Who were the candidates in the presidential election of 1824how was the winner determined?

Answer 1

1824 U.S. Presidential Election - Electoral Votes

• 131 votes (50.19%) - minimum required
• 99 votes (37.93%) - U.S. Sen. Andrew Jackson (DR-TN)
• 84 votes (32.18%) - Sec. of State John Quincy Adams (DR-MA)
• 41 votes (15.71%) - Treasury Sec. William H. Crawford (DR-GA)
• 37 votes (14.18%) - House Speaker Henry Clay (DR-KY)

Following the instructions in the 12th Amendment to the U.S. Constitution, the House of Representatives was required to elect a President from among the three candidates who got the most electoral votes. First they make sure that a quorum is present; in this case it is at least one Representative from at least two thirds, or 16, of the states. Each state delegation has one vote, so the Representatives from each state vote among themselves to determine who will get that state's vote. Even if some states are not present and voting, the votes of a majority of all states is required for a win. In this case, that's at least 13 votes.

1824 U.S. Presidential Election - House Votes (1st Ballot)

• 13 votes (54%) - minimum required
• 13 votes (54%) - Sec. of State John Quincy Adams (DR-MA)
• 7 votes (29%) - U.S. Sen. Andrew Jackson (DR-TN)
• 4 votes (17%) - Treasury Sec. William H. Crawford (DR-GA)