1848 United States presidential election in Iowa

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United States presidential election in Iowa, 1848

November 7, 1848 1852 →
  Lewis Cass crop.jpg Zachary Taylor portrait.jpg Van Buren.jpg
Nominee Lewis Cass Zachary Taylor Martin Van Buren
Party Democratic Whig Free Soil
Home state Michigan Louisiana New York
Running mate William O. Butler Millard Fillmore Charles F. Adams
Electoral vote 4 0 0
Popular vote 11,238 9,930 1,103
Percentage 50.46% 44.59% 4.95%

President before election

James K. Polk
Democratic

Elected President

Zachary Taylor
Whig

The 1848 United States presidential election in Iowa took place on November 7, 1848, as part of the 1848 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.

Iowa voted for the Democratic candidate Lewis Cass in the state's first presidential election, over Whig candidate Zachary Taylor and Free Soil candidate Martin Van Buren. Cass won Iowa by a margin of 5.87%.

Results[edit]

United States presidential election in Iowa, 1848[1][2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Lewis Cass of Michigan William O. Butler of Kentucky 11,238 50.46% 4 100.00%
Whig Zachary Taylor of Louisiana Millard Fillmore of New York 9,930 44.59% 0 0.00%
Free Soil Martin Van Buren of New York Charles F. Adams, Sr. of Massachusetts 1,103 4.95% 0 0.00%
Total '22,271 100.00% 4 100.00%

References[edit]

  1. ^ "1848 Presidential General Election Results - Iowa". U.S. Election Atlas. Retrieved 27 December 2013.
  2. ^ "1848 Presidential Election". The American Presidency Project. University of California Santa Barbara. Retrieved 27 December 2013.