Chapter 2, Problem 2.1P

Interpretation Introduction

Interpretation: If the given samples of carbon monoxide are consistent with law of definite proportions or not should be determined.

Concept introduction: The law of definite proportion states that, all the samples of a given compound, regardless of how they are prepared, have same proportion of their constituents.

Answer to Problem 2.1P

Solution: Decomposition of both the samples of carbon monoxide is consistent with the law of definite proportions.

Explanation of Solution

Given information: The amount of oxygen produced on decomposition of sample 1 of $\text{CO}$is $17.2\text{g}$ .

The amount of carbon produced on decomposition of sample 1 of $\text{CO}$is $12.9\text{g}$ .

The amount of oxygen produced on decomposition of sample 2 of is $\text{CO}$

$10.5\text{g}$ .

The amount of carbon produced on decomposition of sample 2 of $\text{CO}$is $7.88\text{g}$ .

The molecular formula of carbon monoxide is CO thus, carbon and oxygen are present in 1:1 ratio by moles.

To determine if the given samples are consistent with law of definite proportion, the number of moles of C and O in each sample should be calculated. The ratio of C:O should be 1:1.

Sample 1: Mass of carbon is 12.9 g and molar mass of carbon is 12.0 g/mol, the number of moles of carbon can be calculated as follows:

  $n_C=\frac{m}{M}$

Putting the values,

  $n_C=\frac{12.9\text{ g}}{12.0\text{ g/mol}}=1.075\text{ mol}$

Similarly, mass of oxygen is 17.2 g and molar mass of oxygen is 16.0 g/mol, the number of moles of oxygen can be calculated as follows:

  $n_O=\frac{17.2\text{ g}}{16.0\text{ g/mol}}=1.075\text{ mol}$

Now, the ratio of number of moles of C and O will be:

  $\frac{n_C}{n_O}=\frac{1.075}{1.075}=1$

Thus, the ratio is 1:1.

Sample 2: Mass of carbon is 7.88 g and molar mass of carbon is 12.0 g/mol, the number of moles of carbon can be calculated as follows:

  $n_C=\frac{m}{M}$

Putting the values,

  $n_C=\frac{7.88\text{ g}}{12.0\text{ g/mol}}=0.656\text{ mol}$

Similarly, mass of oxygen is 10.5 g and molar mass of oxygen is 16.0 g/mol, the number of moles of oxygen can be calculated as follows:

  $n_O=\frac{10.5\text{ g}}{16.0\text{ g/mol}}=0.656\text{ mol}$

Now, the ratio of number of moles of C and O will be:

  $\frac{n_C}{n_O}=\frac{0.656}{0.656}=1$

Thus, the ratio is 1:1.

Now, the mole ratios are same for the two given samples of carbon monoxide. So, the results are consistent with law of definite proportions.

Conclusion

The decomposition of both the samples of carbon monoxide is consistent with the law of definite proportions.