NCERT Exemplar Some Basic Concepts of Chemistry & Stoichiometry - Chemistry

Table of Contents
1. I. Multiple Choice Questions - (Type-I)
2. II. Multiple Choice Questions (Type-II)
3. III. Short Answer Type
4. IV. Matching Type
5. V. Assertion and Reason Type
6. VI. Long Answer Type
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I. Multiple Choice Questions - (Type-I)

1. Two students performed the same experiment separately and each one of them recorded two readings of mass which are given below. Correct reading of mass is 3.0 g. On the basis of given data, mark the correct option out of the following statements.

(a) Results of both the students are neither accurate nor precise
(b) Results of student A are both precise and accurate
(c) Results of student B are neither precise nor accurate
(d) Results of student B are both precise and accurate
Ans. (b)

Solution.
Average of readings of student A = (3.01 + 2.99)/2 = 3.00
Average of readings of student B = (3.05 + 2.95)/2 = 3.00
Correct reading = 3.00
Both students have average values equal to the correct value; therefore both are accurate.
Student A's readings are close to each other (difference 0.02), so A is precise as well.
Student B's readings differ by 0.10, so B is not precise.
Hence, results of student A are both precise and accurate.

2. A measured temperature on the Fahrenheit scale is 200 °F. What will this reading be on Celsius scale?
(a) 40°C
(b) 94°C
(c) 93.3°C
(d) 30°C
Ans. (c)

Solution.
Use the conversion: °C = (5/9)(°F - 32).
°C = (5/9)(200 - 32) = (5/9)(168) = 93.3°C.

3. What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per 500 mL?
(a) 4 mol L^-1
(b) 20 mol L^-1
(c) 0.2 mol L^-1
(d) 2 mol L^-1
Ans. (c)

Solution.
Molar mass of NaCl = 58.5 g mol^-1.
Number of moles of NaCl = 5.85 / 58.5 = 0.10 mol.
Volume = 500 mL = 0.5 L.
Molarity = moles / volume = 0.10 / 0.5 = 0.20 M.

4. If 500 mL of a 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?
(a) 1.5 M
(b) 1.66 M
(c) 0.017 M
(d) 1.59 M
Ans. (b)

Solution.
M₁V₁ = M₂V₂
5 × 500 = M₂ × 1500
M₂ = (5 x 500)/1500 = 1.66 M

5. The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following element contains the greatest number of atoms?
(a) 4 g He
(b) 46 g Na
(c) 0.40 g Ca
(d) 12 g He
Ans. (d)

Solution.
Molar mass of He = 4 g mol^-1.
12 g He = 12 / 4 = 3 moles.
Number of atoms = 3 × N_A = 3 × 6.022 × 10²³.

6. If the concentration of glucose (C₆H₁₂O₆) in blood is 0.9 g L^-1, what will be the molarity of glucose in blood?
(a) 5 M
(b) 50 M
(c) 0.005 M
(d) 0.5 M
Ans. (c)

Solution.
Molar mass of glucose = 180 g mol^-1.
Moles = 0.9 / 180 = 0.005 mol per litre.
Molarity = 0.005 M.

7. What will be the molality of the solution containing 18.25 g of HCl gas in 500 g of water?
(a) 0.1 m
(b) 1 M
(c) 0.5 m
(d) 1 m
Ans. (d)

Solution.
Molar mass of HCl = 36.5 g mol^-1.
Moles of HCl = 18.25 / 36.5 = 0.50 mol.
Mass of solvent = 500 g = 0.500 kg.
Molality = moles of solute / mass of solvent (kg) = 0.50 / 0.500 = 1.0 m.

8. One mole of any substance contains 6.022 × 10²³ atoms/molecules. Number of molecules of H₂SO₄ present in 100 mL of 0.02 M H₂SO₄ solution is ______.
(a) 12.044 × 10²⁰ molecules
(b) 6.022 × 10²³ molecules
(c) 1 × 10²³ molecules
(d) 12.044 × 10²³ molecules
Ans. (a)

Solution.
Molarity = 0.02 M; Volume = 100 mL = 0.1 L.
Moles = M × V = 0.02 × 0.1 = 2.0 × 10^-3 mol.
Number of molecules = 2.0 × 10^-3 × 6.022 × 10²³ = 12.044 × 10²⁰ molecules.

