O_3$ (mass 159.7). Since $n = \dfrac{159.8}{159.7} \approx 1$, the molecular formula is also $Fe_2O_3$.

Q14. A 4.24 mg sample of butyric acid gives 8.45 mg of $CO_2$ and 3.46 mg of $H_2O$ on complete burning. Find the mass percent of each element, and the empirical and molecular formula if the molecular mass is 88 u.

$$\text{Mass of C} = \frac{12}{44}\times 8.45 = 2.30\ \text{mg}, \qquad \%\,C = \frac{2.30}{4.24}\times 100 = 54.35\%$$ $$\text{Mass of H} = \frac{2}{18}\times 3.46 = 0.384\ \text{mg}, \qquad \%\,H = \frac{0.384}{4.24}\times 100 = 9.07\%$$ $$\%\,O = 100 - 54.35 - 9.07 = 36.58\%$$

Moles (per 100 g): C $= \dfrac{54.35}{12} = 4.53$, H $= 9.07$, O $= \dfrac{36.58}{16} = 2.29$. Dividing by 2.29, the ratio is C : H : O $= 2 : 4 : 1$, so the empirical formula is $C_2H_4O$ (mass 44). Then $n = \dfrac{88}{44} = 2$, and the molecular formula is $C_4H_8O_2$.

Q15. A welding fuel has only carbon and hydrogen. Burning a sample gives 3.38 g of $CO_2$ and 0.690 g of $H_2O$. 10.0 L of the gas at STP weighs 11.6 g. Find the empirical formula, the molar mass and the molecular formula.

$$\text{Mass of C} = \frac{12\times 3.38}{44} = 0.922\ \text{g}, \qquad \text{Mass of H} = \frac{2\times 0.690}{18} = 0.077\ \text{g}$$

Moles: C $= \dfrac{0.922}{12} = 0.077$, H $= \dfrac{0.077}{1} = 0.077$, so the empirical formula is CH (mass 13).

Molar mass $= \dfrac{11.6 \times 22.4}{10.0} = 25.98$ g/mol, and $n = \dfrac{25.98}{13} \approx 2$, so the molecular formula is $C_2H_2$.

Q16. Naphthalene has 93.71% carbon and 6.29% hydrogen, and its molar mass is 128 g/mol. Find its molecular formula.

In 100 g: C $= \dfrac{93.71}{12.0} = 7.81$ mol, H $= \dfrac{6.29}{1.00} = 6.29$ mol. The ratio C : H $= 1.24 \approx \dfrac{5}{4}$, so the empirical formula is $C_5H_4$ (mass 64). Since $n = \dfrac{128}{64} = 2$, the molecular formula is $C_{10}H_8$.

Numericals on Mole Concept: Stoichiometry and Limiting Reagent

Stoichiometry numericals on mole concept always start from the balanced equation, and from the limiting reagent when both amounts are given.

Q17. How many grams of HCl react with 5.0 g of $MnO_2$? $4HCl + MnO_2 \rightarrow 2H_2O + MnCl_2 + Cl_2$ (Mn = 55)

$n(MnO_2) = \dfrac{5.0}{87} = 0.0575$ mol. The mole ratio HCl : $MnO_2$ is $4 : 1$, so $n(HCl) = 4 \times 0.0575 = 0.230$ mol.

$$\text{Mass of HCl} = 0.230 \times 36.5 = 8.39\ \text{g}$$

Q18. In the manufacture of $HNO_3$, how many moles of $NO_2$ produce 7.33 mol of $HNO_3$? $3NO_2 + H_2O \rightarrow 2HNO_3 + NO$

2 mol of $HNO_3$ need 3 mol of $NO_2$, so $n(NO_2) = \dfrac{3}{2}\times 7.33 = 10.995 \approx 11.0$ mol.

Q19. $6.023\times 10^{23}$ molecules of $N_2$ react completely with $H_2$: $N_2 + 3H_2 \rightarrow 2NH_3$. Find the number of $NH_3$ molecules formed.

1 molecule of $N_2$ gives 2 molecules of $NH_3$, so the number of molecules is $2 \times 6.023\times 10^{23} = 1.204\times 10^{24}$.

