See main article: 1824 United States presidential election.
| Election Name: | 1824 United States presidential election in Delaware |
| Country: | Delaware |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1820 United States presidential election in Delaware |
| Previous Year: | 1820 |
| Next Election: | 1828 United States presidential election in Delaware |
| Next Year: | 1828 |
| Election Date: | October 26 – December 2, 1824 |
| Image1: | WilliamHCrawford.jpg |
| Nominee1: | William H. Crawford |
| Party1: | Democratic-Republican Party |
| Home State1: | Georgia |
| Running Mate1: | Nathaniel Macon |
| Electoral Vote1: | 2 |
| Nominee2: | John Quincy Adams |
| Party2: | Democratic-Republican Party |
| Home State2: | Massachusetts |
| Running Mate2: | John C. Calhoun |
| Electoral Vote2: | 1 |
| President | |
| Before Election: | James Monroe |
| Before Party: | Democratic-Republican Party |
| After Election: | John Quincy Adams |
| After Party: | Democratic-Republican Party |
The 1824 United States presidential election in Delaware took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose three 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. Delaware cast two electoral votes for William H. Crawford and one for John Quincy Adams.
| 1824 United States presidential election in Delaware[1] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic-Republican | William H. Crawford | 2 | |||
| Democratic-Republican | John Quincy Adams | 1 | |||
| Democratic-Republican | Henry Clay | 0 | |||
| Democratic-Republican | Andrew Jackson | 0 | |||
| Totals | 3 | ||||