1824 United States presidential election in Ohio

Main article: 1824 United States presidential election

1824 United States presidential election in Ohio

October 26 – December 2, 1824

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 Clay   30-40%   40-50%   50-60%   60-70%   70–80%   80–90% Jackson   40-50%   50-60%   60-70%   70-80% Adams   40-50%   50-60%   60-70%   70–80%   80–90%   90-100% No Votes   No data

President before election James Monroe Democratic-Republican Elected President John Quincy Adams Democratic-Republican

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.

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

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

• United States presidential elections in Ohio

References

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