1792 United States presidential election in Massachusetts

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1792 United States presidential election in Massachusetts
Flag of Massachusetts.svg
  1788–89 November 2 – December 5, 1792 1796  
  Gilbert Stuart Williamstown Portrait of George Washington.jpg Jadams.jpeg
Nominee George Washington John Adams
Party Independent Federalist
Home state Virginia Massachusetts
Electoral vote1616
Popular vote20,343
Percentage100.00%

President before election

George Washington
Independent

Elected President

George Washington
Independent

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]

Contents

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

1792 United States presidential election in Massachusetts [2]
PartyCandidateVotesPercentageElectoral votes
Independent George Washington 20,343100.00%10
Totals20,343100.00%10

See also

References

  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. 1 2 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.