Q20. If 0.30 mol of Zn is added to hydrochloric acid with 0.52 mol of HCl, how many moles of $H_2$ form, and which is the limiting reactant? $Zn + 2HCl \rightarrow ZnCl_2 + H_2$

0.30 mol of Zn needs $2 \times 0.30 = 0.60$ mol of HCl, but only 0.52 mol is present, so HCl is the limiting reactant.

$$n(H_2) = \frac{0.52}{2} = 0.26\ \text{mol}$$

Q21. For $A + B_2 \rightarrow AB_2$, find the limiting reagent in each mixture: (i) 300 atoms of A + 200 molecules of $B_2$, (ii) 2 mol A + 3 mol $B_2$, (iii) 100 atoms of A + 100 molecules of $B_2$, (iv) 5 mol A + 2.5 mol $B_2$, (v) 2.5 mol A + 5 mol $B_2$.

The ratio is $1 : 1$.

Q22. Potassium bromide contains 32.9% potassium by mass. If 6.4 g of bromine reacts with 3.60 g of K, find the number of moles of K that combine with bromine to form KBr.

32.9 g of K combine with 67.1 g of Br, so 3.60 g of K need $\dfrac{67.1}{32.9}\times 3.60 = 7.34$ g of Br. Only 6.4 g of bromine is present, so bromine is the limiting reactant.

$$n(Br_2) = \frac{6.4}{160} = 0.04\ \text{mol}, \qquad 2K + Br_2 \rightarrow 2KBr \;\Rightarrow\; n(K) = 2\times 0.04 = 0.08\ \text{mol}$$

Q23. How many grams of $N_2$ react completely with 0.3 g of $H_2$ to form ammonia, and how much ammonia forms? $N_2 + 3H_2 \rightarrow 2NH_3$

$n(H_2) = \dfrac{0.3}{2} = 0.15$ mol. From the ratio, $n(N_2) = \dfrac{0.15}{3} = 0.05$ mol and $n(NH_3) = \dfrac{2}{3}\times 0.15 = 0.10$ mol.

$$\text{Mass of } N_2 = 0.05\times 28 = 1.4\ \text{g}, \qquad \text{Mass of } NH_3 = 0.10\times 17 = 1.7\ \text{g}$$

Q24. In $2Br^- + Cl_2 \rightarrow 2Cl^- + Br_2$, what volume of 0.050 M $Cl_2$ solution reacts completely with 50.0 mL of 0.060 M NaBr?

$n(Br^-) = 0.0500 \times 0.060 = 0.0030$ mol. The ratio $Br^- : Cl_2$ is $2 : 1$, so $n(Cl_2) = 0.0015$ mol.

$$V = \frac{0.0015}{0.050} = 0.030\ \text{L} = 30\ \text{mL}$$

Q25. 500 mL of 0.25 M $Na_2SO_4$ is added to 15.00 g of $BaCl_2$. Find the moles and the mass of $BaSO_4$ formed. (Ba = 137, S = 32, O = 16)

$n(Na_2SO_4) = 0.500\times 0.25 = 0.125$ mol, and $n(BaCl_2) = \dfrac{15.00}{208.2} = 0.072$ mol. They react $1 : 1$, so $BaCl_2$ is the limiting reagent.

$$n(BaSO_4) = 0.072\ \text{mol}, \qquad \text{mass} = 0.072\times 233.4 = 16.8\ \text{g}$$

Q26. Calculate the lime, $Ca(OH)_2$, needed to remove the hardness of 60,000 L of water that contains 16.2 g of calcium bicarbonate per 100 L. (Ca = 40, C = 12, O = 16, H = 1)

Mass of $Ca(HCO_3)_2 = \dfrac{16.2}{100}\times 60000 = 9720$ g. The molar mass is 162 g/mol, so $n = 60$ mol.

$$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 + 2H_2O$$

The ratio is $1 : 1$, so 60 mol of $Ca(OH)_2$ is needed: $60 \times 74 = 4440$ g, that is, 4.44 kg.

Q27. 20 g of $CH_4$ burns in 64 g of $O_2$: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$. Find the limiting reagent, the mass of $CO_2$ formed and the mass of the reactant left.

