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The 1824 United States presidential election in Kentucky took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose 14 representatives, or electors to the Electoral College, who voted for President and Vice President.

During this election, the Democratic-Republican Party was the only major national party, and four different candidates from this party sought the presidency. Kentucky voted for Henry Clay over Andrew Jackson, John Quincy Adams, and William H. Crawford. Clay won Kentucky, his home state, by a wide margin of 43.46%.[1]

Kentucky was divided into three electoral districts for this election, with electors chosen through block voting. District 1 elected four electors, while Districts 2 and 3 elected five electors each.[1]

Results

1824 United States presidential election in Kentucky
Party Candidate Votes Percentage Electoral votes
Democratic-Republican Henry Clay 17,207 71.73% 14
Democratic-Republican Andrew Jackson 6,658 27.76% 0
Democratic-Republican John Quincy Adams 120 0.50% 0
Democratic-Republican William H. Crawford 3 0.01% 0
Totals 23,988 100.0% 14

Results by electoral district

Results by District[1]
District Henry Clay

Democratic-Republican

Andrew Jackson

Democratic-Republican

John Quincy Adams

Democratic-Republican

William H. Crawford

Democratic-Republican

Margin Total Votes Cast
# % Electors # % Electors # % Electors # % Electors # %
1 3,861 71.47% 4 1,473 27.27% 0 65 1.20% 0 3 0.06% 0 2,320 42.94% 5,402
2 6,165 69.46% 5 2,711 30.54% 0 no candidates no ballots 3,454 38.92% 8,876
3 7,181 73.95% 5 2,474 25.48% 0 55 0.57% 0 no candidates 4,652 47.90% 9,710
Total 17,207 71.73% 14 6,658 27.76% 0 120 0.50% 0 3 0.01% 0 10,426 43.46% 23,988

References

  1. ^ a b c "A New Nation Votes". elections.lib.tufts.edu. Retrieved July 24, 2024.
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