Chapter 3, Problem 72A

(a)

Interpretation Introduction

Interpretation: The mass percent of carbon in 44 g of carbon dioxide needs to be determined.

Concept Introduction: For any compound, the mass percent of any element present in it can be calculated as follows:

  $\%m=\frac{m_{element}}{m_{compound}}\times 100\%$

Answer to Problem 72A

  27.29%

Explanation of Solution

The given mass of carbon dioxide is 44 g. The molar mass of carbon dioxide is 44.01 g/mol. The number of moles can be calculated as follows:

  $n=\frac{m}{M}$

Putting the values,

  $n=\frac{44\text{ g}}{44.01\text{ g/mol}}=0.99=1.0\text{ mol}$

From the formula of carbon dioxide, 1 mol of carbon dioxide contains 1 mol of C thus, the number of moles of C will be 1 mol.

The molar mass of C is 12.01 g/mol thus, the mass of C atom will be:

  $m=n\times M=1\text{ mol}\times 12.01\text{ g/mol=12}\text{.01 g}$

Now, the mass percent of C can be calculated as follows:

  $\%m=\frac{m_{element}}{m_{compound}}\times 100\%$

Putting the values,

  $\%m=\frac{12.01\text{ g}}{44.0\text{ g}}\times 100\%=27.29\%$

(b)

Interpretation Introduction

Interpretation: The mass percent of oxygen in 44 g of carbon dioxide needs to be determined.

Concept Introduction: For any compound, the mass percent of any element present in it can be calculated as follows:

  $\%m=\frac{m_{element}}{m_{compound}}\times 100\%$

Answer to Problem 72A

  72.68%

Explanation of Solution

The given mass of carbon dioxide is 44 g. The molar mass of carbon dioxide is 44.01 g/mol. The number of moles can be calculated as follows:

  $n=\frac{m}{M}$

Putting the values,

  $n=\frac{44\text{ g}}{44.01\text{ g/mol}}=0.99=1.0\text{ mol}$

From the formula of carbon dioxide, 1 mol of carbon dioxide contains 2 mol of O thus, the number of moles of O will be 2 mol.

The molar mass of O is 15.99 g/mol thus, the mass of O atom will be:

  $m=n\times M=2\text{ mol}\times 15.99\text{ g/mol=31}\text{.98 g}$

Now, the mass percent of O can be calculated as follows:

  $\%m=\frac{m_{element}}{m_{compound}}\times 100\%$

Putting the values,

  $\%m=\frac{31.98\text{ g}}{44.0\text{ g}}\times 100\%=72.68\%$