3Chemical calculation

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STOICHIOMETRY
▢ Definition: Calculations based on chemical equation → stoichiometry
▢ Chemical Equation: Symbolic representation of chemical change.
▢ Problem Types:

Table 1: Chemical Equation Based Problems

Type
Relationship
Core Idea
A
Mass–mass
Mass of reactant/product ↔ mass of reactant/product
B
Mass–volume
Mass of solid/liquid ↔ volume of gas
C
Volume–volume
Gas volume ↔ gas volume
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MASS–MASS RELATIONSHIP
▢ Steps:
  1. Write balanced chemical equation.
  2. Write moles below formulas of reactants/products.
  3. Write relative weights / molecular weights below formulas.
  4. Use unitary method → unknown factor.
▢ Percentage Composition:
❖ Formula: \(\%\ element=\frac{Mass\ of\ element\ in\ 1\ mole}{Molar\ mass\ of\ compound}\times100\)
❖ Example: \(CaCO_3\): Molar mass \(=40+12+48=100\); \(Ca=40\%\), \(C=12\%\), \(O=48\%\)
▢ Purity Calculation:
❖ Example: 90% \(H_2SO_4\) needed to neutralize \(60g\ NaOH\)
❖ Reaction: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
❖ Relation: \(80g\ NaOH\rightarrow98g\ H_2SO_4\)
❖ Calculation: \(60g\ NaOH\rightarrow\frac{98}{80}\times60=73.5g\) pure \(H_2SO_4\); 90% acid required \(=73.5\times\frac{100}{90}=81.67g\)
▢ Impure Reactant:
❖ Example: 75% pure \(KClO_3\) needed for \(24g\ O_2\)
❖ Reaction: \(2KClO_3\xrightarrow{heat}2KCl+3O_2\)
❖ Relation: \(2\ mol\ KClO_3\rightarrow96g\ O_2\)
❖ Result: \(24g\ O_2\rightarrow0.5\ mol\ pure\ KClO_3\); impure \(KClO_3=\frac{100}{75}\times0.5=0.667\ mol\)
▢ Residual Mixture / Limiting Reagent:
❖ Example: \(30g\ Mg+30g\ O_2\)
❖ Reaction: \(2Mg+O_2\rightarrow2MgO\)
❖ Stoichiometry: \(48g\ Mg\rightarrow32g\ O_2\rightarrow80g\ MgO\)
❖ Calculation:
  • •
    \(30g\ Mg\) needs \(\frac{32}{48}\times30=20g\ O_2\)
  • •
    \(O_2\) left \(=30-20=10g\)
  • •
    \(MgO\) formed \(=\frac{80}{48}\times30=50g\)
❖ Residual Mixture: \(50g\ MgO+10g\ O_2\)
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MASS–VOLUME RELATIONSHIP
▢ Basic Principle: At NTP/STP, \(1\ mol\ gas=1\ gram\ molecule=22.4L=22400cc\).
▢ NTP: \(0^\circ C\), \(760\ mm\ Hg\), \(1\ atm\).
▢ Example Table:

Table 1: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

Basis
\(Mg\)
\(2HCl\)
\(MgCl_2\)
\(H_2\)
Mole
1
2
1
1
a.m.u.
24
73
95
2
Gram weight
24g
73g
95g
2g
Weight / Volume
24g
73g
95g
22.4L at NTP
▢ Steps:
  1. Write relevant balanced chemical equation.
  2. Write weights of solid reactants/products.
  3. Express gas in volume.
  4. Convert non-NTP volume to NTP by gas laws.
  5. Use ideal gas equation when needed: \(PV=\frac{m}{M}RT\).
  6. Use unitary method.
▢ Numericals:
  • •
    \(S+O_2\rightarrow SO_2\); \(32g\ S\rightarrow22.4L\ O_2\); \(2g\ S\rightarrow1.4L\ O_2\).
  • •
    \(2BCl_3+3H_2\rightarrow2B+6HCl\); \(21.6g\ B\) needs \(67.2L\ H_2\) at NTP.
  • •
    Gas with \(VD=11.2\): \(Mol.wt.=2VD=22.4\); \(22.4g\rightarrow22.4L\); \(11.2g\rightarrow11.2L\).
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VOLUME–VOLUME RELATIONSHIP
▢ Core Principle: Under similar \(T,P\), gas volume ∝ moles; Avogadro hypothesis.
▢ Steps:
  1. Write balanced chemical equation.
  2. Write gas volumes below formulas using \(1\ mol\ gas=22.4L\).
  3. Convert special condition volume to NTP by ideal gas equation.
  4. Use volume ratio directly under same \(T,P\).
▢ Avogadro Example:
❖ Reaction: \(N_2(g)+3H_2(g)\rightarrow2NH_3(g)\)