9. What is the mass percent of carbon in carbon dioxide?
(a) 0.034%
(b) 27.27%
(c) 3.4%
(d) 28.7%
Ans. (b)

Solution.
Molar mass CO₂ = 12 + 2 × 16 = 44 g mol^-1.
Mass percent C = (12 / 44) × 100 = 27.27%.

10. The empirical formula and molecular mass of a compound are CH₂O and 180 g respectively. What will be the molecular formula of the compound?
(a) C₉H₁₈O₉
(b) CH₂O
(c) C₆H₁₂O₆
(d) C₂H₄O₂
Ans. (c)

Solution. Empirical formula = CH₂O
Empirical formula mass = 12 + 1 + 1 + 16 = 30 g
Molecular mass = 180 g
n = (Molar mass  / Empirical formula mass)
= 180/30 = 6
∴ n = 6
So, molecular formula = n × CH₂O = 6 × CH₂O = C₆H₁₂O₆

11. If the density of a solution is 3.12 g mL^-1, the mass of 1.5 mL solution in significant figures is _______.
(a) 4.7 g
(b) 4680 × 10^-3 g
(c) 4.680 g
(d) 46.80 g
Ans. (a)

Solution.
Mass = density × volume = 3.12 g mL^-1 × 1.5 mL = 4.68 g.
Given the volume 1.5 has two significant figures, the result reported to two significant figures is 4.7 g.

12. Which of the following statements about a compound is incorrect?
(a) A molecule of a compound has atoms of different elements
(b) A compound cannot be separated into its constituent elements by physical methods of separation
(c) A compound retains the physical properties of its constituent elements
(d) The ratio of atoms of different elements in a compound is fixed
Ans. (c)

Solution.
A compound has properties different from those of its constituent elements, so statement (c) is incorrect.

13. Which of the following statements is correct about the reaction given below:
4Fe(s) + 3O₂(g) → 2Fe₂O₃(g)
(a) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows law of conservation of mass.
(b) Total mass of reactants = total mass of product; therefore, law of multiple proportions is followed.
(c) Amount of Fe₂O₃ can be increased by taking any one of the reactants (iron or oxygen) in excess.
(d) Amount of Fe₂O₃ produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess
Ans. (a)

Solution.
4Fe + 30₂ → 2Fe₂O₃ follows law of conservation of mass since mass of reactants is equal to mass of products.

14. Which of the following reactions is not correct according to the law of conservation of mass.
(a) 2Mg(s) + O₂(g) → 2MgO(s)
(b) C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(g)
(c) P₄(s) + 5O₂(g) → P₄O₁₀(s)
(d) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
Ans. (b)

Solution.
Option (b) is unbalanced as written; numbers of atoms on reactant and product sides are not equal. Hence it violates conservation of mass in the given form.

15. Which of the following statements indicates that law of multiple proportion is being followed.
(a) Sample of carbon dioxide taken from any source will always have carbon and oxygen in the ratio 1:2.
(b) Carbon forms two oxides namely CO₂ and CO, where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2:1.
(c) When magnesium burns in oxygen, the amount of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formed.
(d) At constant temperature and pressure 200 mL of hydrogen will combine with 100 mL oxygen to produce 200 mL of water vapour.
Ans. (b)

Solution.
For CO₂: 12 parts C combine with 32 parts O. For CO: 12 parts C combine with 16 parts O. The masses of oxygen combining with fixed mass of carbon (12) are 32 and 16, ratio 2:1. This exemplifies the law of multiple proportions.

II. Multiple Choice Questions (Type-II)

In the following questions, two or more options may be correct.

