United States presidential election in Massachusetts, 1860

United States presidential election in Massachusetts, 1860

November 2, 1860

 
Nominee Abraham Lincoln Stephen A. Douglas John Bell
Party Republican Democratic Constitutional Union
Home state Illinois Illinois Tennessee
Running mate Hannibal Hamlin Herschel Vespasian Johnson Edward Everett
Electoral vote 13 0 0
Popular vote 106,684 34,370 22,331
Percentage 62.80% 20.23% 13.15%

President before election

James Buchanan
Democratic

Elected President

Abraham Lincoln
Republican

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[1].

Results

United States presidential election in Massachusetts, 1860[2]
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%

References

  1. "1860 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. "1860 Presidential General Election Results - Massachusetts".
This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.