United States presidential election in Arkansas, 1840
Main article: United States presidential election, 1840
| Elections in Arkansas | |||||||||
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The 1840 United States presidential election in Arkansas took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for President and Vice President.
Arkansas voted for the Democratic candidate, Martin Van Buren, over Whig candidate William Henry Harrison. Van Buren won Arkansas by a margin of 12.84%.
Results
| United States presidential election in Arkansas, 1840[1] | ||||||||
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| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 6,679 | 56.42% | 3 | 100.00% | ||
| Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 5,160 | 43.58% | 0 | 0.00% | ||
| Total | 11,839 | 100.00% | 3 | 100.00% | ||||
References
- ↑ "1840 Presidential General Election Results - Arkansas". U.S. Election Atlas. Retrieved 23 December 2013.
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