Skip to main content
Chemistry LibreTexts

6.7: Using Chemical Formulas as Conversion Factors

  • Page ID
    289387
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    ⚙️ Learning Objectives

    • Use chemical formulas as conversion factors.


    Figure \(\PageIndex{1}\) shows that 2 hydrogen atoms and 1 oxygen atom are required to make one water molecule, 4 hydrogen atoms and 2 oxygen atoms for two water molecules, and 6 hydrogen atoms and 3 oxygen atoms for three water molecules. To make any number of water molecules, the ratio is always the same: 2 hydrogen atoms to 1 oxygen atom.

    Figure \(\PageIndex{1}\): The ratio of hydrogen atoms to oxygen atoms used to make water molecules, H2O is always 2:1, regardless of the number of water molecules.


    By using formulas to indicate how many atoms of each element we have in a substance, we can relate the number of molecules to the number of atoms of each element. The possible relationships for 1 mole of water are shown in last column of Table \(\PageIndex{1}\).
     

    Table \(\PageIndex{1}\): Molecular Relationships for Water
    1 molecule H2O has: 1 dozen H2O molecules has: 1 mol H2O has: Mole Relationships
    2 H atoms 2 dozen H atoms 2 mol H \(\dfrac{2\:\mathrm m\mathrm o\mathrm l\:\mathrm H}{1\:\mathrm m\mathrm o\mathrm l\:{\mathrm H}_2\mathrm O}\) or \(\dfrac{1\:\mathrm m\mathrm o\mathrm l\:{\mathrm H}_2\mathrm O}{2\:\mathrm m\mathrm o\mathrm l\:\mathrm H}\)
    1 O atom 1 dozen O atoms 1 mol O \(\dfrac{1\:\mathrm m\mathrm o\mathrm l\:\mathrm O}{1\:\mathrm m\mathrm o\mathrm l\:{\mathrm H}_2\mathrm O}\) or \(\dfrac{1\:\mathrm m\mathrm o\mathrm l\:{\mathrm H}_2\mathrm O}{1\:\mathrm m\mathrm o\mathrm l\:\mathrm O}\)


    Therefore, with 30.2 mol H2O, there would be:

    \(30.2\;\cancel{\mathrm{mol}\;{\mathrm H}_2\mathrm O}\times\dfrac{2\;\mathrm{mol}\;\mathrm H}{1\;\cancel{\mathrm{mol}\;{\mathrm H}_2\mathrm O}}=\boxed{\;60.4\;\mathrm{mol}\;\mathrm H}\)
         

    \(30.2\;\cancel{\mathrm{mol}\;{\mathrm H}_2\mathrm O}\times\dfrac{1\;\mathrm{mol}\;\mathrm O}{1\;\cancel{\mathrm{mol}\;{\mathrm H}_2\mathrm O}}=\;\boxed{30.2\;\mathrm{mol}\;\mathrm O}\)

    The following example illustrates how we can use the relationships in Table \(\PageIndex{1}\) as conversion factors.
     

    ✅ Example \(\PageIndex{1}\): Ethanol

    How many moles of hydrogen atoms are present in 2.5 mol of ethanol (C2H6O)?

    Solution

    Steps for Problem Solving  

    Identify the "given" information and what the problem is asking you to "find."
     

    Given: 2.5 mol C2H6O

    Find: mol H atoms
     

    List known relationships.
     
    1 mol C2H6O = 6 mol H

    Prepare a concept map and use the proper conversion factor.
     

    \({\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\mathrm{mol}\;{\mathrm C}_2{\mathrm H}_6\mathrm O\;}}\xrightarrow[{1\;\mathrm{mol}\;{\mathrm C}_2{\mathrm H}_6\mathrm O}]{6\;\mathrm{mol}\;\mathrm H}{\color[rgb]{0.0, 0.0, 1.0}\boxed{\;\;\;\;\mathrm{mol}\;\mathrm H\;\;\;\;}}\)
     

    Calculate the answer.
     

    \(2.5\:\cancel{\mathrm{mol}\:{\mathrm C}_2{\mathrm H}_6\mathrm O}\times\dfrac{6\:\mathrm{mol}\:\mathrm H}{1\:\cancel{\mathrm{mol}\:{\mathrm C}_2{\mathrm H}_6\mathrm O}}=\boxed{15\:\mathrm{mol}\:\mathrm H}\)
     

    Think about your result.
     
    There 6 H atoms per C2H6O molecule, so the final answer should be 6 times as large.

     

    ✏️ Exercise \(\PageIndex{1}\)

    How many moles of sodium, sulfur, and oxygen atoms are in a sample containing 6.75 mol of Na2SO4, sodium sulfate?

    Answer
    13.5 mol Na atoms, 6.75 mol S atoms, and 27.0 mol O atoms 


    Summary

    • In any given formula, the ratio of the number of moles of molecules (or formula units) to the number of moles of atoms can be used as a conversion factor. 

     

     


    This page is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Lance S. Lund (Anoka-Ramsey Community College), Anonymous, Marisa Alviar-Agnew, and Henry Agnew. 


    6.7: Using Chemical Formulas as Conversion Factors is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?