Table 1: Volume Ratio

Basis
\(N_2\)
\(H_2\)
\(NH_3\)
Mole
1
3
2
Volume at NTP
22.4L
\(3\times22.4L\)
\(2\times22.4L\)
Volume ratio
1 volume
3 volumes
2 volumes
▢ Numericals:
  • •
    Propane combustion: \(C_3H_8+5O_2\rightarrow3CO_2+4H_2O\); \(1L\ C_3H_8\rightarrow5L\ O_2\); \(20L\ C_3H_8\rightarrow100L\ O_2\).
  • •
    Haber process: \(N_2+3H_2\rightarrow2NH_3\); as per given solution, \(10L\ N_2+30L\ H_2\rightarrow20L\ NH_3\); 50% yield → \(10L\ NH_3\); used \(N_2=5L\), \(H_2=15L\); left \(N_2=25L\), \(H_2=15L\).
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EMPIRICAL FORMULA FROM COMPOSITION
▢ Definition: Simplest whole-number atomic ratio in compound.
▢ Formula: \(No.\ of\ moles=\frac{Wt.\ of\ element}{At.wt.\ of\ element}\)
▢ Steps:
  1. Convert mass/% of each element into moles.
  2. Divide all moles by smallest mole value.
  3. Multiply by suitable integer if ratio is fractional.
  4. Write simplest formula.
▢ Examples:

Table 1: Empirical Formula Examples

Compound Data
Mole Ratio
Empirical Formula
Phosphorus oxide: \(P=43.6\%, O=56.4\%\)
\(P:O=\frac{43.6}{31}:\frac{56.4}{16}=1.41:3.53=1:2.5=2:5\)
\(P_2O_5\)
Hydrocarbon: \(10.5g\ C\) per \(1g\ H\)
\(C:H=\frac{10.5}{12}:\frac{1}{1}=0.875:1=7:8\)
\(C_7H_8\)
64g compound: \(C=24g,H=8g,O=32g\)
\(C:H:O=\frac{24}{12}:\frac{8}{1}:\frac{32}{16}=2:8:2=1:4:1\)
\(CH_4O\)
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Molecular formula from empirical formula
▢ Definition: Actual number of atoms of each element in molecule.
▢ Relation: \(Molecular\ formula=(Empirical\ formula)_n\)
▢ Multiplier: \(n=\frac{Molecular\ weight}{Empirical\ formula\ weight}\)
▢ Example: Carbon–Nitrogen Compound:
❖ Data: \(N=53.8\%, C=46.2\%, VD=25.8\)
❖ Empirical Ratio: \(N:C=\frac{53.8}{14}:\frac{46.2}{12}=3.85:3.85=1:1\) → empirical formula \(CN\).
❖ Molecular Weight: \(Mol.wt.=2VD=2\times25.8=51.6\)
❖ Multiplier: \(n=\frac{51.6}{12+14}=\frac{51.6}{26}=2\)
❖ Molecular Formula: \((CN)_2=C_2N_2\)
▢ Vapour Density Comparison:
❖ Given: Vapour density of volatile substance = 4 times methane \((CH_4=1)\).
❖ Methane: \(Mol.wt.=16\)
❖ Substance: \(Mol.wt.=4\times16=64\)
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Quick numerical facts
▢ Mole / Volume / Atom Relations:
  • •
    \(n(N_2)=\frac{7}{28}=0.25\)
  • •
    \(n(O_2)=\frac{16}{32}=0.5\)
  • •
    \(n(H_2)=\frac{2}{2}=1\)
  • •
    \(n(NO_2)=\frac{16}{46}=0.35\)
  • •
    \(4.4g\ CO_2\) at STP → \(\frac{4.4\times22.4}{44}=2.24L\)
  • •
    \(1\ gram\ atom\ N=1\ mol\ N=\frac{1}{2}\ mol\ N_2=11.2L\) at STP
  • •
    \(16g\ O_2=0.5\ mol\)
  • •
    \(16g\ SO_2=0.25\ mol\)
  • •
    \(32g\ SO_2=0.5\ mol\)
▢ Urea Nitrogen: \(60g\ H_2NCONH_2\) has \(28g\ N\); \(100g\) urea has \(\frac{28}{60}\times100=46\%\ N\).
▢ Carbon-14: In \(12g\) carbon, \(C^{14}=\frac{2}{100}\times12=0.24g\); atoms \(=\frac{0.24\times6.02\times10^{23}}{14}=1.032\times10^{22}\).
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Frequently used reactions

