1824 United States presidential election in Ohio

Summary

The 1824 United States presidential election in Ohio took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose 16 representatives, or electors to the Electoral College, who voted for President and Vice President.

1824 United States presidential election in Ohio

← 1820 October 26 – December 2, 1824 1828 →
 
Nominee Henry Clay Andrew Jackson John Quincy Adams
Party Democratic-Republican Democratic-Republican Democratic-Republican
Home state Kentucky Tennessee Massachusetts
Running mate Nathan Sanford John C. Calhoun John C. Calhoun
Electoral vote 16 0 0
Popular vote 19,255 18,489 12,280
Percentage 38.49% 36.96% 24.55%

County Results

President before election

James Monroe
Democratic-Republican

Elected President

John Quincy Adams
Democratic-Republican

During this election, the Democratic-Republican Party was the only major national party, and four different candidates from this party sought the presidency. Ohio voted for Henry Clay over Andrew Jackson, John Quincy Adams, and William H. Crawford. Clay won Ohio by a narrow margin of 1.53%. This was the first time that Ohio voted for a losing presidential candidate.

Results edit

1824 United States presidential election in Ohio[1]
Party Candidate Votes Percentage Electoral votes
Democratic-Republican Henry Clay 19,255 38.49% 16
Democratic-Republican Andrew Jackson 18,489 36.96% 0
Democratic-Republican John Quincy Adams 12,280 24.55% 0
Totals 50,024 100.0% 16

See also edit

References edit

  1. ^ "1824 Presidential General Election Results - Ohio". U.S. Election Atlas. Retrieved February 27, 2013.