Election in Ohio
1824 United States presidential election in Ohio
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← 1820 | October 26 – December 2, 1824 | 1828 → |
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| | | | 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
See also
References
- ^ "1824 Presidential General Election Results - Ohio". U.S. Election Atlas. Retrieved February 27, 2013.