1840 United States presidential election in Rhode Island

Main article: 1840 United States presidential election

The 1840 United States presidential election in Rhode Island took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.

1840 United States presidential election in Rhode Island

← 1836 October 30 - December 2, 1840 1844 →

Nominee William Henry Harrison Martin Van Buren Party Whig Democratic Home state Ohio New York Running mate John Tyler none Electoral vote 4 0 Popular vote 5,278 3,301 Percentage 61.22% 38.29%

County Results Harrison   50-60%   60-70%   70-80%

President before election Martin Van Buren Democratic Elected President William Henry Harrison Whig

Rhode Island voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Rhode Island by a margin of 22.93%.

With 61.22% of the popular vote, Rhode Island would be Harrison's third strongest state in the 1840 election after Kentucky and Vermont.^[1]

Results

1840 United States presidential election in Rhode Island[2]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Whig	William Henry Harrison of Ohio	John Tyler of Virginia	5,278	61.22%	4	100.00%
	Democratic	Martin Van Buren of New York	Richard M. Johnson of Kentucky	3,301	38.29%	0	0.00%
	Liberty	James G. Birney of New York	Thomas Earle of Pennsylvania	42	0.49%	0	0.00%
Total				8,621	100.00%	4	100.00%

See also

• United States presidential elections in Rhode Island

References

1. "1840 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
2. "1840 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved December 23, 2013.