| Election Name: | 1884 United States presidential election in Missouri |
| Country: | Missouri |
| Type: | presidential |
| Ongoing: | no |
| Image1: | StephenGroverCleveland.jpg |
| Nominee1: | Grover Cleveland |
| Party1: | Democratic Party (United States) |
| Home State1: | New York |
| Running Mate1: | Thomas A. Hendricks |
| Electoral Vote1: | 16 |
| Popular Vote1: | 236,023 |
| Percentage1: | 53.49% |
| Nominee2: | James G. Blaine |
| Party2: | Republican Party (United States) |
| Home State2: | Maine |
| Running Mate2: | John A. Logan |
| Electoral Vote2: | 0 |
| Popular Vote2: | 203,081 |
| Percentage2: | 46.02% |
| Map Size: | 340px |
| President | |
| Before Election: | Chester A. Arthur |
| Before Party: | Republican Party (United States) |
| After Election: | Grover Cleveland |
| After Party: | Democratic Party (United States) |
| Next Election: | 1888 United States presidential election in Missouri |
| Next Year: | 1888 |
| Previous Election: | 1880 United States presidential election in Missouri |
| Previous Year: | 1880 |
| Election Date: | November 4, 1884 |
The 1884 United States presidential election in Missouri took place on November 4, 1884. All contemporary 38 states were part of the 1884 United States presidential election. Voters chose 16 electors to the Electoral College, which selected the president and vice president.[1]
Missouri was won by Governor Grover Cleveland of New York, and Governor Thomas A. Hendricks of Indiana, with 53.49% of the vote, against former Secretary of State and Senator James G. Blaine of Maine and his running mate Senator John A. Logan of Illinois, with 46.02% of the vote.[1]
| 1884 United States presidential election in Missouri | ||||||
|---|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | ||
| Democratic | Grover Cleveland | 236,023 | 53.49% | 16 | ||
| Republican | James G. Blaine | 203,081 | 46.02% | 0 | ||
| Prohibition | John St. John | 2,164 | 0.49% | 0 | ||