See main article: 1792 United States presidential election.
Election Name: | 1792 United States presidential election in Massachusetts |
Country: | Massachusetts |
Type: | presidential |
Ongoing: | no |
Previous Election: | 1788–89 United States presidential election in Massachusetts |
Previous Year: | 1788–89 |
Next Election: | 1796 United States presidential election in Massachusetts |
Next Year: | 1796 |
Election Date: | November 2 – December 5, 1792 |
Image1: | Gilbert Stuart Williamstown Portrait of George Washington.jpg |
Nominee1: | George Washington |
Party1: | Independent (politician) |
Home State1: | Virginia |
Electoral Vote1: | 16 |
Popular Vote1: | 20,343 |
Percentage1: | 100.00% |
Nominee2: | John Adams |
Party2: | Federalist Party |
Home State2: | Massachusetts |
Electoral Vote2: | 16 |
Popular Vote2: | – |
Percentage2: | – |
President | |
Before Election: | George Washington |
Before Party: | Independent (politician) |
After Election: | George Washington |
After Party: | Independent (politician) |
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]
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]
1792 United States presidential election in Massachusetts | |||||
---|---|---|---|---|---|
Party | Candidate | Votes | Percentage | Electoral votes | |
Independent | George Washington | 20,343 | 100.00% | 10 | |
Totals | 20,343 | 100.00% | 10 | ||