Extract numbers from chemical formula

This seems to work just fine:

Formula in B2 is below. Drag across and down

=IFERROR(IFERROR(--(MID($A2,SEARCH(B$1,$A2)+1,3)),IFERROR(--(MID($A2,SEARCH(B$1,$A2)+1,2)),--MID($A2,SEARCH(B$1,$A2)+1,1))),0)

Or a shorter array formula, which must be entered with ctrl+shift+enter

=MAX(IFERROR(--MID($A2,SEARCH(B$1,$A2)+1,ROW($A$1:$A$3)),0))

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If you wanted to keep the VBA super simple, something like this works as well:

Public Function ElementCount(str As String, element As String) As Long
    Dim i As Integer
    Dim s As String

    For i = 1 To 3
        s = Mid(str, InStr(str, element) + 1, i)
        On Error Resume Next
        ElementCount = CLng(s)
        On Error GoTo 0
    Next i
End Function

Use it like so:

=ElementCount(A1,"C")

Use Regular Expressions

This is a good task for regular expressions (regex). Because VBA doesn't support regular expressions out of the box we need to reference a Windows library first.

1. Add reference to regex under Tools then References

2. and selecting Microsoft VBScript Regular Expression 5.5

3. Add this function to a module

    Option Explicit 
   
    Public Function ChemRegex(ByVal ChemFormula As String, ByVal Element As String) As Long
        Dim strPattern As String
        strPattern = "([CNHO])([0-9]*)" 
                     'this pattern is limited to the elements C, N, H and O only.
        Dim regEx As New RegExp
   
        Dim Matches As MatchCollection, m As Match
   
        If strPattern <> "" Then
            With regEx
                .Global = True
                .MultiLine = True
                .IgnoreCase = False
                .Pattern = strPattern
            End With
   
            Set Matches = regEx.Execute(ChemFormula)
            For Each m In Matches
                If m.SubMatches(0) = Element Then
                    ChemRegex = IIf(Not m.SubMatches(1) = vbNullString, m.SubMatches(1), 1) 
                                'this IIF ensures that in CH4O the C and O are count as 1
                    Exit For
                End If
            Next m
        End If
    End Function
   

4. Use the function like this in a cell formula

   E.g. in cell B2: =ChemRegex($A2,B$1) and copy it to the other cells

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Recognize also chemical formulas with multiple occurrences of elements like CH₃OH or CH₂COOH

Note that the code above cannot count something like CH3OH where elements occur more than once. Then only the first H3 is count the last is omitted.

If you need also to recognize formulas in the format like CH3OH or CH2COOH (and summarize the occurrences of the elements) then you need to change the code to recognize these too …

If m.SubMatches(0) = Element Then
    ChemRegex = ChemRegex + IIf(Not m.SubMatches(1) = vbNullString, m.SubMatches(1), 1)
    'Exit For needs to be removed.
End If

Recognize also chemical formulas with 2 letter elements like NaOH or CaCl₂

In addition to the change above for multiple occurrences of elements use this pattern:

strPattern = "([A-Z][a-z]?)([0-9]*)"   'https://regex101.com/r/nNv8W6/2

1. Note that they need to be in the correct upper/lower letter case. CaCl2 works but not cacl2 or CACL2.

2. Note that this doesn't proof if these letter combinations are existing elements of the periodic table. So this will also recognize eg. Xx2Zz5Q as fictive elements Xx = 2, Zz = 5 and Q = 1.

   To accept only combinations that exist in the periodic table use the following pattern:

    strPattern = "([A][cglmrstu]|[B][aehikr]?|[C][adeflmnorsu]?|[D][bsy]|[E][rsu]|[F][elmr]?|[G][ade]|[H][efgos]?|[I][nr]?|[K][r]?|[L][airuv]|[M][cdgnot]|[N][abdehiop]?|[O][gs]?|[P][abdmortu]?|[R][abefghnu]|[S][bcegimnr]?|[T][abcehilms]|[U]|[V]|[W]|[X][e]|[Y][b]?|[Z][nr])([0-9]*)"
    'https://regex101.com/r/Hlzta2/3
    'This pattern includes all 118 elements up to today. 
    'If new elements are found/generated by scientist they need to be added to the pattern.
   

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Recognize also chemical formulas with prenthesis like Ca(OH)₂

Therefore another RegEx is needed to handle the parenthesis and multiply them.

Public Function ChemRegex(ByVal ChemFormula As String, ByVal Element As String) As Long
    Dim regEx As New RegExp
    With regEx
        .Global = True
        .MultiLine = True
        .IgnoreCase = False
    End With
    
    'first pattern matches every element once
    regEx.Pattern = "([A][cglmrstu]|[B][aehikr]?|[C][adeflmnorsu]?|[D][bsy]|[E][rsu]|[F][elmr]?|[G][ade]|[H][efgos]?|[I][nr]?|[K][r]?|[L][airuv]|[M][cdgnot]|[N][abdehiop]?|[O][gs]?|[P][abdmortu]?|[R][abefghnu]|[S][bcegimnr]?|[T][abcehilms]|[U]|[V]|[W]|[X][e]|[Y][b]?|[Z][nr])([0-9]*)"
    
    Dim Matches As MatchCollection
    Set Matches = regEx.Execute(ChemFormula)
    
    Dim m As Match
    For Each m In Matches
        If m.SubMatches(0) = Element Then
            ChemRegex = ChemRegex + IIf(Not m.SubMatches(1) = vbNullString, m.SubMatches(1), 1)
        End If
    Next m
    
    'second patternd finds parenthesis and multiplies elements within
    regEx.Pattern = "(\((.+?)\)([0-9]+)+)+?"
    Set Matches = regEx.Execute(ChemFormula)
    For Each m In Matches
        ChemRegex = ChemRegex + ChemRegex(m.SubMatches(1), Element) * (m.SubMatches(2) - 1) '-1 because all elements were already counted once in the first pattern
    Next m
End Function

This will also recognize parenthesis. Note that it does not recognize nested parenthesis.

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Also have a look at a similar question: Determine total number of atoms in a chemical formula