1840 United States presidential election in Maryland
The 1840 United States presidential election in Maryland took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose 10 representatives, or electors to the Electoral College, who voted for President and Vice President.
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Elections in Maryland |
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Government |
Maryland voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Maryland by a margin of 7.66%.
Results
1840 United States presidential election in Maryland[1] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 33,528 | 53.83% | 10 | 100.00% | ||
Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 28,752 | 46.17% | 0 | 0.00% | ||
Total | 62,280 | 100.00% | 10 | 100.00% |
References
- "1840 Presidential General Election Results - Maryland". U.S. Election Atlas. Retrieved 23 December 2013.
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