1840 United States presidential election in Indiana

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United States presidential election in Indiana, 1840

← 1836 October 30 - December 2, 1840 1844 →
  Harrison crop.jpg Van Buren.jpg
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate John Tyler none
Electoral vote 9 0
Popular vote 65,302 51,604
Percentage 55.86% 44.14%

President before election

Martin Van Buren
Democratic

Elected President

William Henry Harrison
Whig

The 1840 United States presidential election in Indiana took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose nine representatives, or electors to the Electoral College, who voted for President and Vice President.

Indiana voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Indiana by a margin of 11.72%.

Results[edit]

United States presidential election in Indiana, 1840[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio John Tyler of Virginia 65,302 55.86% 9 100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 51,604 44.14% 0 0.00%
Total 116,906 100.00% 9 100.00%

References[edit]

  1. ^ "1840 Presidential General Election Results - Indiana". U.S. Election Atlas. Retrieved 23 December 2013.