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United States presidential election in Missouri, 1948

From Wikipedia, the free encyclopedia

United States presidential election in Missouri, 1948

← 1944 November 2, 1948[1] 1952 →

All 15 Missouri votes to the Electoral College
Harry S. Truman.jpg
Nominee Harry S. Truman Thomas Dewey
Party Democratic Republican
Home state Missouri New York
Running mate Alben W. Barkley Earl Warren
Electoral vote 15 0
Popular vote 917,315 655,039
Percentage 58.1% 41.5%

President before election

Harry S. Truman

Elected President

Harry S. Truman

The 1948 United States presidential election in Missouri took place on November 2, 1948, as part of the 1948 United States presidential election. Missouri voters chose fifteen[2] representatives, or electors, to the Electoral College, who voted for president and vice president.

Incumbent President Harry S. Truman, a native Missourian, won the state with 58.11% of the popular vote, against Republican Governor Thomas Dewey, with 41.49% of the popular vote.[3][4] As of the 2016 presidential election, this is the last election in which Moniteau County and Cole County voted for the Democratic candidate.

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  • The American Presidential Election of 1976
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  • The American Presidential Election of 1868
  • The American Presidential Election of 1972
  • The American Presidential Election of 1944



United States presidential election in Missouri, 1948
Party Candidate Votes %
Democratic Harry S. Truman (inc.) 917,315 58.11%
Republican Thomas Dewey 655,039 41.49%
Progressive Henry A. Wallace 3,998 0.25%
Write-in 2,276 0.14%
Total votes 1,578,628 100%


  1. ^ "United States Presidential election of 1948 - Encyclopædia Britannica". Retrieved October 26, 2017.
  2. ^ "1948 Election for the Forty-First Term (1949-53)". Retrieved October 26, 2017.
  3. ^ "1948 Presidential General Election Results - Missouri". Retrieved October 26, 2017.
  4. ^ "The American Presidency Project - Election of 1948". Retrieved October 26, 2017.

This page was last edited on 12 November 2018, at 10:45
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