Table 1: Stoichiometry Reactions

Reaction
Use
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Acid-base neutralisation
\(2KClO_3\xrightarrow{heat}2KCl+3O_2\)
Oxygen preparation / purity
\(2Mg+O_2\rightarrow2MgO\)
Residual mixture / limiting reagent
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
Mass–volume relation
\(S+O_2\rightarrow SO_2\)
Gas volume from mass
\(2BCl_3+3H_2\rightarrow2B+6HCl\)
Hydrogen volume calculation
\(N_2+3H_2\rightarrow2NH_3\)
Volume–volume / Haber process
\(C_3H_8+5O_2\rightarrow3CO_2+4H_2O\)
Combustion volume
\(CaCO_3\rightarrow CaO+CO_2\)
Quicklime / \(CO_2\) calculation
\(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)
Precipitation calculation
\(CO_2+C\rightarrow2CO\)
Gas volume doubling
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Hydrogen displacement
\(Zn+2NaOH\rightarrow Na_2ZnO_2+H_2\)
Hydrogen with alkali
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MULTIPLE CHOICE QUESTIONS
Q1.
The formula of metallic hydroxide with equivalent weight 150 is \(M(OH)_2\cdot xH_2O\). If atomic weight of metal is 176, value of \(x\) will be
Q2.
Equivalent weight of calcium phosphate \(Ca_3(PO_4)_2\) is \((Mol.wt.=310)\)
Q3.
Percentage of \(P_2O_5\) in diammonium phosphate \((NH_4)_2HPO_4\) is
Q4.
Amount of zinc \((At.wt.=65)\) required to produce 224 ml \(H_2\) at STP on treatment with dilute \(H_2SO_4\) will be
Q5.
Two elements, X \((At.wt.=75)\) and Y \((At.wt.=16)\), combine to give a compound having 75.8% of X. Formula of compound is
Q6.
Weight of quicklime obtained by strongly heating 25 gm marble is
Q7.
How much quicklime can be obtained from 25 gm of \(CaCO_3\)?
Q8.
4.4 g of an unknown gas occupies 2.24 litres of volume at standard temperature and pressure. The gas may be
Q9.
Amount of barium sulphate precipitated from 200 ml of \(\frac{N}{10}\) sulphuric acid is
Q10.
When same amount of zinc is treated separately with excess \(H_2SO_4\) and excess \(NaOH\), ratio of volumes of \(H_2\) evolved is
Q11.
If 30 g Mg and 30 g oxygen are reacted, residual mixture contains
Q12.
Vapour density of volatile substance is 4 in comparison to methane \((CH_4=1)\). Molecular weight will be
Q13.
Formula of metal chloride is \(MCl_3\). It contains 20% of metal. Atomic weight of metal is
Q14.
Hydrogen phosphate of certain metal has formula \(MHPO_4\). Formula of metal chloride would be
Q15.
How many litres of \(CO_2\) at STP will be formed when 100 ml of 0.1 M \(H_2SO_4\) reacts with excess of \(Na_2CO_3\)?
Q16.
Maximum amount of \(BaSO_4\) precipitated on mixing \(BaCl_2\) (0.5 M) with \(H_2SO_4\) (1 M) will correspond to
Q17.
3 g of carbon is completely burnt in large excess of oxygen. Weight of \(CO_2\) formed is
Q18.
Decomposition of certain mass of \(CaCO_3\) gave 11.2 \(dm^3\) of \(CO_2\) at STP. Mass of KOH required to completely neutralise the gas is
Q19.
26 cc \(CO_2\) is passed over red hot coke. Volume of CO evolved is
Q20.
Number of moles of KCl in 1000 ml of 3 molar solution is
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COMPETITIVE EXAM MCQS
Q1.
Trivalent metal having equivalent weight 20 forms an oxide. Molecular weight of that oxide is
Q2.
Volume occupied by 1.008 mg of hydrogen at STP is
Q3.
When 8 gm of \(H_2\) and 8 gm of \(O_2\) combine, amount of water formed is
Q4.
Of the following, only empirical formula is
Q5.
