United States presidential election in Arkansas, 1836

United States presidential election in Arkansas, 1836
Arkansas
November 3 – December 7, 1836

 
Nominee Martin Van Buren Hugh White
Party Democratic Whig
Home state New York Tennessee
Running mate Richard Johnson John Tyler
Electoral vote 3 0
Popular vote 2,380 1,334
Percentage 64.08% 35.92%

President before election

Andrew Jackson
Democratic

Elected President

Martin Van Buren
Democratic

The 1836 United States presidential election in Arkansas took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for President and Vice President.

Arkansas voted for the Democratic candidate, Martin Van Buren, over Whig candidate Hugh White. Van Buren won Arkansas by a margin of 28.16%.

Results

United States presidential election in Arkansas, 1836[1]
Party Candidate Votes Percentage Electoral votes
Democratic Martin Van Buren 2,380 64.08% 3
Whig Hugh White 1,334 35.92% 0
Totals 3,714 100.0% 3

References

  1. "1836 Presidential General Election Results - Arkansas". U.S. Election Atlas. Retrieved 4 August 2012.
This article is issued from Wikipedia - version of the 9/8/2016. The text is available under the Creative Commons Attribution/Share Alike but additional terms may apply for the media files.