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1836 United States presidential election in Kentucky

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Main article:1836 United States presidential election
1836 United States presidential election in Kentucky
← 1832
November 7–9, 1836
1840 →
← MI
NY →
 
NomineeWilliam Henry HarrisonMartin Van Buren
PartyWhigDemocratic
Home stateOhioNew York
Running mateFrancis GrangerRichard Mentor Johnson
Electoral vote150
Popular vote36,86133,229
Percentage52.59%47.41%

County Results

Harrison

  50–60%
  60–70%
  70–80%
  80–90%

Van Buren

  50–60%
  60–70%
  70–80%
  80–90%

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The1836 United States presidential election in Kentucky was held between November 7 and 9, 1836 as part of the1836 United States presidential election.[1] Voters chose 15 representatives, or electors to theElectoral College, who voted forPresident andVice President.

Kentucky voted forWhig candidateWilliam Henry Harrison over theDemocratic candidate,Martin Van Buren. Harrison won Kentucky by a margin of 5.18%.

Results

[edit]
1836 United States presidential election in Kentucky[2]
PartyCandidateVotesPercentageElectoral votes
WhigWilliam Henry Harrison36,86152.59%15
DemocraticMartin Van Buren33,22947.41%0
Totals70,090100.0%15

References

[edit]
  1. ^"Presidential Elections".Weekly Messenger. November 12, 1836.
  2. ^"1836 Presidential General Election Results - Kentucky". U.S. Election Atlas. RetrievedAugust 4, 2012.
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