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1828 United States presidential election in Vermont

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Main article:1828 United States presidential election
1828 United States presidential election in Vermont

← 1824
October 31 – December 2, 1828
1832 →
 
NomineeJohn Quincy AdamsAndrew Jackson
PartyNational RepublicanDemocratic
Home stateMassachusettsTennessee
Running mateRichard RushJohn C. Calhoun
Electoral vote70
Popular vote25,3638,350
Percentage75.23%24.77%

County Results
Adams
  50–60%
  60–70%
  70–80%
  80–90%

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The1828 United States presidential election in Vermont took place between October 31 and December 2, 1828, as part of the1828 United States presidential election. Voters chose seven representatives, or electors to theElectoral College, who voted forPresident andVice President.

Vermont voted for theNational Republican candidate,John Quincy Adams, over theDemocratic candidate,Andrew Jackson. Adams won Vermont by a margin of 50.46%.

With 75.23% of the popular vote, Adam's victory in Vermont was his third strongest state in the 1828 election afterRhode Island andMassachusetts.[1] No Democrat was to win Vermont untilLyndon B. Johnson did so in1964. This election marks the first time Vermont chose its electors by popular vote rather than appointment by the state legislature.[2]

Results

[edit]
1828 United States presidential election in Vermont[3]
PartyCandidateVotesPercentageElectoral votes
National RepublicanJohn Quincy Adams (incumbent)25,36375.23%7
DemocraticAndrew Jackson8,35024.77%0
Totals33,713100.0%7

See also

[edit]

References

[edit]
  1. ^"1828 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. RetrievedMarch 5, 2018.
  2. ^Moore, John L., ed. (1985).Congressional Quarterly's Guide to U.S. Elections (2nd ed.). Washington, D.C.: Congressional Quarterly, Inc. pp. 254–56.
  3. ^"1828 Presidential General Election Results - Vermont". U.S. Election Atlas. RetrievedFebruary 28, 2013.
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