1788–89 United States presidential election in Massachusetts

Main article: 1788–89 United States presidential election

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

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

• United States presidential elections in Massachusetts

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