16. One mole of oxygen gas at STP is equal to _______.
(a) 6.022 × 10²³ molecules of oxygen
(b) 6.022 × 10²³ atoms of oxygen
(c) 16 g of oxygen
(d) 32 g of oxygen
Ans. (a, d)

Solution.
One mole of O₂ contains 6.022 × 10²³ molecules and has mass equal to the molecular weight: 32 g.

17. Sulphuric acid reacts with sodium hydroxide as follows:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
When 1 L of 0.1M sulphuric acid solution is allowed to react with 1 L of 0.1 M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is
(a) 0.1 mol L^-1
(b) 7.10 g
(c) 0.025 mol L^-1
(d) 3.55 g
Ans. (b, c)

Solution.
Moles H₂SO₄ = 0.1 L × 0.1 M = 0.10 mol.
Moles NaOH = 0.1 L × 0.1 M = 0.10 mol.
Reaction requires 2 mol NaOH per 1 mol H₂SO₄, so NaOH is limiting.
0.10 mol NaOH will react with 0.05 mol H₂SO₄ to give 0.05 mol Na₂SO₄.
Molar mass Na₂SO₄ = 142 g mol^-1.
Mass = 0.05 × 142 = 7.10 g.
Total volume after mixing = 2.0 L, so molarity = 0.05 mol / 2.0 L = 0.025 mol L^-1.

18. Which of the following pairs have the same number of atoms?
(a) 16 g of O₂(g) and 4 g of H₂(g)
(b) 16 g of O₂ and 44 g of CO₂
(c) 28 g of N₂ and 32 g of O₂
(d) 12 g of C(s) and 23 g of Na(s)
Ans. (c, d)

Solution.
28 g N₂ = 1 mol N₂ = 2 × N_A atoms; 32 g O₂ = 1 mol O₂ = 2 × N_A atoms. Thus (c) is correct.
12 g C = 1 mol C and 23 g Na = 1 mol Na; both contain N_A atoms. Thus (d) is correct.

19. Which of the following solutions have the same concentration?
(a) 20 g of NaOH in 200 mL of solution
(b) 0.5 mol of KCl in 200 mL of solution
(c) 40 g of NaOH in 100 mL of solution
(d) 20 g of KOH in 200 mL of solution
Ans. (a, b)

Solution.
20 g NaOH = 20 / 40 = 0.50 mol in 0.200 L → M = 0.50 / 0.200 = 2.5 M.
0.5 mol KCl in 0.200 L → M = 0.5 / 0.200 = 2.5 M.
Thus (a) and (b) have equal concentrations.

20. 16 g of oxygen has same number of molecules as in
(a) 16 g of CO
(b) 28 g of N₂
(c) 14 g of N₂
(d) 1.0 g of H₂
Ans. (c, d)

Solution.
(a) Number of molecules of oxygen in 16g of oxygen = 16/32 × 6.023 × 10²³
= 0.5 × 6.023 × 10²³
(b) 0.5 moles is present in 14g of nitrogen and in 1.0 g H₂. Hence they will also have 0.5 × 6.023 × 10²³
(c) Number of molecules of N₂ = 14/28 × 6.023 × 10²³ = 0.5 × 6.023 × 10²³
(d) Number of molecules of H₂ = 1/2 × 6.023 × 10²³ = 0.5 × 6.023 × 10²³

21. Which of the following terms are unitless?
(a) Molality
(b) Molarity
(c) Mole fraction
(d) Mass percent
Ans. (c, d)

Solution.
Mass percent= (Mass of solute / Mass of solution) × 100
Mole Fraction: It is the ratio of number of moles of a particular component to the total number of moles of the solution.

22. One of the statements of Dalton's atomic theory is given below: "Compounds are formed when atoms of different elements combine in a fixed ratio" Which of the following laws is not related to this statement?
(a) Law of conservation of mass
(b) Law of definite proportions
(c) Law of multiple proportions
(d) Avogadro law
Ans. (a, d)

Solution.
The statement directly relates to laws that describe fixed composition (law of definite proportions) and multiple compounds with simple ratios (law of multiple proportions).
Law of conservation of mass is more general; Avogadro's law concerns equal volumes of gases and is not related to the stated Dalton statement.

