<page_title> 1844 United States presidential election in Vermont </page_title> <section_title> Results </section_title> <table> <cell> Henry Clay of Kentucky <col_header> United States presidential election in Vermont, 1844 </col_header> <col_header> Candidate </col_header> <col_header> Count </col_header> </cell> <cell> 54.84% <col_header> United States presidential election in Vermont, 1844 </col_header> <col_header> Popular vote </col_header> </cell> </table>
Henry Clay had 54.84% of the popular vote in the 1844 United States presidential election in Vermont.