1788–89 United States presidential election in Massachusetts

The 1788–89 United States presidential election in Massachusetts took place on January 7, 1789, as part of the 1788–1789 United States presidential election to elect the first President. Massachusetts was entitled to 10 electors, with two being appointed by the state legislature and the rest being chosen by state legislature from the two most popular candidates in each U.S. House district. Each elector voted once for President and again for Vice President.[1]

1788–89 United States presidential election in Massachusetts

January 7, 1789 1792 →
 
Nominee George Washington John Adams
Party Independent Federalist
Home state Virginia Massachusetts
Electoral vote 10 10
Popular vote 4,609
Percentage 100.00%

President before election

Office established

Elected President

George Washington
Independent

Massachusetts unanimously voted for independent candidate and commander-in-chief of the Continental Army, George Washington. The total vote was composed of 4,477 for Federalist electors and 132 for Anti-Federalist electors, all of whom were supportive of Washington.[2] Several candidates of unknown affiliation also received votes

Results

edit
1788-1789 United States presidential election in Massachusetts
Party Candidate Votes Percentage Electoral votes
Independent George Washington 4,609 100.00% 10
Totals 17,740 100.00% 10

See also

edit

References

edit
  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 New Nation Votes". elections.lib.tufts.edu. Retrieved July 16, 2024.