Which has highest mass?
Q6.
100 ml of gaseous hydrocarbon consumes 300 ml of oxygen for complete combustion. Hydrocarbon is
Q7.
Excess of 2 moles/\(dm^3\) HCl is treated with 2.8 gm iron as shown in the reaction: \(Fe+HCl\rightarrow FeCl_2+H_2\). Volume of HCl consumed is
Q8.
1.5 mole of \(CH_4\) is completely burnt in excess of air. Gram of \(O_2\) consumed is
Q9.
Specific heat of bivalent metal is 0.07. Equivalent weight is
Q10.
6 moles of Al reacts with HCl to give how many grams of \(H_2\)?
Q11.
Value of \(X\) in organic compound with molecular formula \(C_XH*{12}\), having vapour density 42, is
Q12.
How many moles of ammonia can be produced from 8.00 moles of hydrogen reacting with nitrogen?
Q13.
2 gm \(O_2\) at NTP has volume
Q14.
If vapour density of compound \(X_nO\) is 14, X has valency half that of oxygen. Atomic weight of X is
Q15.
Which of the following contains smallest number of \(CO_2\) molecules?
Q16.
Which of the following statements is true?
Q17.
A compound has 87.75% nitrogen and 12.25% hydrogen by weight. Empirical formula of compound is
Q18.
One ml hydrogen gas at NTP contains about
Q19.
1 gm ordinary sample of limestone dissolved in 16.6 cc of 0.92N HCl leaving sandy residue. Percentage of pure \(CaCO_3\) in sample is
Q20.
Compound has molecular weight 78 and empirical formula CH. Molecular formula is
Q21.
6 g carbon is completely burnt in large excess oxygen. Volume of \(CO_2\) formed at STP is
Q22.
In organic compound 78.6% carbon, 8% hydrogen, 12% nitrogen. Empirical formula of compound is
Q23.
How many moles of HCl reacts with one mole of \(KMnO_4\)?
Q24.
3.34 g of an oxide X of atomic weight 127 contains 2.54 g of X. Formula should be
Q25.
0.6 g compound occupies 224 \(cm^3\) at STP. It consists of 6.67% hydrogen, 40% carbon and rest oxygen. Molecular formula of compound is
Q26.
Assuming density of water to be 1 g/cc, volume occupied by one molecule of water is
Q27.
Number of molecules in 36 mg of water is
Q28.
A group of atom can give single valency, illustrated by
Q29.
6 gm carbon when oxidized completely at STP forms what volume of \(CO_2\)?
Q30.
Which of the following method is used for determination of molecular weight?
Q31.
An element has atomic weight A and atomic number Z. Number of neutrons in the element is
Q32.
Atomic weight is defined as
Q33.
2.2 gm \(CO_2\) at NTP occupies
Q34.
Compound having C and H has 20% hydrogen. Molecular formula of compound is
Q35.
Number of molecules in 36 mg of water is
Q36.
A sample of pure water, irrespective of source, contains 88.89% oxygen and 11.11% hydrogen by mass. This data supports
Q37.
An organic compound containing C, H and N gave analysis C = 40 g, H = 3.33 g, N = 56.64 g. Chemical composition of compound is
Q38.
1 gram hydrogen contains how many molecules?
Q39.
If standard reference for calculation of molecular weight is taken as \(\frac{1}{6}\) of C, molecular weight of water will be
Q40.
Which of the following weighs the least?
Q41.
2.7 g of Al reacts with how many grams of \(O_2\)?
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BRUSH UP CHEMICAL CALCULATION MCQS
Q1.
Natural abundances of oxygen isotopes are \(^{16}O=99.76\%\), \(^{17}O=0.04\%\), \(^{18}O=0.20\%\). Atomic mass of oxygen is
Q2.