III. Short Answer Type

23. What will be the mass of one atom of C-12 in grams?
Ans.

Mass of 6.022 × 10²³ atoms of C = 12 g.
Mass of one atom = 12 / (6.022 × 10²³) = 1.992648 × 10^-23 g.

24. How many significant figures should be present in the answer of the following calculations?

Ans.

Significant figures = 2

25. What is the symbol for SI unit of mole? How is the mole defined?
Ans.

Symbol for the SI unit of mole is mol.
One mole is defined as the amount of a substance that contains as many particles (atoms, molecules, ions, etc.) as there are atoms in exactly 12 g (0.012 kg) of the carbon-12 isotope.

26. What is the difference between molality and molarity?
Ans.

Molarity (M): Number of moles of solute per litre of solution.
M = moles of solute / volume of solution in litres.
Molarity depends on temperature because volume changes with temperature.
Molality (m): Number of moles of solute per kilogram of solvent.
m = moles of solute / mass of solvent in kg.
Molality is independent of temperature because it depends on mass, not volume.

27. Calculate the mass percent of calcium, phosphorus and oxygen in calcium phosphate Ca₃(PO₄)₂.
Ans.

Molecular mass of Ca₃(PO₄) = 3 x 40 + 2 × 31 + 8 × 16 = 310
Mass percent of Ca = ((3 × 40)/310) × 100 = 38.71 %
Mass percent of P = (( 2 × 31)/310) × 100 = 20%
Mass percent of O = ((8 × 16)/310) × 100 = 41.29 %

28. 45.4 L of dinitrogen reacted with 22.7 L of dioxygen and 45.4 L of nitrous oxide was formed. The reaction is given below:
2N₂(g) + O₂(g) → 2N₂O(g)
Which law is being obeyed in this experiment? Write the statement of the law?
Ans.

Gay-Lussac's Law of Gaseous Volumes is obeyed. Statement: When gases react together at constant temperature and pressure, the volumes of gaseous reactants and products are in simple whole-number ratios to one another (provided volumes are measured at same T and P).

29. If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in whole number ratio.
(a) Is this statement true? 
(b) If yes, according to which law? 
(c) Give one example related to this law
Ans.

(a) Yes, it is true.
(b) This is the law of multiple proportions.
(c) Example: Carbon forms CO and CO₂. For 12 g carbon, oxygen masses are 16 g (in CO) and 32 g (in CO₂), ratio 16:32 = 1:2.

30. Calculate the average atomic mass of hydrogen using the following data:

Ans. 

Average Atomic mass

31. Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc. The following reaction takes place.
Zn + 2HCl → ZnCl₂ + H₂
Calculate the volume of hydrogen gas liberated at STP when 32.65 g of zinc reacts with HCl. 1 mol of a gas occupies 22.7 L volume at STP; atomic mass of Zn = 65.3 u.
Ans. 65.3 g Zn corresponds to 1 mol and produces 1 mol H₂ (22.7 L) at STP.
Therefore, 32.65 g Zn (which is 0.5 mol) will produce 0.5 × 22.7 L = 11.35 L of H₂ at STP.

32. The density of 3 molal solution of NaOH is 1.110 g mL^-1. Calculate the molarity of the solution.
Ans.

3 m solution means 3 mol NaOH in 1000 g solvent.
Mass of NaOH = 3 × 40 = 120 g.
Mass of solution = 1000 + 120 = 1120 g.
Volume of solution = mass / density = 1120 g / 1.110 g mL^-1 = 1009.009... mL ≈ 1.009 L.
Molarity = moles / volume (L) = 3 / 1.009 ≈ 2.97 M.

