Chapter 10.3, Problem 36SP

Interpretation Introduction

Interpretation: The percent composition of ethane should be calculated.

Concept Introduction: The percent composition represents the relative amount of the elements in a compound. It is represented as the percent by mass of each element in the compound.

Answer to Problem 36SP

The percent composition of C is 80.0% and H is 20.0% in ethane.

Explanation of Solution

The percent by mass of each element in the compound represents the percent composition as:

  $\text{percent by mass of an element = }\frac{\text{mass of an element}}{\text{mass of the compound}}\times 100\%$

The percent composition can be determined by the molecular formula of the compound as well as using the relation:

  $\text{percent by mass of an element = }\frac{\text{mass of an element in 1 mol compound}}{\text{molar mass of the compound}}\times 100\%$ …. (1)

The molar mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$

is determined as:

The moles of C and H are converted to grams of C and H as:

  $\begin{array}{l}\text{2 mol C }\times \frac{12.0\text{ g C}}{1\text{ mol C}}\text{ = 24}\text{.0 g C}\\ \text{6 mol H }\times \frac{\text{1}\text{.0 g H}}{1\text{ mol H}}\text{ = 6}\text{.0 g H}\end{array}$

The mass of 1 mol of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ is:

  $\begin{array}{l}{\text{mass of 1 mol of C}}_{\text{2}}{\text{H}}_{\text{6}}\text{ = 24}\text{.0 g N + 6}\text{.0 g H}=\text{ 30}\text{.0 g}\\ {\text{molar mass of C}}_{\text{2}}{\text{H}}_{\text{6}}=30.\text{0 g/mol}\end{array}$

Now, determining the percent composition of $\text{C}$ using relation (1):

  $\begin{array}{l}\text{\%C = }\frac{{\text{mass of C in 1 mol of C}}_{\text{2}}{\text{H}}_{\text{6}}}{{\text{molar mass of C}}_{\text{2}}{\text{H}}_{\text{6}}}\times 100\%\\ \text{\%C}=\frac{\text{24}\text{.0 g}}{\text{30}\text{.0 g}}\times 100\%\\ \text{\%C = }80.0\%\end{array}$

Now, determining the percent composition of $\text{H}$ using relation (1):

  $\begin{array}{l}\text{\%H = }\frac{{\text{mass of H in 1 mol of C}}_{\text{2}}{\text{H}}_{\text{6}}}{{\text{molar mass of C}}_{\text{2}}{\text{H}}_{\text{6}}}\times 100\%\\ \text{\%H}=\frac{\text{6}\text{.0 g}}{\text{30}\text{.0 g}}\times 100\%\\ \text{\%H = }20.0\%\end{array}$

Hence, the percent composition of C is 80.0% and H is 20.0% in ethane.

Interpretation Introduction

Interpretation: The percent composition of sodium hydrogen sulfate should be calculated.

Concept Introduction: The percent composition represents the relative amount of the elements in a compound. It is represented as the percent by mass of each element in the compound.

Answer to Problem 36SP

The percent composition of Na is 19.2.0%, H is 0.83%, S is 26.7% and O is 53.3% in sodium hydrogen sulfate.

Explanation of Solution

The percent by mass of each element in the compound represents the percent composition as:

  $\text{percent by mass of an element = }\frac{\text{mass of an element}}{\text{mass of the compound}}\times 100\%$

The percent composition can be determined by the molecular formula of the compound as well as using the relation:

  $\text{percent by mass of an element = }\frac{\text{mass of an element in 1 mol compound}}{\text{molar mass of the compound}}\times 100\%$ …. (1)

The molar mass of ${\text{NaHSO}}_{\text{4}}$

is determined as:

The moles of Na, H, S, and O are converted to grams of Na, H, S, and O as:

  $\begin{array}{l}\text{1 mol Na }\times \frac{23.0\text{ g Na}}{1\text{ mol Na}}\text{ = 23}\text{.0 g Na}\\ \text{1 mol H }\times \frac{\text{1}\text{.0 g H}}{1\text{ mol H}}\text{ = 1}\text{.0 g H}\\ \text{1 mol S }\times \frac{\text{32}\text{.0 g S}}{1\text{ mol S}}\text{ = 32}\text{.0 g S}\\ \text{4 mol O }\times \frac{\text{16}\text{.0 g O}}{1\text{ mol O}}\text{ = 64}\text{.0 g O}\end{array}$

The mass of 1 mol of ${\text{NaHSO}}_{\text{4}}$ is:

  $\begin{array}{l}{\text{mass of 1 mol of NaHSO}}_{\text{4}}\text{ = 23}\text{.0 g Na + 1}\text{.0 g H + 32}\text{.0 g S + 64}\text{.0 g O}=\text{ 120}\text{.0 g}\\ {\text{molar mass of NaHSO}}_{\text{4}}=120.0\text{ g/mol}\end{array}$

Now, determining the percent composition of $\text{Na}$ using relation (1):

  $\begin{array}{l}\text{\%Na = }\frac{{\text{mass of Na in 1 mol of NaHSO}}_{\text{4}}}{{\text{molar mass of NaHSO}}_{\text{4}}}\times 100\%\\ \text{\%Na}=\frac{\text{23}\text{.0 g}}{\text{120}\text{.0 g}}\times 100\%\\ \text{\%Na = }19.2\%\end{array}$

Now, determining the percent composition of $\text{H}$ using relation (1):

  $\begin{array}{l}\text{\%H = }\frac{{\text{mass of H in 1 mol of NaHSO}}_{\text{4}}}{{\text{molar mass of NaHSO}}_{\text{4}}}\times 100\%\\ \text{\%H}=\frac{\text{1}\text{.0 g}}{\text{120}\text{.0 g}}\times 100\%\\ \text{\%H = }0.83\%\end{array}$

Now, determining the percent composition of $\text{S}$ using relation (1):

  $\begin{array}{l}\text{\%S = }\frac{{\text{mass of S in 1 mol of NaHSO}}_{\text{4}}}{{\text{molar mass of NaHSO}}_{\text{4}}}\times 100\%\\ \text{\%S}=\frac{\text{32}\text{.0 g}}{\text{120}\text{.0 g}}\times 100\%\\ \text{\%S = 26}\text{.7}\%\end{array}$

Now, determining the percent composition of $\text{O}$ using relation (1):

  $\begin{array}{l}\text{\%O= }\frac{{\text{mass of O in 1 mol of NaHSO}}_{\text{4}}}{{\text{molar mass of NaHSO}}_{\text{4}}}\times 100\%\\ \text{\%O}=\frac{\text{64}\text{.0 g}}{\text{120}\text{.0 g}}\times 100\%\\ \text{\%O = 53}\text{.3}\%\end{array}$

Hence, the percent composition of Na is 19.2.0%, H is 0.83%, S is 26.7% and O is 53.3% in sodium hydrogen sulfate.