1840 United States presidential election in Maine
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Elections in Maine |
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The 1840 United States presidential election in Maine took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose ten representatives, or electors to the Electoral College, who voted for president and vice president.
Maine voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won the state by a very narrow margin of 0.46%.
Maine was typically a Democratic state during the Second Party System, however, with Harrison narrowly winning the state, this would be the only time that a Whig presidential candidate would win Maine.
Results
[edit]1840 United States presidential election in Maine[1] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 46,612 | 50.23% | 10 | 100.00% | ||
Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 46,190 | 49.77% | 0 | 0.00% | ||
Total | 92,802 | 100.00% | 10 | 100.00% |
See also
[edit]References
[edit]- ^ "1840 Presidential General Election Results - Maine". U.S. Election Atlas. Retrieved December 23, 2013.