Some Basic Concepts of Chemistry
1.1 – 1.3 Molar Mass & Formulas
1.1 Calculate Molar Mass: H₂O, CO₂, CH₄
(i) $H_2O$:
Atomic masses: $H = 1.008 \text{ u}, O = 16.00 \text{ u}$.
Molar Mass $= 2(1.008) + 16.00 = 2.016 + 16.00 = \mathbf{18.016 \text{ g/mol}}$.
(ii) $CO_2$:
Atomic masses: $C = 12.01 \text{ u}, O = 16.00 \text{ u}$.
Molar Mass $= 12.01 + 2(16.00) = 12.01 + 32.00 = \mathbf{44.01 \text{ g/mol}}$.
(iii) $CH_4$:
Atomic masses: $C = 12.01 \text{ u}, H = 1.008 \text{ u}$.
Molar Mass $= 12.01 + 4(1.008) = 12.01 + 4.032 = \mathbf{16.042 \text{ g/mol}}$.
1.2 Mass percent of elements in Sodium Sulphate (Na₂SO₄)
Step 1: Calculate Molar Mass of $Na_2SO_4$.
$Na=22.99, S=32.06, O=16.00$.
$M = 2(22.99) + 32.06 + 4(16.00) = 45.98 + 32.06 + 64.00 = 142.04 \text{ g/mol}$.
Step 2: Calculate Mass Percentages.
$\text{Mass \%} = \frac{\text{Mass of Element in compound}}{\text{Molar Mass}} \times 100$
Sulphur (S): $\frac{32.06}{142.04} \times 100 \approx \mathbf{22.57\%}$
Oxygen (O): $\frac{64.00}{142.04} \times 100 \approx \mathbf{45.06\%}$
1.3 Empirical Formula of Iron Oxide (69.9% Fe, 30.1% O)
| Element | % Mass | Atomic Mass | Relative Moles | Simple Ratio | Whole Number |
|---|---|---|---|---|---|
| Fe | 69.9 | 55.85 | $69.9/55.85 = 1.25$ | $1.25/1.25 = 1$ | $1 \times 2 = 2$ |
| O | 30.1 | 16.00 | $30.1/16.00 = 1.88$ | $1.88/1.25 = 1.5$ | $1.5 \times 2 = 3$ |
Empirical Formula: $Fe_2O_3$ (Hematite).
1.4 – 1.8 Stoichiometry & Solutions
1.4 Amount of CO₂ produced.
Reaction: $C(s) + O_2(g) \rightarrow CO_2(g)$
(i) 1 mole C burnt in air (excess $O_2$):
1 mol C produces 1 mol $CO_2$. Mass = $44 \text{ g}$.
(ii) 1 mole C burnt in 16 g $O_2$:
Moles of $O_2 = 16/32 = 0.5 \text{ mol}$.
Stoichiometry is 1:1. Here $O_2$ is the limiting reagent (0.5 < 1).
$CO_2$ produced = 0.5 mol = $0.5 \times 44 = \mathbf{22 \text{ g}}$.
(iii) 2 moles C burnt in 16 g $O_2$:
Moles $O_2 = 0.5 \text{ mol}$. $O_2$ is still limiting.
$CO_2$ produced = 0.5 mol = $\mathbf{22 \text{ g}}$.
1.5 Mass of Sodium Acetate for 0.375 M Solution.
Given: Molarity ($M$) = 0.375 mol/L. Volume ($V$) = 500 mL = 0.5 L.
Molar Mass ($MM$) of $CH_3COONa = 82.0245 \text{ g/mol}$.
Mass $= n \times MM = 0.1875 \times 82.0245 = \mathbf{15.380 \text{ g}}$.
1.6 Concentration of Nitric Acid.
Given: Density $d = 1.41 \text{ g/mL}$. Mass % = 69% ($HNO_3$).
Molar mass of $HNO_3 = 1 + 14 + 3(16) = 63 \text{ g/mol}$.
Consider 1 Litre (1000 mL) of solution.
Mass of solution $= V \times d = 1000 \times 1.41 = 1410 \text{ g}$.
Mass of $HNO_3 = 69\%$ of $1410 = 0.69 \times 1410 = 972.9 \text{ g}$.
Moles of $HNO_3 = 972.9 / 63 = 15.44 \text{ mol}$.
Molarity: $15.44 \text{ mol} / 1 \text{ L} = \mathbf{15.44 \text{ M}}$.
1.7 Copper from 100 g Copper Sulphate.
Molar mass of $CuSO_4$:
$Cu(63.5) + S(32) + 4O(64) = 159.5 \text{ g/mol}$.
159.5 g of $CuSO_4$ contains 63.5 g of Cu.
100 g of $CuSO_4$ contains: $\frac{63.5}{159.5} \times 100 = \mathbf{39.81 \text{ g}}$.
1.8 Molecular Formula of Iron Oxide.
From Q1.3, the Empirical Formula is $Fe_2O_3$.
Empirical Formula Mass $= 2(55.85) + 3(16.00) = 159.7 \text{ g/mol}$.
