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United States presidential election in Connecticut, 1824

From Wikipedia, the free encyclopedia

United States presidential election in Connecticut, 1824

← 1820 October 26 – December 2, 1824 1828 →

 
JohnQAdams.jpg
WilliamHCrawford.jpg
Nominee John Quincy Adams William H. Crawford
Party Democratic-Republican Democratic-Republican
Home state Massachusetts Georgia
Running mate John C. Calhoun Nathaniel Macon
Electoral vote 8 0
Popular vote 7,494 1,965
Percentage 70.39% 18.46%

President before election

James Monroe
Democratic-Republican

Elected President

John Quincy Adams
Democratic-Republican

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 eight 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. Connecticut voted for John Quincy Adams over William H. Crawford, Andrew Jackson, and Henry Clay. Adams won Connecticut by a margin of 51.93%.

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Transcription

Results

United States presidential election in Connecticut, 1824[1]
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

References

  1. ^ "1824 Presidential General Election Results - Connecticut". U.S. Election Atlas. Retrieved 27 February 2013.
This page was last edited on 15 October 2018, at 17:56
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