Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination(Solved)

Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination

• Law of conservation of mass
• Law of constant proportion
• Empirical formula

1. If 10 grams of CaCO₃ on heating gave 4.4g of CO₂ and 5.6g of CaO, show that these observations are in agreement with the law of conservation of mass.(Based on Law of conservation of mass)

Solution:  Mass of the reactants = 10g ; 

Mass of the products = 4.4 + 6.6g = 10g

Since the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass.

2. 1.375 g of cupric oxide was reduced by heating and the weight of copper that remained was 1.098g. In another experiment 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate was converted into cupric oxide by ignition . The weight of cupric oxide formed was 1.476 g. which law of chemical combinations does this data state?

Solution: in first experiment: Copper oxide = 1.375g ; 

 Copper left = 1.098g ;  

Oxygen present = 1.375 - 1.098 = 0.277g

Percentage of oxygen in CuO = (0.277x 100)/1.375= 20.14

Second Experiment: Copper taken = 1.179g 

Copper oxide formed = 1.476g 

Oxygen present = 1.476 - 1.179 = 0.297g 

Percentage of oxygen in CuO = (0.297x 100)/1.47= 20.12

Since percentage of oxygen is the same in both the above cases, so the law of constant composition is illustrated.

3. Hydrogen and oxygen are known to form two compounds H₂O and H₂O₂. The hydrogen content in one is 11.2 % while in the other it is 5.93 %. Show that this data illustrates the law of multiple proportions.

Solution: Law of multiple proportions - It states that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.

In water - H => 11.2 %, then O => 88.8 %    

Ratio of O:H = 88.8/11.2 = 7.9 part

In hydrogen peroxide -  H => 5.93 %, 

then O => 94.07 %       

Ratio of O:H = 94.07/5.93 = 15.86 part

Ratio of the masses of oxygen that combine with fixed mass of hydrogen: 15.86 : 7.9 = 2:1. 

This is constant with the law of multiple proportions.

4. In an experiment, 1.288g of copper oxide was obtained from 1.03 g of copper. In another experiment 3.672 g of copper oxide gave, on reduction, 2.938 g of copper. Show that these figures verify the law of constant proportions.

Sol: In first experiment: Mass of copper = 1.03 g 

then Mass of oxygen = 1.288 – 1.03 = 0.285

Cu : O = 1.03 : 0.285 = (1.03 / 0.285) : 1 = 3.99 :1 = 4:1

In 2nd experiment: Mass of copper = 2.938g then  

Mass of oxygen = 3.672 – 2.938 = 0.734

Cu : O = 2.938 : 0.734  = (2.938 /0.734) : 1 =  4:1

Since Ratio of Cu and O remain same in both experiment this shows low of constant proportion.

5. A 0.24 gm sample of compound of oxygen and boron was found by analysis to contain 0.096 gm of boron and 0.144 gm of oxygen. Calculate the percentage composition of the compound by weight.

Ans: % of boron in sample = (0.096 4- 0.24) x 100 = 40% ;  

 % of oxygen In sample = (0.144 ÷ 0.24) x 100 = 60%

The sample of compound contains 40% boron and 60% oxygen by weight.

6.  When 3.0 gm of carbon s burnt in 8.0 gm oxygen, 11.0 gm of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.0 gm of carbon is burnt in 50.00 gm of oxygen ? Which law of chemical combination will govern your answer ?

Ans: When 3.0 gm of carbon s burnt in 8.0 gm oxygen, 11.0 gm of carbon dioxide is produced. 

It means all of carbon and oxygen are combined in the ratio of 3 : 8 to form carbon dioxide. 

Thus when there is 3 gm carbon and 50 gm oxygen, then also only 8 gm of oxygen will be used and 11 gm of carbon dioxide will be formed. The remaining oxygen is not used. 

This indicates law of definite proportions which says that in compounds, the combining elements are present in definite proportions by mass.

7. In a reaction, 5.3 gm of sodium carbonate reacted with 6 gm of ethanoic acid. The products were 2.2 gm of carbon dioxide, 0.9 gm water and 8.2 gm sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.

Ans: Mass of reactants = Mass of sodium carbonate + Mass of ethanoic acid 

                                       = 5.3 1 + 6.0 = 11.3 gm

Mass of products = Mass of carbon dioxide + Mass of water + Mass of sodium ethanoate 

                              = 2.2 + 0.9 +  8.2 = 11.3 gm 

The mass of products is equal to the mass of reactants. Thus, mass is neither created nor lost during the given chemical change which is in agreement with the law of conservation of mass. 

8. Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 gm of hydrogen gas ?

Ans: Since hydrogen and oxygen combine in the ratio of 1:8 oy mass, it means that 1 gm of hydrogen and 8 gm of oxygen will be required to form water.

So, Oxygen required to react with 3 gm of hydrogen to form water = 3 x 8 = 24 gm

9. Find the percentage of water of crystallization in feSO₄ .7H₂O.

Ans: The the molecular mass of feSO₄ .7H₂O. = 55.9 + 32.0 + 4 x 16 + 7(18) 

= 55.9 + 32 + 64 + 126 = 277.9 g/mol.

277.9 g/mol of FeSO4 contain 126g of water

Hence, 100g of crystal will contain (126/100) x277.9 g 

= 45.34 of water of crystallization

The amount of water of crystallization in feSO₄ .7H₂0. = 45.34% by mass

10. 42g of copper gave 3.025g of a black oxide of copper, 6. 49g of a black oxide, on reduction with hydrogen, gave 5.192g of copper. Show that these figures are in accordance with law of constant proportion?

Solution: The percentage of copper is first oxide (2.43 x 100)x 3.025 = 80.0

The percentage of copper is second oxide = (5.192x 100)x6.49= 80.02

As the percentage of copper in both the oxides is same, thence law of constant composition is verified.

11. A compound was found to have the following percentage composition by mass Zn = 22.65%,          S = 11.15%. H = 4.88%, 0 = 61.32%. The relative molecular mass is 287g/mol. Find the molecular formula of the compound, assuming that all the hydrogen in the compound is present in water of crystallizations.

Solution: Zn : S : O : H = ( 22.65/65): (11.15/32): (61.32/16) : (4.88/1) 

                                          = 0.348: 0.3484 :3.833: 4.88

To obtain an integral ratio(whole number), we divide by smallest number 0.3484

Zn : S : O : H  = 1 : 1 : 11 : 14   

Now, Empirical formula is Zn SO₁₁H₁₄

Relative molecular mass = {65 + 32 + (1 1x 16) +14} = 287

Moles =(n)= 287/287 = 1  then.  Molecular formula is Zn 80111-114

12. What is Empirical formula?

Ans: Empirical formula is the formula of a compound, which shows the simplest whole number ratio between the atoms of the elements in the compound. It does not indicate the actual number of atoms of the elements present but the simplest whole number ratio. Example:

Organic compound Benzene:  Empirical formula =  C1 H1              

molecular formula = C 6 H6

Glucose  : Empirical formula  =C₁ H ₂ O₁                              

molecular formula  = C₆ H ₁₂ O₆

13. How can we differentiate a molecular formula from an empirical formula?

Ans: If the subscripts in the formula have a common divisor, it is usually a molecular formula. Generally the empirical formula is multiplied by this common divisor to get the molecular formula.

Example: Molecular formula is CH₃COOH

Empirical formula of acetic acid is CH₂O

Molecular formula = [{C4:H2:O2}/2]  =  CH2O

14. An oxide of iron contain 72.41% of Fe Calculate empirical formula of iron oxide?

Solution : Relative no. of fe = 72.41 /56 = 1.29 ; Relative no. of  oxygen = 27.59/16= 1.72

Simple ratio = Fe: O = (1.29/1.29) : (1.72/1.29)  = 1 : 1.33 = (1x3) : (1.33 x3) = 3:4

Therefore Empirical formula = Fe₃O₄.

15. A sample of ammonia contains 28 g nitrogen and 6 g hydrogen. Another sample contains 15 g hydrogen. Find the amount of nitrogen in the second sample.

Answer: The ratio of hydrogen and nitrogen in the first sample = 6:28. 

According to the law of definite proportion, 

The ratio of hydrogen and nitrogen in the second sample = 3 :14. 

Let weight of nitrogen in the second sample be x, Hence, 3:14 = 15:x 

Or, x = (15x14)/3 = 70 g. 

Thus weight of nitrogen = 70 g.

16. Hydrogen and oxygen combine in the ratio 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas ?

Answer : Ratio of hydrogen and oxygen = 1:8 

Let x g of oxygen is required, Hence, 3:x = 1:8 

Or, x = 3x8 = 24 g 

Thus mass of oxygen gas required to react with 3 g of hydrogen is 24 g. 

For more solved Question visit:  Atoms and molecules class 09 Chemistry