Percent Composition by Mass
Chapter 2
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Overview
The mass fraction (w) of a substance within a mixture or compound is the unitless ratio of the mass of the substance (m) and the total mass of the mixture or compound (mtot) given as
\[w(X) = \dfrac{m(X)}{m_{\mathrm{tot}}}\]
The percent composition (by mass) (alternatively, ‘% by mass’ or ‘mass %’ given as x(X)% ) of a substance in a mixture or compound is the mass fraction of each substance in a compound times 100%.
\[w(X)\% = w(X) \times 100~\%\]
Example 1: Percent by mass of H and O in H2O2
Determine the percent composition by mass of hydrogen and in hydrogen peroxide (H2O2) (to one decimal place).
\[\begin{align*} w(\mathrm{H})\% &= w(\mathrm{H}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{m(\mathrm{H})}{m(\mathrm{H_2O_2})}\right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{2~\!A_{\mathrm{r}}^{\circ}(\mathrm{H})}{A_{\mathrm{r}}^{\circ}(\mathrm{H_2O_2})} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{2 \times 1.01}{34.02} \right ) \times 100~\% \\[1.5ex] &= 5.\bar{9}376~\% \\[1.5ex] &= 5.9~\% \\[3.0ex] \end{align*}\]
\[\begin{align*} w(\mathrm{O})\% &= w(\mathrm{O}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{m(\mathrm{O})}{m(\mathrm{H_2O_2})}\right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{2~\!A_{\mathrm{r}}^{\circ}(\mathrm{H})}{A_{\mathrm{r}}^{\circ}(\mathrm{H_2O_2})} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{2 \times 16.00}{34.02} \right ) \times 100~\% \\[1.5ex] &= 94.\bar{0}62~\%\\[1.5ex] &= 94.1~\% \\[1.5ex] \textrm{or}\\[1.5ex] w(\mathrm{O})\% &= 100~\% - w(\mathrm{H})\% \\[0.5ex] &= 100~\% - 5.\bar{9}376~\% \\[0.5ex] &= 94.\bar{0}62~\% \\[0.5ex] &= 94.1~\% \end{align*}\]
Example 2: Chemical formula from % by mass
We can determine a chemical formula from a compound by considering the % by mass of the elements in that compound. Consider the following prompt:
“A 352 g compound containing chromium and oxygen decomposes completely to give 240.84 g of Cr and 111.16 g O. Using % by mass, determine if the compound is Cr2O3.”
To solve this, we must:
- Determine the % by mass of Cr and O in the stated sample
- Determine the % by mass of Cr and O in Cr2O3
- Compare (1) and (2). If they match, the compound is Cr2O3. If they do not reasonably match, the compound is not Cr2O3.
Step 1: Determine the % by mass of Cr and O in the 352 g sample.
\[\begin{align*} w(\mathrm{Cr})\% &= w(\mathrm{Cr}) \times 100~\% \\[1.5ex] &= \dfrac{m(\textrm{Cr})}{m_{\textrm{tot}}} \times 100~\% \\[1.5ex] &= \dfrac{240.84~\mathrm{g}}{352~\mathrm{g}} \times 100~\% \\[1.5ex] &= 68.\bar{4}20~\% \end{align*}\]
\[\begin{align*} w(\mathrm{O})\% &= w(\mathrm{O}) \times 100~\% \\[1.5ex] &= \dfrac{m(\textrm{O})}{m_{\textrm{tot}}} \times 100~\% \\[1.5ex] &= \dfrac{111.16~\mathrm{g}}{352~\mathrm{g}} \times 100~\% \\[1.5ex] &= 31.\bar{5}79~\% \end{align*}\]
Step 2: Determine the % by mass of Cr and O in Cr2O3.