Average atomic weight of copper is 63.546 amu. Natural copper consists of two isotopes \(^{63}Cu\) and \(^{65}Cu\). Their natural abundance are 69.09% and 30.91% respectively. If atomic weight of \(^{63}Cu\) is 62.9296 amu, atomic weight of \(^{65}Cu\) isotope is
Q3.
Equivalent weight of Mg is 12 means
Q4.
When 31.6 g \(Mn_xO_y\) is heated with carbon, 13.2 g \(CO_2\) gas is produced. Formula of \(Mn_xO_y\) is
Q5.
Which shows law of multiple proportion?
Q6.
Which shows law of reciprocal proportion?
Q7.
Atomic mass of an element having \(3\times10^{21}\) particles which weigh 0.2 g is
Q8.
Molecular mass of gas where 1.12 l weighs 2 g is
Q9.
Equivalent weight of \(H_3PO_3\) is
Q10.
For the reaction \(2Na_2S_2O_3+I_2\rightarrow2NaI+Na_2S_4O_6\), equivalent mass of \(Na_2S_2O_3\) is
Q11.
Equivalent mass of metal containing 20% oxygen is
Q12.
3 g of metal oxide is completely converted into its chloride and 5 g of metal chloride is produced. Equivalent mass of metal is
Q13.
Vapour density is
Q14.
What is V.D., if density of air is 0.000129 g/\(cm^3\)?
Q15.
If V.D. of \(N_2O_4\) at \(20^\circ C\) is 38.3, degree of dissociation \((\alpha)\) is
Q16.
All of the following are chemical changes except
Q17.
Mass of 1 litre gas at NTP is 3.165 g, then atomic mass of gas is; given \(C_v=0.082\), \(C_p=0.115\)
Q18.
A metal sulphate is isomorphous with \(ZnSO_4\) and it contains 20% of metal. Atomic mass of metal is
Q19.
Molecular mass of chloride if equivalent mass of triatomic metal is 33.3 is
Q20.
Minimum molecular mass of compound if it contains 16% of \(O_2\) is
Q21.
0.03 mol atom of \(Na_2CO_3\) has
Q22.
Total number of atoms in 1000 ml of 0.2 M iodine solution is
Q23.
0.5 litre of \(H_2O\) contains
Q24.
Total number of valence electrons in 4.2 g of \(N^{3-}\) ion is
Q25.
3.34 g of an oxide \((X_2O_n)\) contains 2.54 g of X (atomic mass of X = 127). Formula of oxide is
Q26.
Quicklime contains 71.47% calcium. How much calcium is present in a sample of quicklime which contains 16 g oxygen?
Q27.
10 l of nitrogen gas and 10 l of hydrogen gas are introduced into an evacuated flask of 10 litre capacity. It is then heated to 700 K and 3 atm pressure. Volume of ammonia at NTP produced is
Q28.
Mass of atom of silver is
Q29.
One million silver atoms weigh \(1.79\times10^{-16}\) g. Atomic mass of silver is
Q30.
Number of Na atoms in 5.3 g of \(Na_2CO_3\) is
Q31.
Number of gold atoms in 0.3 g of 20 carat gold is
Q32.
An organic compound contains C = 33.8%, H = 4.7%, N = 13.2%, Cl = 33.4%, O = rest. Empirical formula is
Q33.
Volume of air containing 21% oxygen by volume required to completely burn 10 g of sulphur having purity level 98% is
Q34.
5 g impure sample of sodium bicarbonate when heated strongly gave 600 ml \(CO_2\) measured at \(27^\circ C\) and 760 mm Hg pressure. Percentage purity of sample is
Q35.
Percentage yield of sodium sulphate made from 4.19 g of \(Na_2CO_3\), if 5.61 g was obtained, is
Q36.
Symbolic representation of a molecule of an element or compound is called
Q37.
Four 1-litre flasks are separately filled with gases \(H_2\), He, \(O_2\) and \(O_3\) at same temperature and pressure. Ratio of number of atoms of these gases would be
Q38.
Equivalent weight of \(MnSO_4\) is half of its molecular mass when it is converted to
Q39.
0.5 mole of \(BaCl_2\) is mixed with 0.2 moles of \(Na_3PO_4\). Maximum number of moles of \(Ba_3(PO_4)_2\) formed is
Q40.
Which law directly explains law of conservation of mass?