Chapter 10.3, Problem 48LC

Interpretation Introduction

Interpretation: The grams of hydrogen in 124 g of calcium acetate, ${\text{Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}$ 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 48LC

The grams of hydrogen in 124 g of ${\text{Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}$ is $\text{4}\text{.7 g H}$ .

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 ${\text{Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}$ is determined as:

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

  $\begin{array}{l}\text{1 mol Ca }\times \frac{40.0\text{ g Ca}}{1\text{ mol Ca}}\text{ = 40}\text{.0 g Ca}\\ \text{4 mol C }\times \frac{\text{12}\text{.0 g C}}{1\text{ mol C}}\text{ = 48}\text{.0 g C}\\ \text{6 mol H }\times \frac{\text{1}\text{.0 g H}}{1\text{ mol H}}\text{ = 6}\text{.0 g H}\\ \text{4 mol O }\times \frac{\text{16}\text{.0 g O}}{1\text{ mol O}}\text{ = 64}\text{.0 g O}\end{array}$

The mass of 1 mol of ${\text{Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}$ is:

  $\begin{array}{l}{\text{mass of 1 mol of Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}\text{ = 40}\text{.0 g Ca + 48}\text{.0 g C + 6}\text{.0 g H + 64}\text{.0 g O}=\text{ 158}\text{.0 g}\\ {\text{molar mass of Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}=158.\text{0 g/mol}\end{array}$

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

  $\begin{array}{l}\text{\%H = }\frac{{\text{mass of H in 1 mol of Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}}{{\text{molar mass of Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}}\times 100\%\\ \text{\%H}=\frac{\text{6}\text{.0 g}}{\text{158}\text{.0 g}}\times 100\%\\ \text{\%H = 3}\text{.8}\%\end{array}$

The conversion factor to convert the mass of ${\text{Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}$ to mass of $\text{H}$ is:

  $\frac{\text{3}\text{.8 g H}}{100{\text{ g Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}}$

Now, to calculate the mass of $\text{H}$ in 124 g of ${\text{Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}$ , the required conversion factor is:

  $\frac{\text{3}\text{.8 g H}}{100{\text{ g Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}}{\text{×124 g Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}\text{ = 4}\text{.7 g H}$

Hence, the grams of hydrogen in 124 g of ${\text{Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}$ is $\text{4}\text{.7 g H}$ .