33. Volume of a solution changes with change in temperature, then, will the molality of the solution be affected by temperature? Give reason for your answer.
Ans. No, molality of a solution does not change with temperature since mass remains unaffected by temperature.
Molality, m = moles of solute/weight of solvent (in g) × 1000

34. If 4 g of NaOH dissolves in 36 g of H₂O, calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is 1 g mL^-1).
Ans. Number of moles H₂O = 36 / 18 = 2.00 mol.
Number of moles NaOH = 4 / 40 = 0.10 mol.
Total moles = 2.00 + 0.10 = 2.10 mol.
Mole fraction H₂O = 2.00 / 2.10 = 0.952.
Mole fraction NaOH = 0.10 / 2.10 = 0.048.
Mass of solution = 36 + 4 = 40 g → volume = 40 mL (density = 1 g mL^-1).
Molarity = moles solute / volume (L) = 0.10 / 0.040 = 2.5 M.

35. The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction 2A + 4B → 3C + 4D, when 5 moles of A react with 6 moles of B, then
(i) which is the limiting reagent?
(ii) calculate the amount of C formed?
Ans. 2A + 4B → 3C + 4D
According to the above equation, 2 mols of 'A' require 4 mols of 'B' for the reaction.
(i) Hence, for 5 mols of 'A', the moles of 'B' required
 = 5 mol of A × 4 mol of B/2 mol of A = 10 mol of B
But we have only 6 mols of 'B', hence, 'B' is the limiting reagent
(ii) Amount of 'C' formed is determined by amount of 'B'.
Since 4 mols of 'B' give 3 mols of 'C'. Hence 6 mols of 'B' will give
6 mol of B × 3 mol of C/4 mol of C = 4.5 mol of C

IV. Matching Type

36. Match the following:

Ans. (i) → (b); (ii) → (c); (iii) → (a); (iv) → (e); (v) → (d)

37. Match the following physical quantities with units

Ans. (i) → (e); (ii) → (d); (iii) → (b); (iv) → (g); (v) → (c), (h); (vi) → (f); (vii) → (a); (viii) → (i)

(i) Molarity = mol L^-1
(ii) Mole fraction = no units
(iii) Mole = mol
(iv) Molality = mol kg^-1
(v) Pressure = Pascal or N m^-2
(vi) Luminous intensity = candela
(vii) Density = g mL^-1
(viii) Mass = kg

V. Assertion and Reason Type

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

38. Assertion (A): The empirical mass of ethene is half of its molecular mass.
Reason (R): The empirical formula represents the simplest whole number ratio of various atoms present in a compound.
(a) Both A and R are true and R is the correct explanation of A
(b) A is true but R is false
(c) A is false but R is true
(d) Both A and R are false
Ans. (a)

Explanation.
Empirical formula of ethene is CH₂ (mass 14). Molecular formula C₂H₄ has mass 28; empirical mass is half the molecular mass. The empirical formula indeed represents the simplest whole-number ratio.

39. Assertion (A): One atomic mass unit is defined as one-twelfth of the mass of one carbon-12 atom.
Reason (R): Carbon-12 isotope is the most abundant isotope of carbon and has been chosen as standard.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) Both A and R are false.
Ans. (b)

Explanation.
One atomic mass unit (1 u) is defined as exactly one-twelfth of the mass of one carbon-12 atom. Carbon-12 was chosen as the standard (not necessarily because it is the most abundant) to provide a consistent mass scale; thus R is true but is not the precise reason for the definition.

40. Assertion (A): Significant figures for 0.200 is 3 whereas for 200 it is 1.
Reason (R): Zero at the end or right of a number are significant provided they are not on the right side of the decimal point.
(a) Both A and R are true and R is correct explanation of A
(b) Both A and R are true but R is not a correct explanation of A
(c) A is true but R is false
(d) Both A and R are false
Ans. (c)

Explanation.
Significant figures for 0.200 = 3
Significant figure for 200 = 1
Zeros at the end of a number without decimal-point, may or may not be significant depending on the accuracy of measurement.

