| Elections in Kentucky |
|---|
 |
 Government |
The 1840 United States presidential election in Kentucky took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose 15 representatives, or electors to the Electoral College, who voted for President and Vice President.
Kentucky voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Kentucky by a margin of 28.4%.
With 64.20% of the popular vote, Kentucky would prove to be Harrison's strongest state in the 1840 election.[1]
Results
| United States presidential election in Kentucky, 1840[2] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 58,488 | 64.20% | 15 | 100.00% | ||
| Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 32,616 | 35.80% | 0 | 0.00% | ||
| Total | 91,104 | 100.00% | 15 | 100.00% | ||||
References
- ^ "1840 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
- ^ "1840 Presidential General Election Results - Kentucky". U.S. Election Atlas. Retrieved December 23, 2013.
State and district results of the 1840 United States presidential election | ||
|---|---|---|
 | ||