1804 United States presidential election in Rhode Island
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A presidential election was held in Rhode Island on November 21, 1804, as part of the 1804 United States presidential election.[1] The Democratic-Republican Party's ticket of incumbent president Thomas Jefferson and former New York governor George Clinton was elected unanimously.[2] The Federalist Party did not nominate electors. Jefferson won the national election in a landslide over the de facto Federalist candidate, Charles Cotesworth Pinckney.[3]
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November 21, 1804
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General election
Summary
Rhode Island chose four electors on a statewide general ticket. Nineteenth-century election laws required voters to elect the members of the Electoral College individually, rather than as a block. This sometimes resulted in small differences in the number of votes cast for electors pledged to the same presidential nominee, if some voters did not vote for all the electors nominated by a party.[4] The following table shows the result for the leading Democratic-Republican elector to give an approximate sense of the statewide popular vote.
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Democratic-Republican | Thomas Jefferson George Clinton |
1,312 | 100.00 | |
| Total votes | 1,312 | 100.00 | ||
Results
| Party | Candidate | Votes | |
|---|---|---|---|
| Democratic-Republican | James Aldrich | 1,312 | |
| Democratic-Republican | James Helme | 1,311 | |
| Democratic-Republican | Benjamin Remington | 1,309 | |
| Democratic-Republican | Constant Taber | 1,308 | |
Total votes |
≈1,312 | ||
Results by county
| County | Thomas Jefferson Democratic-Republican |
Total | |
|---|---|---|---|
| Votes | % | ||
| Bristol | 38 | 100.00 | 38 |
| Kent | 138 | 100.00 | 138 |
| Newport | 345 | 100.00 | 345 |
| Providence | 575 | 100.00 | 575 |
| Washington | 216 | 100.00 | 216 |
| TOTAL | 1,312 | 100.00 | 1,312 |