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.
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
County Results
| ||||||||||||||||||||||||||
|
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%.
Results
edit1840 United States presidential election in Illinois[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% |
See also
editReferences
edit- ^ "1840 Presidential General Election Results - Illinois". U.S. Election Atlas. Retrieved December 23, 2013.