41. Assertion (A): Combustion of 16 g of methane gives 18 g of water.
Reason (R): In the combustion of methane, water is one of the products.
(a) Both A and R are true but R is not the correct explanation of A
(b) A is true but R is false
(c) A is false but R is true
(d) Both A and R are false
Ans. (c)

Explanation.
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
16 g of CH₄ on complete combustion will give 36 g of water. Thus, A is false; R is true.

VI. Long Answer Type

42. A vessel contains 1.6 g of dioxygen at STP (273.15 K, 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate
(i) volume of the new vessel
(ii) number of molecules of dioxygen
Ans.

Solution. p₁ = 1 atm, T₁ = 273 K, V₁ =?
32 g of oxygen occupies 22.4 L of volume at STP*
Hence, 1.6 g of O₂ will occupy,
1.6g O₂ × 22.4L/32g O₂ = 1.12 L
V₁ = 1.12 L
p₂ = p₁/2 = 1/2 =0.5 atm
V₂ = ?
p₁V₁ = p₂V₂
V₂ = p₁ ×V₁/p₂ = 1 atm × 1.12 L/0.5 atm
= 2.24 L
(ii) Number of molecules of O₂ in the vessel
= 6.022 × 10²³ × 1.6/32 = 3.011 × 10²²

43. Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction given below:
CaCO₃ (s) + 2HCl (aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)
What mass of CaCl₂ will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of CaCO₃? Name the limiting reagent. Calculate the number of moles of CaCl₂ formed in the reaction.
Ans.

Solution.
Number of moles of HCl

Mass of CaCO₃ = 1000 g
Number of moles of CaCO₃ = 1000g/100g = 10mol
According to given equation 1 mol of CaCO₃(s) requires 2 mol of HCl (aq).
Hence, for the reaction of 10 mol of CaCO₃(s) number of moles of HCl required would be:

But we have only 0.19 mol HCl (aq), hence, HCl (aq) is limiting reagent. So, amount of
CaCl₂ formed will depend on the amount of HCl available. Since, 2 mol HCl (aq) forms 1 mol of CaCl₂, therefore, 0.19 mol of HCl (aq)would give:

Or 0.095 × molar mass of CaCl₂ = 0.095 × 111 = 10.54 g

44. Define the law of multiple proportions. Explain it with two examples. How does this law point to the existence of atoms?
Ans. The law of multiple proportions states that, "If two elements can combine to form more than one compound, the mass of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers."
For example, hydrogen combines with oxygen to form two compounds, namely, water and hydrogen peroxide.
Hydrogen + Oxygen → Water
 2g                   16g              18g
Hydrogen + Oxygen → Hydrogen Peroxide
  2g                   32g                          34g
Here, the masses of oxygen (i.e. 16 g and 32 g) which combine with a fixed mass of hydrogen (2g) bear a simple ratio, i.e., 16 : 32 or 1 : 2.
This law shows that there are constituents which combine in a definate proportion. These constituents are atoms.

45. A box contains some identical red coloured balls, labelled as A, each weighing 2 grams. Another box contains identical blue coloured balls, labelled as B, each weighing 5 grams. Consider the combinations AB, AB₂, A₂B and A₂B₃ and show that law of multiple proportions is applicable.
Ans. Calculate mass of B combined with 1 g of A for each formula (using A = 2 g, B = 5 g):
For AB: 1 unit A (2 g) combines with 1 unit B (5 g). For 1 g A, B mass = 5 / 2 = 2.5 g.
For AB₂: 1 unit A (2 g) combines with 2 units B (10 g). For 1 g A, B mass = 10 / 2 = 5.0 g.
For A₂B: 2 units A (4 g) combine with 1 unit B (5 g). For 1 g A, B mass = 5 / 4 = 1.25 g.
For A₂B₃: 2 units A (4 g) combine with 3 units B (15 g). For 1 g A, B mass = 15 / 4 = 3.75 g.
These B masses with 1 g A are: 2.5 : 5.0 : 1.25 : 3.75 → dividing by 1.25 gives 2 : 4 : 1 : 3, a simple whole-number ratio. Thus the law of multiple proportions is satisfied.