$n(CH_4) = \dfrac{20}{16} = 1.25$ mol and $n(O_2) = \dfrac{64}{32} = 2.0$ mol. Moles divided by the coefficient: $CH_4$ gives 1.25 and $O_2$ gives $\dfrac{2.0}{2} = 1.0$, so $O_2$ is the limiting reagent.

$$n(CO_2) = 1.0\ \text{mol} = 44\ \text{g}, \qquad CH_4\ \text{left} = (1.25 - 1.0)\times 16 = 4\ \text{g}$$

Q28. 100 mL of 0.2 M $Al_2(SO_4)_3$ is mixed with 200 mL of 0.05 M $NH_4OH$. Find the mass of $Al(OH)_3$ precipitated and the molarity of each ion in the resulting solution. (Treat the reaction as complete.)

$$Al_2(SO_4)_3 + 6NH_4OH \rightarrow 2Al(OH)_3 + 3(NH_4)_2SO_4$$

$n(Al_2(SO_4)_3) = 0.100\times 0.2 = 0.020$ mol and $n(NH_4OH) = 0.200\times 0.05 = 0.010$ mol. $NH_4OH$ is the limiting reagent, so

$$n(Al(OH)_3) = \frac{2}{6}\times 0.010 = 0.00333\ \text{mol}, \qquad \text{mass} = 0.00333\times 78 = 0.26\ \text{g}$$

The total volume is 300 mL. $Al^{3+}$ left $= 0.040 - 0.00333 = 0.0367$ mol, so $[Al^{3+}] = 0.122$ M. $SO_4^{2-} = 0.060$ mol, so $[SO_4^{2-}] = 0.20$ M. $NH_4^+ = 0.010$ mol, so $[NH_4^+] = 0.033$ M. The normality is molarity $\times$ charge: $Al^{3+}$ 0.367 N, $SO_4^{2-}$ 0.40 N and $NH_4^+$ 0.033 N.

Numericals on Mole Concept: Molarity, Molality and Mole Fraction

These numericals on mole concept need careful units: litres of solution for molarity, and kilograms of solvent for molality.

Q29. What is the concentration of sugar ($C_{12}H_{22}O_{11}$) in mol/L if 20 g is dissolved in enough water to make 2 L of solution?

$M_B = 342$ g/mol, so $n = \dfrac{20}{342} = 0.0585$ mol.

$$M = \frac{0.0585}{2} = 0.0292\ \text{mol/L}$$

Q30. How many moles and how many grams of NaCl are present in 250 cm³ of 0.50 M NaCl? (Na = 23, Cl = 35.5)

$$n = M\times V = 0.50\times 0.250 = 0.125\ \text{mol}, \qquad \text{mass} = 0.125\times 58.5 = 7.31\ \text{g}$$

Q31. 18.25 g of NaOH is dissolved in water to make 200 cm³ of solution. Find the molarity.

$$M = \frac{18.25}{40}\times\frac{1000}{200} = 2.28\ \text{mol/L}$$

Q32. A sample of $NaNO_3$ weighing 0.83 g is placed in a 50 mL flask and made up to the mark. Find the molarity. (Na = 23, N = 14, O = 16)

$M_B = 85$ g/mol.

$$M = \frac{0.83}{85}\times\frac{1000}{50} = 0.195\ \text{M}$$

Q33. (i) 38 g of NaOH is dissolved to make 50.0 mL of solution. Find the molarity. (ii) How many moles of NaOH are in 27 mL of 0.15 M NaOH?

(i) $M = \dfrac{38}{40}\times\dfrac{1000}{50} = 19$ M.

(ii) $n = 0.15\times 0.027 = 4.05\times 10^{-3}$ mol.

Q34. 0.38 g of NaOH is dissolved to make 50.0 cm³ of solution. Find the molarity.

$$M = \frac{0.38}{40}\times\frac{1000}{50} = 0.19\ \text{mol/L}$$

Q35. How many mL of 0.150 M NaOH contain 0.184 g of NaOH?

$n = \dfrac{0.184}{40} = 0.0046$ mol, so $V = \dfrac{0.0046}{0.150} = 0.0307$ L $= 30.7$ mL.