\[\begin{align*} w(\mathrm{Cr})\% &= w(\mathrm{Cr}) \times 100~\% \\[1.5ex] &= \dfrac{2~\!A_{\textrm{r}}^{\circ}(\textrm{Cr})} {A_{\textrm{r}}^{\circ}(\mathrm{Cr_2O_3})} \times 100~\% \\[1.5ex] &= \dfrac{\left ( 2 \times 52.00 \right )} {152.00} \times 100~\% \\[1.5ex] &= 68.4\bar{2}10~\% \end{align*}\]
\[\begin{align*} w(\mathrm{O})\% &= w(\mathrm{O}) \times 100~\% \\[1.5ex] &= \dfrac{3~\!A_{\textrm{r}}^{\circ}(\textrm{O})} {A_{\textrm{r}}^{\circ}(\mathrm{Cr_2O_3})} \times 100~\% \\[1.5ex] &= \dfrac{\left ( 3 \times 16.00 \right )} {152.00} \times 100~\% \\[1.5ex] &= 31.5\bar{7}89~\% \end{align*}\]
Step 3: Compare the % by mass in the sample to the compound
\[\begin{align*} w(\mathrm{Cr})\% ~ \textrm{in sample} = 68.42 \% \\[1.5ex] w(\mathrm{Cr})\% ~ \textrm{in compound} = 68.42 \% \\[3ex] w(\mathrm{O})\% ~ \textrm{in sample} = 31.58 \% \\[1.5ex] w(\mathrm{O})\% ~ \textrm{in compound} = 31.58 \% \end{align*}\]
Because these values reasonably match, the compound is Cr2O3.
Empirical and Molecular Formulas
Reversing the process considered in determining percent composition by mass leads us to determining the empirical or molecular formula for a compound by starting with the percent composition quantities.
Example: CO2
The percent composition by mass of carbon and oxygen in carbon dioxide (M = 44.01 g mol–1) is 27.29 % and 72.71 %, respectively. Assume an exact 100 g sample of CO2 and determine the molecular formula for carbon dioxide.
Convert mass percent to mass fraction
Divide the mass % of each substance by 100 %.
\[\begin{align*} w(\mathrm{C}) &= \dfrac{w(\mathrm{C})\%}{100~\%} \\[1.5ex] &= \dfrac{27.29~\%}{100~\%} \\[1.5ex] &= 0.2728 \end{align*}\]
\[\begin{align*} w(\mathrm{O}) &= \dfrac{w(\mathrm{O})\%}{100~\%} \\[1.5ex] &= \dfrac{72.71~\%}{100~\%} \\[1.5ex] &= 0.7271 \end{align*}\]
Determine the mass of each substance
The problem states an exact 100 g sample of CO2. Multiply the mass fraction of each substance by the total mass of the sample.
\[\begin{align*} m(\mathrm{C}) &= w(\mathrm{C}) ~ m(\mathrm{CO_2}) \\[1.5ex] &= 0.2728 \left ( 100~\mathrm{g} \right ) \\[1.5ex] &= 27.28~\mathrm{g} \end{align*}\]
\[\begin{align*} m(\mathrm{O}) &= w(\mathrm{O}) ~ m(\mathrm{CO_2}) \\[1.5ex] &= 0.7271 \left ( 100~\mathrm{g} \right ) \\[1.5ex] &= 72.71~\mathrm{g} \end{align*}\]
Convert mass of each substance to moles
Recall from earlier that to convert mass to moles, divide mass by the molar mass.
\[\begin{align*} n(\mathrm{C}) &= m(\mathrm{C}) ~ M(\mathrm{C})^{-1} \\[1.5ex] &= 27.28~\mathrm{g} \left ( \dfrac{1~\mathrm{mol}}{12.01~\mathrm{g}} \right ) \\[1.5ex] &= 2.27\bar{1}4~\mathrm{mol} \end{align*}\]
\[\begin{align*} n(\mathrm{O}) &= m(\mathrm{O}) ~ M(\mathrm{O})^{-1} \\[1.5ex] &= 72.71~\mathrm{g} \left ( \dfrac{1~\mathrm{mol}}{16.00~\mathrm{g}} \right ) \\[1.5ex] &= 4.54\bar{4}3~\mathrm{mol} \end{align*}\]
Determine mole ratio of substances
Take the mole ratio of carbon and oxygen (in any order). The ratio should be very close to clean, whole numbers.
\[r(\mathrm{C, O}) = \dfrac{2.27\bar{1}4~\mathrm{mol~C}}{4.54\bar{4}3~\mathrm{mol~O}} \approx \dfrac{1~\mathrm{mol~C}}{2~\mathrm{mol~O}}\]
We read this as 1 mole of carbon for every 2 moles of oxygen.
Determine the molecular formula
Use the mole ratio to determine the molecular formula for carbon dioxide.
\[\mathrm{C_1O_2} \rightarrow \mathrm{CO_2}\]
Determine the empirical formula
The empirical formula for a compound is the simplest, whole-number atom ratio of atoms in a formula. It conveys only the stoichiometric proportion of a compound. Therefore, in the example above, the empirical formula for CO2 is identical to its molecular formula.
