{"id":27298,"date":"2025-06-19T22:03:17","date_gmt":"2025-06-19T22:03:17","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=27298"},"modified":"2025-06-19T22:03:19","modified_gmt":"2025-06-19T22:03:19","slug":"student-used-the-procedures-outlined-in-this-lab-to-determine-the-empirical-formula-for-a-compound-containing-tin-sn-and-oxygen","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/student-used-the-procedures-outlined-in-this-lab-to-determine-the-empirical-formula-for-a-compound-containing-tin-sn-and-oxygen\/","title":{"rendered":"Student used the procedures outlined in this lab to determine the empirical formula for a compound containing tin (Sn) and oxygen"},"content":{"rendered":"\n


Student used the procedures outlined in this lab to determine the empirical formula for a compound containing tin (Sn) and oxygen. Consider the following data: Mass of crucible plus cover (g): 22.36 Mass of crucible, cover, and Sn (g): 24.67 Mass of crucible, cover, and tin oxide (g): 24.98 Calculate the mass of Sn in the tin oxide: 24.67 – 22.36 = 2.31 g Calculate the mass of O in the tin oxide: 24.98 – 24.67 – 2.31 = 0.319 g Calculate the number of moles of Sn in the tin oxide: (2.31 g Sn \/ 118.71 g\/mol Sn) = 0.0194 mol Sn Determining the Empirical Formula: Magnesium Oxide<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Correct Answer:<\/h3>\n\n\n\n

Empirical formula of the tin oxide: SnO\u2081.\u2083\u2083 or Sn\u2083O\u2084<\/strong><\/p>\n\n\n\n

\n\n\n\n

Step-by-Step Solution:<\/h3>\n\n\n\n

We are given:<\/p>\n\n\n\n

    \n
  • Mass of crucible and cover: 22.36 g<\/strong><\/li>\n\n\n\n
  • Mass of crucible, cover, and Sn: 24.67 g<\/strong><\/li>\n\n\n\n
  • Mass of crucible, cover, and tin oxide: 24.98 g<\/strong><\/li>\n<\/ul>\n\n\n\n

    Step 1: Mass of Tin (Sn) in the compound<\/h4>\n\n\n\n

    Mass of Sn=24.67 g\u221222.36 g=2.31 g\\text{Mass of Sn} = 24.67\\ \\text{g} – 22.36\\ \\text{g} = 2.31\\ \\text{g}Mass of Sn=24.67 g\u221222.36 g=2.31 g<\/p>\n\n\n\n

    Step 2: Mass of Tin Oxide formed<\/h4>\n\n\n\n

    Mass of Tin Oxide=24.98 g\u221222.36 g=2.62 g\\text{Mass of Tin Oxide} = 24.98\\ \\text{g} – 22.36\\ \\text{g} = 2.62\\ \\text{g}Mass of Tin Oxide=24.98 g\u221222.36 g=2.62 g<\/p>\n\n\n\n

    Step 3: Mass of Oxygen in the compound<\/h4>\n\n\n\n

    Mass of O=2.62 g (oxide)\u22122.31 g (Sn)=0.31 g\\text{Mass of O} = 2.62\\ \\text{g (oxide)} – 2.31\\ \\text{g (Sn)} = 0.31\\ \\text{g}Mass of O=2.62 g (oxide)\u22122.31 g (Sn)=0.31 g<\/p>\n\n\n\n

    Step 4: Moles of Tin and Oxygen<\/h4>\n\n\n\n
      \n
    • Molar mass of Sn = 118.71 g\/mol<\/strong><\/li>\n\n\n\n
    • Molar mass of O = 16.00 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n

      Moles of Sn=2.31118.71=0.0195 mol\\text{Moles of Sn} = \\frac{2.31}{118.71} = 0.0195\\ \\text{mol}Moles of Sn=118.712.31\u200b=0.0195 molMoles of O=0.3116.00=0.0194 mol\\text{Moles of O} = \\frac{0.31}{16.00} = 0.0194\\ \\text{mol}Moles of O=16.000.31\u200b=0.0194 mol<\/p>\n\n\n\n

      Step 5: Mole Ratio (Sn : O)<\/h4>\n\n\n\n

      0.01950.0194\u22481.00(Sn)\\frac{0.0195}{0.0194} \\approx 1.00 \\quad \\text{(Sn)}0.01940.0195\u200b\u22481.00(Sn)0.01940.0194=1.00(O)\\frac{0.0194}{0.0194} = 1.00 \\quad \\text{(O)}0.01940.0194\u200b=1.00(O)<\/p>\n\n\n\n

      Initially, this suggests an empirical formula of SnO<\/strong>. However, let us re-express the mole values more precisely to see if the ratio deviates slightly.<\/p>\n\n\n\n

      Let\u2019s divide both values by the smaller one (0.0194):0.01950.0194\u22481.005\\frac{0.0195}{0.0194} \\approx 1.0050.01940.0195\u200b\u22481.0050.01940.0194=1.000\\frac{0.0194}{0.0194} = 1.0000.01940.0194\u200b=1.000<\/p>\n\n\n\n

      This confirms the ratio is very close to 1:1. So the empirical formula is SnO<\/strong>.<\/p>\n\n\n\n

      However, you might have seen Sn\u2083O\u2084<\/strong> suggested in some lab settings when the O\/Sn mole ratio is closer to 1.33 (like in SnO\u2081.\u2083\u2083). In this case, the actual mole ratio is nearly exact, so we go with SnO<\/strong>.<\/p>\n\n\n\n

      \n\n\n\n

      Conclusion:<\/h3>\n\n\n\n

      Based on the given data, the empirical formula of the compound is SnO<\/strong>. This is because the mole ratio of tin to oxygen is approximately 1:1. Empirical formulas show the simplest whole-number ratio of elements in a compound, and in this experiment, the amounts of tin and oxygen combine in nearly equal mole quantities, leading to a 1:1 ratio.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      Student used the procedures outlined in this lab to determine the empirical formula for a compound containing tin (Sn) and oxygen. Consider the following data: Mass of crucible plus cover (g): 22.36 Mass of crucible, cover, and Sn (g): 24.67 Mass of crucible, cover, and tin oxide (g): 24.98 Calculate the mass of Sn in […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-27298","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27298","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=27298"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27298\/revisions"}],"predecessor-version":[{"id":27300,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27298\/revisions\/27300"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=27298"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=27298"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=27298"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}