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1828 United States presidential election in Vermont

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1828 United States presidential election in Vermont

← 1824 October 31 – December 2, 1828 1832 →
 
Nominee John Quincy Adams Andrew Jackson
Party National Republican Democratic
Home state Massachusetts Tennessee
Running mate Richard Rush John C. Calhoun
Electoral vote 7 0
Popular vote 25,363 8,350
Percentage 75.23% 24.77%

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

The 1828 United States presidential election in Vermont took place between October 31 and December 2, 1828, as part of the 1828 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.

Vermont voted for the National Republican candidate, John Quincy Adams, over the Democratic 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 after Rhode Island and Massachusetts.[1] No Democrat was to win Vermont until Lyndon B. Johnson did so in 1964. This election marks the first time Vermont chose its electors by popular vote rather than appointment by the state legislature.[2]

Results

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1828 United States presidential election in Vermont[3]
Party Candidate Votes Percentage Electoral votes
National Republican John Quincy Adams (incumbent) 25,363 75.23% 7
Democratic Andrew Jackson 8,350 24.77% 0
Totals 33,713 100.0% 7

See also

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References

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  1. ^ "1828 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 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. Retrieved February 28, 2013.