1824 United States presidential election in Connecticut
The 1824 United States presidential election in Connecticut took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose 8 representatives, or electors to the Electoral College, who voted for President and Vice President.
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Elections in Connecticut | ||||||||||
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During this election, the Democratic-Republican Party was the only major national party, and 4 different candidates from this party sought the Presidency. Connecticut voted for John Quincy Adams over William H. Crawford, Andrew Jackson, and Henry Clay. Adams won Connecticut by a margin of 51.93%.
Results
1824 United States presidential election in Connecticut[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic-Republican | John Quincy Adams | 7,494 | 70.39% | 8 | |
Democratic-Republican | William H. Crawford | 1,965 | 18.46% | 0 | |
N/A | Other | 1,188 | 11.16% | 0 | |
Totals | 10,647 | 100.0% | 8 | ||
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References
- "1824 Presidential General Election Results - Connecticut". U.S. Election Atlas. Retrieved 27 February 2013.
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