Q36. Calculate the mass of sodium acetate ($CH_3COONa$, molar mass 82.0245 g/mol) needed to make 500 mL of 0.375 M solution.

$$n = 0.375\times 0.500 = 0.1875\ \text{mol}, \qquad \text{mass} = 0.1875\times 82.0245 = 15.38\ \text{g}$$

Q37. If the density of methanol is 0.793 kg/L, what volume is needed to make 2.5 L of 0.25 M solution?

$n = 0.25\times 2.5 = 0.625$ mol, so the mass is $0.625\times 32 = 20$ g. The density is 0.793 g/mL.

$$V = \frac{20}{0.793} = 25.2\ \text{mL}$$

Q38. Calculate the molality of ethanol in water when the mole fraction of ethanol is 0.040.

For 1 mol of solution: 0.040 mol of ethanol and 0.960 mol of water, which is $0.960\times 18 = 17.28$ g $= 0.01728$ kg.

$$m = \frac{0.040}{0.01728} = 2.31\ \text{mol/kg}$$

Q39. Drinking water has 15 ppm (by mass) of chloroform, $CHCl_3$. Express this as (i) percent by mass and (ii) molality.

(i) 15 ppm means 15 g in $10^6$ g, so the percent by mass is $\dfrac{15}{10^6}\times 100 = 1.5\times 10^{-3}\%$.

(ii) $n = \dfrac{15}{119.5} = 0.1255$ mol in about $10^6$ g $= 10^3$ kg of water.

$$m = \frac{0.1255}{1000} = 1.25\times 10^{-4}\ \text{mol/kg}$$

Q40. A 10% (w/w) glucose solution has a density of 1.20 g/mL. Find (i) the molality, (ii) the molarity and (iii) the mole fraction of each component.

Take 100 g of solution: 10 g of glucose ($n = \dfrac{10}{180} = 0.0556$ mol) and 90 g of water ($n = \dfrac{90}{18} = 5.0$ mol).

(i) $m = \dfrac{0.0556}{0.090} = 0.618$ mol/kg.

(ii) $V = \dfrac{100}{1.20} = 83.3$ mL, so $M = \dfrac{0.0556}{0.0833} = 0.667$ M.

(iii) $x_{glucose} = \dfrac{0.0556}{5.0556} = 0.011$ and $x_{water} = 0.989$.

Bonus: Molarity of concentrated nitric acid

Find the molarity of nitric acid that has a density of 1.41 g/mL and 69% $HNO_3$ by mass. ($M_B = 63$)

$$M = \frac{10\times 1.41\times 69}{63} \approx 15.44\ \text{M}$$

Practice Numericals on Mole Concept

Try these numericals on mole concept without looking at the answers.

1. How many molecules are present in 5.5 g of $CO_2$? (C = 12, O = 16)
2. Find the mass of $NH_3$ formed from 14 g of $N_2$ and excess $H_2$. $N_2 + 3H_2 \rightarrow 2NH_3$
3. A compound has 40% C, 6.7% H and 53.3% O, and its molar mass is 180 g/mol. Find its molecular formula.
4. Find the molarity of a solution made by dissolving 5.85 g of NaCl in 500 mL of solution. (Na = 23, Cl = 35.5)

Answers: 1. $n = \dfrac{5.5}{44} = 0.125$ mol, so $N = 0.125\times 6.022\times 10^{23} = 7.53\times 10^{22}$ molecules. 2. $n(N_2) = 0.5$ mol, so $n(NH_3) = 1.0$ mol and the mass is 17 g. 3. The ratio C : H : O $= 3.33 : 6.7 : 3.33 = 1 : 2 : 1$, so the empirical formula is $CH_2O$ (mass 30), $n = 6$, and the molecular formula is $C_6H_{12}O_6$. 4. $n = \dfrac{5.85}{58.5} = 0.1$ mol, so $M = \dfrac{0.1}{0.5} = 0.2$ M.

Common Mistakes in Numericals on Mole Concept

These mistakes cost marks in numericals on mole concept.

Related Topics

For more help with numericals on mole concept, use these eSaral pages.

• Mole Concept notes for Class 11, JEE and NEET
• NCERT Solutions for Class 11 Chemistry Chapter 1
• Molarity formula
• Molality formula