1845 United States House of Representatives election in Florida
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Florida's single seat to the United States House of Representatives | |||||||||||||||||||
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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
| 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.
| 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]