8.5: Mass Percent Composition from a Chemical Formula
- Page ID
- 451544
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Given a chemical formula calculate the percent composition of each component element.
The percent composition of a compound can also be determined from the formula of the compound. The subscripts in the formula are first used to calculate the mass of each element in one mole of the compound. This is divided by the molar mass of the compound and multiplied by \(100\%\).
\[\% \: \text{by mass} = \frac{\text{mass of element in} \: 1 \: \text{mol}}{\text{molar mass of compound}} \times 100\%\]
The percent composition of a given compound is always the same, given that the compound is pure.
Dichlorine heptoxide \(\left( \ce{Cl_2O_7} \right)\) is a highly reactive compound used in some organic synthesis reactions. Calculate the percent composition of dichlorine heptoxide.
Solution
Steps for Problem Solving | Calculate the percent composition of dichlorine heptoxide \(\left( \ce{Cl_2O_7} \right)\). |
---|---|
Identify the "given" information and what the problem is asking you to "find." |
Given : Cl2O7 Find: % Composition (% Cl and %O) |
List other known quantities. |
Mass of Cl in 1 mol Cl2O7 , 2 Cl : 2 x 35.45 g = 70.90 g Mass of O in 1 mol Cl2O7 , 7 O: 7 x 16.00 g = 112.00 g Molar mass of Cl2O7 = 182.90 g/mol |
Cancel units and calculate. |
\[\% \ce{Cl} = \frac{70.90 \: \text{g} \: \ce{Cl}}{182.90 \: \text{g}} \times 100\% = 38.76\% \: \ce{Cl} \nonumber\] \[\% \: \ce{O} = \frac{112.00 \: \text{g} \: \ce{O}}{182.90 \: \text{g}} \times 100\% = 61.24\% \: \ce{O} \nonumber\] Calculate the percent by mass of each element by dividing the mass of that element in 1 mole of the compound by the molar mass of the compound and multiplying by \(100\%\). |
Think about your result. | The percentages add up to \(100\%\). |
Percent composition can also be used to determine the mass of a certain element that is contained in any mass of a compound. In the previous sample problem, it was found that the percent composition of dichlorine heptoxide is \(38.76\% \: \ce{Cl}\) and \(61.24\% \: \ce{O}\). Suppose that you needed to know the masses of chlorine and oxygen present in a \(12.50 \: \text{g}\) sample of dichlorine heptoxide. You can set up a conversion factor based on the percent by mass of each element.
\[12.50 \: \text{g} \: \ce{Cl_2O_7} \times \frac{38.76 \: \text{g} \: \ce{Cl}}{100 \: \text{g} \: \ce{Cl_2O_7}} = 4.845 \: \text{g} \: \ce{Cl}\]
\[12.50 \: \text{g} \: \ce{Cl_2O_7} \times \frac{61.24 \: \text{g} \: \ce{O}}{100 \: \text{g} \: \ce{Cl_2O_7}} = 7.655 \: \text{g} \: \ce{O}\]
The sum of the two masses is \(12.50 \: \text{g}\), the mass of the sample size.
Barium fluoride is a transparent crystal that can be found in nature as the mineral frankdicksonite. Determine the percent composition of barium fluoride.
- Answer a:
- 78.32% Ba and 21.67% F
Contributions & Attributions
This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality:
Henry Agnew (UC Davis)