See main article: 1844 United States presidential election.
| Election Name: | 1844 United States presidential election in Rhode Island |
| Country: | Rhode Island |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1840 United States presidential election in Rhode Island |
| Previous Year: | 1840 |
| Next Election: | 1848 United States presidential election in Rhode Island |
| Next Year: | 1848 |
| Election Date: | November 1 - December 4, 1844 |
| Image1: | Clay 1848.jpg |
| Nominee1: | Henry Clay |
| Party1: | Whig Party (United States) |
| Home State1: | Kentucky |
| Running Mate1: | Theodore Frelinghuysen |
| Electoral Vote1: | 4 |
| Popular Vote1: | 7,322 |
| Percentage1: | 59.55% |
| Nominee2: | James K. Polk |
| Party2: | Democratic Party (United States) |
| Home State2: | Tennessee |
| Running Mate2: | George M. Dallas |
| Electoral Vote2: | 0 |
| Popular Vote2: | 4,867 |
| Percentage2: | 39.58% |
| Map Size: | 250px |
| President | |
| Before Election: | John Tyler |
| Before Party: | Independent (politician) |
| After Election: | James K. Polk |
| After Party: | Democratic Party (United States) |
The 1844 United States presidential election in Rhode Island took place between November 1 and December 4, 1844, as part of the 1844 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.
Rhode Island voted for the Whig candidate, Henry Clay, over Democratic candidate James K. Polk. Clay won Rhode Island by a margin of 19.97%.
With 59.55% of the popular vote, Rhode Island would prove to be Henry Clay's strongest state in the nation.[1]