See main article: 1788–89 United States presidential election.
| Election Name: | 1788–89 United States presidential election in Massachusetts |
| Country: | Massachusetts |
| Type: | presidential |
| Ongoing: | no |
| Next Election: | 1792 United States presidential election in Massachusetts |
| Next Year: | 1792 |
| Election Date: | January 7, 1789 |
| Image1: | Gilbert Stuart Williamstown Portrait of George Washington.jpg |
| Nominee1: | George Washington |
| Party1: | Independent (politician) |
| Home State1: | Virginia |
| Electoral Vote1: | 10 |
| Popular Vote1: | 4,609 |
| Percentage1: | 100.00% |
| Nominee2: | John Adams |
| Party2: | Federalist Party |
| Home State2: | Massachusetts |
| Electoral Vote2: | 10 |
| Popular Vote2: | – |
| Percentage2: | – |
| President | |
| Before Election: | Office established |
| After Election: | George Washington |
| After Party: | Independent (politician) |
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]
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
| 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 | ||