Question

Express the composition of each of the following compounds as the mass percent of itselements.
a. formaldehyde, CH₂O
b. glucose, C₆H₁₂O₆
c. acetic acid, HC₂H₃O₂
Considering your answers, which type of formula-empirical or molecular-can be obtained from elemental analysis that gives mass percent composition? Explain.

Short answer

The type of empirical or molecular formula can be same (CH₂O) but have different molecular formula. The percentage composition can be same for these compounds.

Step-by-step solution

Finding percentage composition of formaldehyde

The molar mass of formaldehyde can be denoted as,

$$\begin{gathered} =1\mathrm{mol}\mathrm{C}\left(\frac{12.01\mathrm{g}}{1\mathrm{mol}\mathrm{C}}\right)+2\mathrm{mol}\mathrm{H}\left(\frac{1.008\mathrm{g}}{1\mathrm{mol}\mathrm{H}}\right)+1\mathrm{mol}\mathrm{O}\left(\frac{16\mathrm{g}}{1\mathrm{mol}\mathrm{O}}\right) \\ =30.03\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right. \end{gathered}$$

To calculate the percentage composition, mass percentage of carbon, hydrogen and oxygen.

$\begin{array}{rcl}\mathrm{Percent}\mathrm{C}& =& \frac{12.01\mathrm{gH}}{30.03{\mathrm{gCH}}_2\mathrm{O}}\times 100\mathrm{percent}\\ & =& 40.002\mathrm{PercentC}\\ \mathrm{Percent}\mathrm{H}& =& \frac{2.016\mathrm{gC}}{30.03{\mathrm{gCH}}_2\mathrm{O}}\times 100\mathrm{percent}\\ & =& 6.1735\mathrm{PercentH}\\ \mathrm{Percent}\mathrm{O}& =& \frac{16\mathrm{gC}}{30.03{\mathrm{gCH}}_2\mathrm{O}}\times 100\mathrm{percent}\\ & =& 53.28\mathrm{PercentO}\\ & & \end{array}$

The composition of CH₂O is$40.002\mathrm{PercentC},6.7135\mathrm{PercentH}\mathrm{and}53.28\mathrm{percentO}$

Finding percentage composition of glucose

The molar mass can be denoted as,

$$\begin{gathered} =6\mathrm{mol}\mathrm{C}\left(\frac{12.01\mathrm{g}}{1\mathrm{mol}\mathrm{C}}\right)+12\mathrm{mol}\mathrm{H}\left(\frac{1.008\mathrm{g}}{1\mathrm{mol}\mathrm{H}}\right)+6\mathrm{mol}\mathrm{O}\left(\frac{16\mathrm{g}}{1\mathrm{mol}\mathrm{O}}\right) \\ =180.2\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right. \end{gathered}$$

To calculate the percentage composition, mass percentage of carbon, hydrogen and oxygen.

$\begin{array}{rcl}\mathrm{Percent}\mathrm{C}& =& \frac{12.01\mathrm{gH}}{180.155{\mathrm{gCH}}_2\mathrm{O}}\times 100\mathrm{percent}\\ & =& 40.002\mathrm{PercentC}\\ \mathrm{Percent}\mathrm{H}& =& \frac{2.016\mathrm{gC}}{180.155{\mathrm{gCH}}_2\mathrm{O}}\times 100\mathrm{percent}\\ & =& 6.1735\mathrm{PercentH}\\ \mathrm{Percent}\mathrm{O}& =& \frac{16\mathrm{gC}}{180.155{\mathrm{gCH}}_2\mathrm{O}}\times 100\mathrm{percent}\\ & =& 53.28\mathrm{PercentO}\end{array}$

The composition of C₆H₁₂O₆ is $40.002\mathrm{PercentC},6.7135\mathrm{PercentH}\mathrm{and}53.28\mathrm{percentO}$

Finding percentage composition of acetic acid

The molar mass can be denoted as,

$$\begin{gathered} =2\mathrm{mol}\mathrm{C}\left(\frac{12.01\mathrm{g}}{1\mathrm{mol}\mathrm{C}}\right)+4\mathrm{mol}\mathrm{H}\left(\frac{1.008\mathrm{g}}{1\mathrm{mol}\mathrm{H}}\right)+2\mathrm{mol}\mathrm{O}\left(\frac{16\mathrm{g}}{1\mathrm{mol}\mathrm{O}}\right) \\ =60.052\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right. \end{gathered}$$

To calculate the percentage composition, mass percentage of carbon, hydrogen and oxygen.

$\begin{array}{rcl}\mathrm{Percent}\mathrm{C}& =& \frac{12.01\mathrm{gH}}{60.052{\mathrm{gCH}}_2\mathrm{O}}\times 100\mathrm{percent}\\ & =& 40.002\mathrm{PercentC}\\ \mathrm{Percent}\mathrm{H}& =& \frac{2.016\mathrm{gC}}{60.052{\mathrm{gCH}}_2\mathrm{O}}\times 100\mathrm{percent}\\ & =& 6.1735\mathrm{PercentH}\\ \mathrm{Percent}\mathrm{O}& =& \frac{16\mathrm{gC}}{60.052{\mathrm{gCH}}_2\mathrm{O}}\times 100\mathrm{percent}\\ & =& 53.28\mathrm{PercentO}\end{array}$

The composition of HC₂H₃O₂ is $40.002\mathrm{PercentC},6.7135\mathrm{PercentH}\mathrm{and}53.28\mathrm{percentO}$

The type of empirical or molecular formula can be same (CH₂O) but have different molecular formula. The percentage composition can be same for these compounds.