The problem gives the same percentage data as Q1.3. For simple ionic oxides like this, the molecular formula unit is typically the same as the empirical formula unless specified otherwise (e.g., molecular mass given).
Answer: $Fe_2O_3$.
1.9 – 1.12 Moles & Molarity
1.9 Average Atomic Mass of Chlorine.
$= \frac{2649.59 + 895.68}{100} = \frac{3545.27}{100} = \mathbf{35.453 \text{ u}}$.
1.10 Calculations for 3 moles of Ethane ($C_2H_6$).
(i) Moles of Carbon atoms:
1 mol $C_2H_6$ has 2 mol C. $\therefore 3 \times 2 = \mathbf{6 \text{ moles}}$.
(ii) Moles of Hydrogen atoms:
1 mol $C_2H_6$ has 6 mol H. $\therefore 3 \times 6 = \mathbf{18 \text{ moles}}$.
(iii) Number of molecules:
$3 \times 6.022 \times 10^{23} = \mathbf{1.8066 \times 10^{24} \text{ molecules}}$.
1.11 Concentration of Sugar Solution.
Molar mass of sugar ($C_{12}H_{22}O_{11}$):
$12(12.01) + 22(1.008) + 11(16.00) = 342.3 \text{ g/mol}$.
Moles of sugar $= 20 \text{ g} / 342.3 \text{ g/mol} = 0.0584 \text{ mol}$.
Volume $= 2 \text{ L}$.
Molarity $= 0.0584 / 2 = \mathbf{0.0292 \text{ mol/L}}$.
1.12 Volume of Methanol needed.
Target: 2.5 L of 0.25 M solution.
Moles of methanol required $= M \times V = 0.25 \times 2.5 = 0.625 \text{ mol}$.
Molar mass $CH_3OH = 12 + 4(1) + 16 = 32 \text{ g/mol}$.
Mass needed $= 0.625 \times 32 = 20 \text{ g} = 0.020 \text{ kg}$.
Density $= 0.793 \text{ kg/L}$.
Volume $= \text{Mass} / \text{Density} = 0.020 / 0.793 = \mathbf{0.0252 \text{ L}}$ or $\mathbf{25.2 \text{ mL}}$.
1.13 – 1.20 Units & Significant Figures
1.13 Calculate Pressure in Pascals.
Mass $m = 1034 \text{ g} = 1.034 \text{ kg}$. Area $A = 1 \text{ cm}^2 = 10^{-4} \text{ m}^2$.
Force $F = mg = 1.034 \times 9.8 = 10.1332 \text{ N}$.
1.14 SI Unit of Mass.
The SI unit of mass is the **kilogram (kg)**. It is defined by taking the fixed numerical value of the Planck constant $h$ to be $6.62607015 \times 10^{-34}$ when expressed in the unit J s, which is equal to kg m² s⁻¹.
1.15 Match Prefixes.
- (i) micro $\rightarrow 10^{-6}$
- (ii) deca $\rightarrow 10$
- (iii) mega $\rightarrow 10^6$
- (iv) giga $\rightarrow 10^9$
- (v) femto $\rightarrow 10^{-15}$
1.17 Contamination of Chloroform (15 ppm).
(i) Percent by Mass:
15 ppm means 15 parts in $10^6$ parts.
$\% = \frac{15}{10^6} \times 100 = \mathbf{1.5 \times 10^{-3} \%}$.
(ii) Molality:
Consider 1000 g (1 kg) of water sample. Mass of solute ($CHCl_3$) $\approx 15 \text{ mg}$ (approx for 1L/1kg dilute). Exact: 15g in $10^6$g solution.
In 1000g solution, mass solute = $0.015$ g. Mass solvent $\approx 1000$ g = 1 kg.
Molar mass $CHCl_3 = 12 + 1 + 3(35.5) = 119.5 \text{ g/mol}$.
Moles $= 0.015 / 119.5 = 1.255 \times 10^{-4} \text{ mol}$.
Molality $= 1.255 \times 10^{-4} / 1 \text{ kg} = \mathbf{1.255 \times 10^{-4} \text{ m}}$.
1.19 Count Significant Figures.
- (i) 0.0025: 2 (2, 5)
- (ii) 208: 3
- (iii) 5005: 4
- (iv) 126,000: 3 (Trailing zeros without decimal usually insignificant)
- (v) 500.0: 4 (Zeros after decimal are significant)
- (vi) 2.0034: 5
1.20 Rounding off.
- (i) 34.216 $\to$ 34.2
- (ii) 10.4107 $\to$ 10.4
- (iii) 0.04597 $\to$ 0.0460 (Leading zeros don’t count)
- (iv) 2808 $\to$ 2810
1.21 – 1.27 Laws & Reactions
1.21 Law of Chemical Combination.
(a) Law Obeyed: Law of Multiple Proportions.
Fix mass of Dinitrogen at 28 g:
(i) 14g N : 16g O $\rightarrow$ 28g N : 32g O
(ii) 14g N : 32g O $\rightarrow$ 28g N : 64g O
(iii) 28g N : 32g O
(iv) 28g N : 80g O
Masses of Oxygen reacting with fixed mass (28g) of Nitrogen are: 32, 64, 32, 80.
