1848 United States presidential election in Rhode Island

Main article: 1848 United States presidential election

The 1848 United States presidential election in Rhode Island took place on November 7, 1844, as part of the 1848 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.

1848 United States presidential election in Rhode Island

← 1844 November 7, 1848 1852 →

Nominee Zachary Taylor Lewis Cass Martin Van Buren Party Whig Democratic Free Soil Home state Louisiana Michigan New York Running mate Millard Fillmore William O. Butler Charles Francis Adams, Sr. Electoral vote 4 0 0 Popular vote 6,779 3,646 730 Percentage 60.77% 32.68% 6.54%

County Results Taylor   50–60%   60–70%

President before election James K. Polk Democratic Elected President Zachary Taylor Whig

Rhode Island voted for the Whig candidate, Zachary Taylor, over Democratic candidate Lewis Cass and Free Soil candidate former president Martin Van Buren. Taylor won the state by a margin of 28.09%.

With 60.77% of the popular vote, Rhode Island would prove to be Taylor's strongest state in the election in terms of percentage in the popular vote.^[1]

Results

1848 United States presidential election in Rhode Island[2]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Whig	Zachary Taylor of Louisiana	Millard Fillmore of New York	6,779	60.77%	4	100.00%
	Democratic	Lewis Cass of Michigan	William O. Butler of Kentucky	3,646	32.68%	0	0.00%
	Free Soil	Martin Van Buren of New York	Charles Francis Adams, Sr. of Massachusetts	730	6.54%	0	0.00%
Total				11,155	100.00%	4	100.00%

See also

• United States presidential elections in Rhode Island