United States presidential election in Ohio, 1840

United States presidential election in Ohio, 1840
Ohio
October 30 - December 2, 1840
 
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate John Tyler none
Electoral vote 21 0
Popular vote 148,157 124,782
Percentage 54.10 45.57
President before election

Martin Van Buren
Democratic

 Elected President

William Henry Harrison
Whig

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

Ohio voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Ohio by a margin of 8.53%.

Results

United States presidential election in Ohio, 1840[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio John Tyler of Virginia 148,157 54.10% 21 100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 124,782 45.57% 0 0.00%
Liberty James G. Birney of New York Thomas Earle of Pennsylvania 903 0.33% 0 0.00%
Total 273,842 100.00% 21 100.00%

References

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