1824 United States presidential election in Connecticut

Summary

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.

1824 United States presidential election in Connecticut

← 1820 October 26 – December 2, 1824 1828 →
 
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%

County Results
Adams
  50-60%
  70-80%
  80-90%
  90-100%


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. 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 edit

1824 United States presidential election in Connecticut[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

See also edit

References edit

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