1998 United States Senate election in Iowa

Summary

The 1998 United States Senate election in Iowa was held November 3, 1998. Incumbent Republican United States Senator Chuck Grassley sought re-election to a fourth term in the United States Senate, running against former State Representative David Osterberg, who won the Democratic nomination unopposed. Grassley had not faced a competitive election since 1980; this year proved no different, and Grassley defeated Osterberg in a landslide.

1998 United States Senate election in Iowa

← 1992 November 3, 1998 2004 →
 
Nominee Chuck Grassley David Osterberg
Party Republican Democratic
Popular vote 648,480 289,049
Percentage 68.41% 30.49%

County results
Grassley:      50-60%      60-70%      70-80%      80-90%      >90%

U.S. senator before election

Chuck Grassley
Republican

Elected U.S. Senator

Chuck Grassley
Republican

Democratic primary edit

Candidates edit

Results edit

Democratic primary results[1]
Party Candidate Votes %
Democratic David Osterberg 86,064 99.45%
Democratic Write-ins 476 0.55%
Total votes 86,540 100.00%

Republican primary edit

Candidates edit

Results edit

Republican primary results[1]
Party Candidate Votes %
Republican Chuck Grassley (Incumbent) 149,943 99.72%
Republican Write-ins 419 0.28%
Total votes 150,362 100.00%

General election edit

Results edit

United States Senate election in Iowa, 1998[2]
Party Candidate Votes % ±%
Republican Chuck Grassley (Incumbent) 648,480 68.41% -1.20%
Democratic David Osterberg 289,049 30.49% +3.29%
Natural Law Susan Marcus 7,561 0.80% -0.47%
Socialist Workers Margaret Trowe 2,542 0.27% +0.16%
Write-ins 275 0.03%
Majority 359,431 37.92% -4.50%
Turnout 947,907
Republican hold Swing

See also edit

References edit

  1. ^ a b "Results" (PDF). sos.iowa.gov. 1998. Retrieved October 15, 2019.
  2. ^ "Election info" (PDF). clerk.house.gov. 1998. Retrieved October 15, 2019.