Chapter 10.3, Problem 38SP

a.

Interpretation Introduction

Interpretation: The mass of hydrogen in 350 g 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 38SP

The mass of hydrogen in 350 g of ethane is $\text{7}0.0\text{ g H}$ .

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 $\text{H}$ using relation (1):

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

The conversion factor to convert the mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ to mass of $\text{H}$ is:

  $\frac{20.0\text{ g H}}{100{\text{ g C}}_{\text{2}}{\text{H}}_{\text{6}}}$

Now, to calculate the mass of $\text{H}$ in 350 g of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$

, the required conversion factor is:

  $\frac{20.0\text{ g H}}{100{\text{ g C}}_{\text{2}}{\text{H}}_{\text{6}}}\times 350{\text{ g C}}_{\text{2}}{\text{H}}_{\text{6}}\text{ = 7}0.0\text{ g H}$

Hence, the mass of hydrogen in 350 g of ethane is $\text{7}0.0\text{ g H}$ .

b.

Interpretation Introduction

Interpretation: The mass of hydrogen in 20.2 g 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 38SP

The mass of hydrogen in 20.2 g of sodium hydrogen sulfate is $\text{ 0}\text{.17 g H}$ .

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 $\text{H}$ using relation (1):

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

The conversion factor to convert the mass of ${\text{NaHSO}}_{\text{4}}$ to mass of $\text{H}$ is:

  $\frac{0.83\text{ g H}}{100{\text{ g NaHSO}}_{\text{4}}}$

Now, to calculate the mass of $\text{H}$ in 20.2 g of ${\text{NaHSO}}_{\text{4}}$ , the required conversion factor is:

  $\frac{0.83\text{ g H}}{100{\text{ g NaHSO}}_{\text{4}}}\times 20.2{\text{ g NaHSO}}_{\text{4}}\text{ = 0}\text{.17 g H}$

Hence, the mass of hydrogen in 20.2 g of sodium hydrogen sulfate is $\text{ 0}\text{.17 g H}$ .