United States presidential election in Pennsylvania, 1840
Main article: United States presidential election, 1840
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The 1840 United States presidential election in Pennsylvania took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose 30 representatives, or electors to the Electoral College, who voted for President and Vice President.
Pennsylvania voted for the Whig candidate, William Henry Harrison, over the Democratic candidate, Martin Van Buren. Harrison won Pennsylvania by a margin of 0.12%.
Results
United States presidential election in Pennsylvania, 1840[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Whig | William Henry Harrison | 144,010 | 50.00% | 30 | |
Democratic | Martin Van Buren | 143,676 | 49.88% | 0 | |
Liberty | James G. Birney | 340 | 0.12% | 0 | |
Totals | 288,026 | 100.0% | 30 | ||
References
- ↑ "1840 Presidential General Election Results - Pennsylvania". U.S. Election Atlas. Retrieved 4 August 2012.
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