1792 United States presidential election in Massachusetts

The 1792 United States presidential election in Massachusetts took place between November 2 and December 5, 1792, as part of the 1792 United States presidential election. In this election, two Congressional districts chose five electors each, the remaining two districts chose three electors. Each elector chosen by majority vote of voters in Congressional district. If an insufficient number of electors are chosen by majority vote from a Congressional district, remaining electors would be appointed by the state legislature.[1]

1792 United States presidential election in Massachusetts

← 1788–89 November 2 – December 5, 1792 1796 →
 
Nominee George Washington John Adams
Party Independent Federalist
Home state Virginia Massachusetts
Electoral vote 16 16
Popular vote 20,343
Percentage 100.00%

President before election

George Washington
Independent

Elected President

George Washington
Independent

Massachusetts unanimously voted for independent candidate and incumbent president, George Washington. The total vote is composed of 20,343 for Federalist electors, all of whom were supportive of Washington.[2]

Results

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1792 United States presidential election in Massachusetts[2]
Party Candidate Votes Percentage Electoral votes
Independent George Washington 20,343 100.00% 10
Totals 20,343 100.00% 10

See also

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References

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  1. ^ "The Electoral Count for the Presidential Election of 1789". The Papers of George Washington. Archived from the original on September 14, 2013. Retrieved May 4, 2005.
  2. ^ a b Dubin, Michael J. (2002). United States Presidential Elections, 1788-1860: The Official Results by County and State. Jefferson: McFarland & Company. pp. 4–5. ISBN 9780786410170.