1860 United States presidential election in Massachusetts
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Turnout | 65.8%[1] 4.0 pp | ||||||||||||||||||||||||||||||||
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Elections in Massachusetts |
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Massachusetts portal |
The 1860 United States presidential election in Massachusetts took place on November 2, 1860, as part of the 1860 United States presidential election. Voters chose 13 electors of the Electoral College, who voted for president and vice president.
Massachusetts was won by Republican candidate Abraham Lincoln, who won the state by 42.57%.
With 62.80% of the popular vote, Massachusetts would prove to be Lincoln's third strongest state in the 1860 election in terms of popular vote percentage after neighboring Vermont and Minnesota.[2]
Results
1860 United States presidential election in Massachusetts[3] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Republican | Abraham Lincoln of Illinois | Hannibal Hamlin of Maine | 106,684 | 62.80% | 13 | 100.00% | ||
Democratic | Stephen A. Douglas of Illinois | Herschel Vespasian Johnson of Georgia | 34,370 | 20.23% | 0 | 0.00% | ||
Constitutional Union | John Bell of Tennessee | Edward Everett of Massachusetts | 22,331 | 13.15% | 0 | 0.00% | ||
Southern Democratic | John C. Breckinridge of Kentucky | Joseph Lane of Oregon | 6,163 | 3.63% | 0 | 0.00% | ||
N/A | Others | Others | 328 | 0.19% | 0 | 0.00% | ||
Total | 169,876 | 100.00% | 13 | 100.00% |
See also
References
- ^ Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
- ^ "1860 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- ^ "1860 Presidential General Election Results - Massachusetts".