Ratio: 32:64:32:80 = 2:4:2:5 (Simple whole numbers).
(b) Conversions:
(i) 1 km = $10^6$ mm = $10^{15}$ pm
(ii) 1 mg = $10^{-6}$ kg = $10^6$ ng
(iii) 1 mL = $10^{-3}$ L = $10^{-3}$ dm$^3$
1.23 Limiting Reagent Identification.
Reaction: $A + B_2 \rightarrow AB_2$ (1 atom A reacts with 1 molecule B).
- (i) 300 atoms A + 200 molecules B: B is limiting (200 < 300).
- (ii) 2 mol A + 3 mol B: A is limiting (2 < 3).
- (iii) 100 atoms A + 100 molecules B: No limiting reagent (Stoichiometric).
- (iv) 5 mol A + 2.5 mol B: B is limiting.
- (v) 2.5 mol A + 5 mol B: A is limiting.
1.24 Ammonia Synthesis (Limiting Reagent).
Eq: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$
Given: 2000 g $N_2$, 1000 g $H_2$.
Moles $N_2 = 2000/28 = 71.43$ mol.
Moles $H_2 = 1000/2 = 500$ mol.
(i) Identify Limiting:
1 mol $N_2$ needs 3 mol $H_2$.
71.43 mol $N_2$ needs $71.43 \times 3 = 214.29$ mol $H_2$.
We have 500 mol $H_2$ (Excess). Thus, $N_2$ is limiting.
Calculate $NH_3$:
1 mol $N_2 \rightarrow$ 2 mol $NH_3$.
71.43 mol $N_2 \rightarrow 142.86$ mol $NH_3$.
Mass $NH_3 = 142.86 \times 17 = \mathbf{2428.6 \text{ g}}$.
(ii) Unreacted Reactant: $H_2$.
(iii) Mass Unreacted:
Used $H_2 = 214.29$ mol.
Left $H_2 = 500 – 214.29 = 285.71$ mol.
Mass $= 285.71 \times 2 = \mathbf{571.4 \text{ g}}$.
1.25 Difference: 0.50 mol vs 0.50 M Na₂CO₃
0.50 mol $Na_2CO_3$: Refers to a specific amount of substance (mass = $0.50 \times 106 = 53$ g).
0.50 M $Na_2CO_3$: Refers to a concentration. It means 0.50 moles of $Na_2CO_3$ are dissolved in 1 Litre of solution.
1.28 – 1.36 Advanced Problems
1.28 Largest Number of Atoms.
Calculate moles ($n = m/M$), then atoms ($n \times N_A$).
- (i) 1 g Au: $1/197$ mol.
- (ii) 1 g Na: $1/23$ mol.
- (iii) 1 g Li: $1/7$ mol.
- (iv) 1 g $Cl_2$: $1/71$ mol (molecules) $\rightarrow 2/71$ mol (atoms).
Comparing denominators: $1/7$ is the largest value.
(iii) 1 g Li has the largest number of atoms.
1.29 Molarity from Mole Fraction ($X = 0.040$).
Let total solution contain 1 mole of components.
$n_{ethanol} = 0.04$. $n_{water} = 0.96$.
Mass of water = $0.96 \times 18 = 17.28$ g.
Mass of ethanol = $0.04 \times 46 = 1.84$ g.
Total Mass = $19.12$ g.
Assuming density of solution $\approx 1$ g/mL (given):
Volume of solution = $19.12$ mL = $0.01912$ L.
1.34 Welding Gas Analysis.
Step 1: Empirical Formula
From combustion:
C: $3.38$ g $CO_2 \times (12/44) = 0.9218$ g C.
H: $0.690$ g $H_2O \times (2/18) = 0.0767$ g H.
Moles C: $0.9218/12 = 0.0768$.
Moles H: $0.0767/1 = 0.0767$.
Ratio 1:1. Empirical Formula: CH.
Step 2: Molar Mass
10.0 L at STP weighs 11.6 g.
At STP, 22.4 L = 1 mole.
Molar Mass $= \frac{11.6}{10.0} \times 22.4 = \mathbf{26.0 \text{ g/mol}}$.
Step 3: Molecular Formula
Empirical Mass (CH) = $12+1 = 13$.
$n = \text{Molar Mass} / \text{Emp Mass} = 26/13 = 2$.
Molecular Formula = $(CH)_2 = \mathbf{C_2H_2}$ (Acetylene).
1.36 Chlorine Preparation (MnO₂ + HCl).
Reaction: $4HCl + MnO_2 \rightarrow 2H_2O + MnCl_2 + Cl_2$.
Given: 5.0 g $MnO_2$.
Molar mass $MnO_2 = 54.9 + 2(16) = 86.9 \text{ g/mol}$.
Moles $MnO_2 = 5.0 / 86.9 = 0.0575 \text{ mol}$.
Stoichiometry: 1 mol $MnO_2$ reacts with 4 mol $HCl$.
Moles $HCl$ needed $= 4 \times 0.0575 = 0.230 \text{ mol}$.
Mass $HCl = 0.230 \times 36.5 = \mathbf{8.40 \text{ g}}$.