Consider the following series of molecules: glucose (C6H12O6), acetic acid (C2H4O2), and formaldehyde (CH2O). While these three molecules are inherently different, they all have the same empirical formula of CH2O. This empirical formula tells us that these molecules have the same ratio of carbon atom to two hydrogen atoms to one oxygen atom.
Practice
What is the empirical formula for the following molecules?
- C2H6
- C6H14
- C6H12O6
Solution
- CH3
- C3H7
- CH2O
Practice
Determine the molecular formula for a compound composed of, by mass, 48.64 % C, 8.16 % H, and 43.20 % O.
Solution
Assume an exact 100 g sample so the percent quantities for each element become a quantity in grams.
\[\begin{align*} m(\mathrm{C}) &= 48.64~\mathrm{g} \\[1.5ex] m(\mathrm{H}) &= 8.16~\mathrm{g} \\[1.5ex] m(\mathrm{O}) &= 43.20~\mathrm{g} \end{align*}\]
Find the number of moles of each element by dividing by the element’s molar mass.
\[\begin{align*} n(\mathrm{C}) &= m(\mathrm{C}) ~ M(\mathrm{C})^{-1} \\[1.5ex] &= \mathrm{48.64~\mathrm{g}} \left ( \dfrac{\mathrm{mol}}{12.01~\mathrm{g}} \right ) \\[1.5ex] &= 4.04\bar{9}9~\mathrm{mol} \\[3ex] n(\mathrm{H}) &= m(\mathrm{H}) ~ M(\mathrm{C})^{-1}\\[1.5ex] &= \mathrm{8.16~\mathrm{g}} \left ( \dfrac{\mathrm{mol}}{1.01~\mathrm{g}} \right ) \\[1.5ex] &= 8.07\bar{9}2~\mathrm{mol} \\[3ex] n(\mathrm{O}) &= m(\mathrm{O}) ~ M(\mathrm{C})^{-1}\\[1.5ex] &= \mathrm{43.20~\mathrm{g}} \left ( \dfrac{\mathrm{mol}}{16.00~\mathrm{g}} \right ) \\[1.5ex] &= 2.70\bar{0}0~\mathrm{mol} \end{align*}\]
Write out the molecular formula and divide each subscript by the smallest number and multiply by an integer to get whole number ratios.
\[ \begin{equation} \begin{gathered} \mathrm{C}_{4.04\bar{9}9}\mathrm{H}_{8.07\bar{9}2}\mathrm{O}_{2.70\bar{0}0} \\[2ex] \mathrm{C}_{\frac{4.04\bar{9}9}{2.70\bar{0}0}}\mathrm{H}_{\frac{8.07\bar{9}2}{2.70\bar{0}0}}\mathrm{O}_{\frac{{2.70\bar{0}0}}{2.70\bar{0}0}} \\[2ex] \mathrm{C}_{1.5}\mathrm{H}_{3}\mathrm{O}_{1} \\[2ex] \mathrm{C}_{1.5\times 2}\mathrm{H}_{3\times 2}\mathrm{O}_{1\times 2} \\[4ex] \mathrm{C}_{3}\mathrm{H}_{6}\mathrm{O}_{2} \end{gathered} \end{equation} \]
Here we multiply the subscripts by 2 to get whole numbers.
Practice
What is the molar mass of a 13.43 mol sample of unknown compound if the mass of the sample was 1.278 68 kg?
Solution
Molar mass is defined as being the mass (in g) of a substance per 1 mole of that substance.
\[\begin{align*} M(\mathrm{unknown}) &= \dfrac{m(\mathrm{unknown})}{n(\mathrm{unknown})} \\[1.5ex] &= \dfrac{1.278~68~\mathrm{kg}}{13.43~\mathrm{mol}} \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \\[1.5ex] &= 95.21~\mathrm{g~mol^{-1}} \\[1.5ex] &= 95.21~\mathrm{g~mol^{-1}} \end{align*}\]
Extend
The sample was further analyzed and found to have the following formula, MCl2, where M is a metal. What is the identity of the metal?
Solution
The molar mass of the unknown is 95.21 g mol–1. The molar mass of two chlorine atoms is 70.9 g. Find the remaining molar mass to determine the metal.
\[\begin{align*} M(\mathrm{M}) &= M(\mathrm{unknown}) - 2~\!M(\mathrm{Cl}) \\[1.5ex] &= 95.21~\mathrm{g~mol^{-1}} - 2(35.45~\mathrm{g~mol^{-1}})\\[1.5ex] &= 24.31~\mathrm{g~mol^{-1}} \end{align*}\]
Finding the molar mass on the periodic table reveals that the metal is magnesium.