United States House of Representatives election in Florida, 1845

United States House of Representatives election in Florida, 1845
Florida
May 26, 1845
Florida's single seat to the United States House of Representatives
  Majority party Minority party
 
Party Democratic Whig
Seats won 1 0
Popular vote 3,608 2,373
Percentage 60.3% 39.7%

The election to the United States House of Representatives in Florida for the 29th Congress was held on May 26, 1845, the first Congressional election in Florida's history.

Background

Florida was admitted to the Union on March 3, 1845, the last day of the 28th Congress, along with Iowa.[1] Neither state was represented in the 28th Congress. As Florida was represented by a single Representative, it was not divided into districts, elections being held at-large until 1874.

Election results

1845 United States House election results[2]
Democratic Whig
David Levy Yulee 3,608 60.3% Benjamin A. Putnam 2,373 39.7%

Yulee had previously served as delegate for Florida Territory. Yulee did not serve as Representative, being subsequently elected one of the first two Senators for Florida.

Special election

A special election was held October 6, 1845[3] to choose a new Representative after Yulee's election to the Senate.

1845 Special election results
Democratic Whig
William H. Brockenbrough 2,472 49.5% Edward C. Cabell 2,523 50.5%

Brockenbrough challenged Cabell's election. And after a recount, Brockenbrough was declared the winner 2,669 - 2,472[4] and was seated January 24, 1846.[5]

See also

References

  1. ↑ Text of the Act admitting Florida and Iowa
  2. ↑ Election details from Ourcampaigns.com
  3. ↑ Election details from Ourcampaigns.com
  4. ↑ Recount details from Ourcampaigns.com
  5. ↑ 29th Congress membership roster
This article is issued from Wikipedia - version of the Friday, August 08, 2014. The text is available under the Creative Commons Attribution/Share Alike but additional terms may apply for the media files.