1836 United States presidential election in Massachusetts
The 1836 United States presidential election in Massachusetts took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose 14 representatives, or electors to the Electoral College, who voted for President and Vice President.
Main article: 1836 United States presidential election
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Turnout | 43.4%[1] ![]() | |||||||||||||||||||||||||
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Massachusetts voted for Whig candidate and state native Daniel Webster over the Democratic candidate, Martin Van Buren. Webster won Massachusetts by a margin of 10.32%.
Results
1836 United States presidential election in Massachusetts[2] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Whig | Daniel Webster | 41,201 | 55.13% | 14 | |
Democratic | Martin Van Buren | 33,486 | 44.81% | 0 | |
N/A | Other | 45 | 0.06% | 0 | |
Totals | 74,732 | 100.0% | 14 | ||
References
- Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
- "1836 Presidential General Election Results - Massachusetts". U.S. Election Atlas. Retrieved 4 August 2012.
State and district results of the 1836 United States presidential election | ||
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Select elections in Massachusetts | |
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General | |
Mass. Senate | |
Mass. House | |
Governor (with winners) |
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Gov.'s Council |
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U.S. President | |
U.S. Senate Class 1 (with winners) |
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U.S. Senate Class 2 (with winners) |
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U.S. House |
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"s/" = Special election |
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