1840 United States presidential election in Maine
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Elections in Maine |
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A presidential election was held in Maine on November 2, 1840 as part of the 1840 United States presidential election.[1] 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[2] | ||||||||
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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]- ^ Dubin, Michael J. (2002). United States Presidential Elections, 1788–1860: The Official Results by County and State. Jefferson, NC: McFarland & Co. p. xvi.
- ^ "1840 Presidential General Election Results - Maine". U.S. Election Atlas. Retrieved December 23, 2013.