1840 United States presidential election in Illinois

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United States presidential election in Illinois, 1840

← 1836 October 30 – December 2, 1840 1844 →
  Van Buren.jpg Harrison crop.jpg
Nominee Martin Van Buren William Henry Harrison
Party Democratic Whig
Home state New York Ohio
Running mate Richard Mentor Johnson John Tyler
Electoral vote 5 0
Popular vote 47,441 45,574
Percentage 50.92% 48.91%

President before election

Martin Van Buren

Elected President

William Henry Harrison

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

Illinois voted for the Democratic candidate, Martin Van Buren, over Whig candidate William Henry Harrison. Van Buren won Illinois by a margin of 2.01%.


United States presidential election in Illinois, 1840[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 47,441 50.92% 5 100.00%
Whig William Henry Harrison of Ohio John Tyler of Virginia 45,574 48.91% 0 0.00%
Liberty James G. Birney of New York Thomas Earle of Pennsylvania 160 0.17% 0 0.00%
Total 93,175 100.00% 5 100.00%


  1. ^ "1840 Presidential General Election Results - Illinois". U.S. Election Atlas. Retrieved 23 December 2013.

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